2017 Problems
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- .gitattributes +4 -0
- 2017/finals/fox_patrols.html +82 -0
- 2017/finals/holes.cpp +237 -0
- 2017/finals/holes.html +69 -0
- 2017/finals/holes.in +3 -0
- 2017/finals/holes.md +82 -0
- 2017/finals/holes.out +30 -0
- 2017/finals/moles.cpp +247 -0
- 2017/finals/moles.html +88 -0
- 2017/finals/moles.in +0 -0
- 2017/finals/moles.md +75 -0
- 2017/finals/moles.out +30 -0
- 2017/finals/patrols.cpp +123 -0
- 2017/finals/patrols.html +82 -0
- 2017/finals/patrols.in +31 -0
- 2017/finals/patrols.md +72 -0
- 2017/finals/patrols.out +30 -0
- 2017/finals/poles.cpp +202 -0
- 2017/finals/poles.html +55 -0
- 2017/finals/poles.in +3 -0
- 2017/finals/poles.md +73 -0
- 2017/finals/poles.out +30 -0
- 2017/finals/strolls.cpp +170 -0
- 2017/finals/strolls.html +73 -0
- 2017/finals/strolls.in +3 -0
- 2017/finals/strolls.md +70 -0
- 2017/finals/strolls.out +30 -0
- 2017/finals/tolls.cpp +210 -0
- 2017/finals/tolls.html +100 -0
- 2017/finals/tolls.in +3 -0
- 2017/finals/tolls.md +89 -0
- 2017/finals/tolls.out +30 -0
- 2017/quals/592385891651966.jpg +3 -0
- 2017/quals/fightingthezombie.html +86 -0
- 2017/quals/fightingthezombie.in +0 -0
- 2017/quals/fightingthezombie.java +107 -0
- 2017/quals/fightingthezombie.md +58 -0
- 2017/quals/fightingthezombie.out +2005 -0
- 2017/quals/lazyloading.html +78 -0
- 2017/quals/lazyloading.in +0 -0
- 2017/quals/lazyloading.java +46 -0
- 2017/quals/lazyloading.md +63 -0
- 2017/quals/lazyloading.out +1005 -0
- 2017/quals/progresspie.html +72 -0
- 2017/quals/progresspie.in +2006 -0
- 2017/quals/progresspie.java +46 -0
- 2017/quals/progresspie.md +56 -0
- 2017/quals/progresspie.out +2005 -0
- 2017/round1/268425761045248.jpg +3 -0
- 2017/round1/336634350719298.jpg +3 -0
.gitattributes
CHANGED
@@ -68,3 +68,7 @@ saved_model/**/* filter=lfs diff=lfs merge=lfs -text
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2016/finals/maximinimax_flow.in filter=lfs diff=lfs merge=lfs -text
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2016/finals/rainbow_string.in filter=lfs diff=lfs merge=lfs -text
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2016/finals/rng.in filter=lfs diff=lfs merge=lfs -text
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2016/finals/maximinimax_flow.in filter=lfs diff=lfs merge=lfs -text
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2016/finals/rainbow_string.in filter=lfs diff=lfs merge=lfs -text
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2016/finals/rng.in filter=lfs diff=lfs merge=lfs -text
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+
2017/finals/holes.in filter=lfs diff=lfs merge=lfs -text
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2017/finals/poles.in filter=lfs diff=lfs merge=lfs -text
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2017/finals/strolls.in filter=lfs diff=lfs merge=lfs -text
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2017/finals/tolls.in filter=lfs diff=lfs merge=lfs -text
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2017/finals/fox_patrols.html
ADDED
@@ -0,0 +1,82 @@
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<p>
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A certain well-hidden valley is home to a thriving population of mysterious creatures — Foxen!
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However, keeping the valley safe from outsiders (such as humans) is a necessity.
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To that end, a group of Foxen have been sent out to patrol the border.
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</p>
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<p>
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On their patrol route, the Foxen know that they're going to pass by an interesting, rectangular forest.
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When viewed from above, the forest can be modeled as a grid of cells
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with <strong>R</strong> rows and <strong>C</strong> columns. The rows are numbered from 1 to <strong>R</strong> from North to South, while the column are numbered from 1 to <strong>C</strong> from West to East.
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One tree is growing in the center of each cell, and each tree's height (in metres) is some positive integer no larger than <strong>H</strong>.
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</p>
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<p>
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If the Foxen were to look at the forest from the North side, all of the trees in any given column of cells would obscure each other and blend together.
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In fact, the Foxen would really only be able make out the overall shape of the forest's "skyline" when viewed from that direction.
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This Northern skyline can be expressed as a sequence of <strong>C</strong> positive integers,
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with the <em>i</em>th one being the largest of the <strong>R</strong> tree heights in the <em>i</em>th column.
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</p>
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<p>
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Similarly, if they were to look at the forest from the West side, they would only be able to make out the shape of its skyline from that direction.
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This Western skyline is a sequence of <strong>R</strong> positive integers,
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with the <em>i</em>th one being the largest of the <strong>C</strong> tree heights in the <em>i</em>th row.
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</p>
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<p>
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On their way to the forest, the Foxen find themselves wondering about what it might look like.
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They've done their research and are aware of its dimensions <strong>R</strong> and <strong>C</strong>,
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as well as the maximum possible height of its trees <strong>H</strong>, but they don't know the actual heights of any of its trees.
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They'd like to determine how many different, distinct-looking forests they might end up finding.
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A forest is a set of heights for all <strong>R</strong>x<strong>C</strong> trees,
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and two forests are considered to be distinct-looking from one another if their Northern skyline sequences differ and/or their Western skyline sequences differ.
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</p>
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<p>
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Please help the Foxen determine the number of possible different, distinct-looking forests! As this quantity may be quite large, they're only interested in its value when taken modulo 1,000,000,007.
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</p>
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<h3>Input</h3>
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<p>
|
44 |
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Input begins with an integer <strong>T</strong>, the number of different forests visited by the Foxen.
|
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For each forest, there is a single line containing the three space-separated integers
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46 |
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<strong>R</strong>, <strong>C</strong>, and <strong>H</strong>.
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</p>
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<h3>Output</h3>
|
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|
51 |
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<p>
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For the <em>i</em>th forest, print a line containing "Case #<strong>i</strong>: "
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followed by the number of possible different, distinct-looking forests modulo 1,000,000,007.
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</p>
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|
56 |
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<h3>Constraints</h3>
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<p>
|
59 |
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1 ≤ <strong>T</strong> ≤ 30 <br />
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1 ≤ <strong>R</strong>, <strong>C</strong>, <strong>H</strong> ≤ 500,000 <br />
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61 |
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</p>
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62 |
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63 |
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<h3>Explanation of Sample</h3>
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|
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<p>
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In the first case, there are 10 possible different, distinct-looking forests which consist of a 2x2 grid of trees, with each tree being either 1m or 2m tall.
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For example, the following 2 forests look different (even though their Western skylines are equal, their Northern skylines differ), so both should be counted:
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68 |
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</p>
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<pre>
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71 |
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1 2 2 1
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72 |
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1 1 1 1
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73 |
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</pre>
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75 |
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<p>
|
76 |
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On the other hand, the following 2 forests look identical to one another from both the North and the West, so only one of them should be counted:
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77 |
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</p>
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78 |
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79 |
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<pre>
|
80 |
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1 2 2 2
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2 1 2 1
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</pre>
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2017/finals/holes.cpp
ADDED
@@ -0,0 +1,237 @@
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1 |
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// Hacker Cup 2017
|
2 |
+
// Final Round
|
3 |
+
// Fox Holes
|
4 |
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// Jacob Plachta
|
5 |
+
|
6 |
+
#include <algorithm>
|
7 |
+
#include <functional>
|
8 |
+
#include <numeric>
|
9 |
+
#include <iostream>
|
10 |
+
#include <iomanip>
|
11 |
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#include <cstdio>
|
12 |
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#include <cmath>
|
13 |
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#include <complex>
|
14 |
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#include <cstdlib>
|
15 |
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#include <ctime>
|
16 |
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#include <cstring>
|
17 |
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#include <cassert>
|
18 |
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#include <string>
|
19 |
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#include <vector>
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20 |
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#include <list>
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21 |
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#include <map>
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22 |
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#include <set>
|
23 |
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#include <deque>
|
24 |
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#include <queue>
|
25 |
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#include <stack>
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26 |
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#include <bitset>
|
27 |
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#include <sstream>
|
28 |
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using namespace std;
|
29 |
+
|
30 |
+
#define LL long long
|
31 |
+
#define LD long double
|
32 |
+
#define PR pair<int,int>
|
33 |
+
|
34 |
+
#define Fox(i,n) for (i=0; i<n; i++)
|
35 |
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#define Fox1(i,n) for (i=1; i<=n; i++)
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36 |
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#define FoxI(i,a,b) for (i=a; i<=b; i++)
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37 |
+
#define FoxR(i,n) for (i=(n)-1; i>=0; i--)
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38 |
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#define FoxR1(i,n) for (i=n; i>0; i--)
|
39 |
+
#define FoxRI(i,a,b) for (i=b; i>=a; i--)
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40 |
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#define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++)
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41 |
+
#define Min(a,b) a=min(a,b)
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42 |
+
#define Max(a,b) a=max(a,b)
|
43 |
+
#define Sz(s) int((s).size())
|
44 |
+
#define All(s) (s).begin(),(s).end()
|
45 |
+
#define Fill(s,v) memset(s,v,sizeof(s))
|
46 |
+
#define pb push_back
|
47 |
+
#define mp make_pair
|
48 |
+
#define x first
|
49 |
+
#define y second
|
50 |
+
|
51 |
+
template<typename T> T Abs(T x) { return(x<0 ? -x : x); }
|
52 |
+
template<typename T> T Sqr(T x) { return(x*x); }
|
53 |
+
|
54 |
+
const int INF = (int)1e9;
|
55 |
+
const LD EPS = 1e-9;
|
56 |
+
const LD PI = acos(-1.0);
|
57 |
+
|
58 |
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bool Read(int &x)
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59 |
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{
|
60 |
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char c,r=0,n=0;
|
61 |
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x=0;
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62 |
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for(;;)
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63 |
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{
|
64 |
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c=getchar();
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65 |
+
if ((c<0) && (!r))
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66 |
+
return(0);
|
67 |
+
if ((c=='-') && (!r))
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68 |
+
n=1;
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69 |
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else
|
70 |
+
if ((c>='0') && (c<='9'))
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71 |
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x=x*10+c-'0',r=1;
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72 |
+
else
|
73 |
+
if (r)
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74 |
+
break;
|
75 |
+
}
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76 |
+
if (n)
|
77 |
+
x=-x;
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78 |
+
return(1);
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79 |
+
}
|
80 |
+
|
81 |
+
#define MOD 1000000007
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82 |
+
#define LIM 1100000
|
83 |
+
|
84 |
+
int L[LIM],A[LIM],B[LIM];
|
85 |
+
int sz,treeM[LIM],treeC[LIM],lazy[LIM];
|
86 |
+
set<int> SL,SR;
|
87 |
+
|
88 |
+
void Prop(int i)
|
89 |
+
{
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90 |
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int v=lazy[i],j=i<<1,k=j+1;
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91 |
+
lazy[i]=0;
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92 |
+
treeM[i]+=v;
|
93 |
+
if (i<sz)
|
94 |
+
lazy[j]+=v,lazy[k]+=v;
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95 |
+
}
|
96 |
+
|
97 |
+
void Update(int i,int r1,int r2,int a,int b,int v)
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98 |
+
{
|
99 |
+
if (a>b)
|
100 |
+
return;
|
101 |
+
Prop(i);
|
102 |
+
if ((a<=r1) && (r2<=b))
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103 |
+
{
|
104 |
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lazy[i]+=v;
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105 |
+
return;
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106 |
+
}
|
107 |
+
int m=(r1+r2)>>1,j=i<<1,k=j+1;
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108 |
+
if (a<=m)
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109 |
+
Update(j,r1,m,a,b,v);
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110 |
+
if (b>m)
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111 |
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Update(k,m+1,r2,a,b,v);
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112 |
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Prop(j),Prop(k);
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113 |
+
treeM[i]=min(treeM[j],treeM[k]);
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114 |
+
treeC[i]=0;
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115 |
+
if (treeM[i]==treeM[j])
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116 |
+
treeC[i]+=treeC[j];
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117 |
+
if (treeM[i]==treeM[k])
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118 |
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treeC[i]+=treeC[k];
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119 |
+
}
|
120 |
+
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121 |
+
int Query(int i,int r1,int r2,int a,int b)
|
122 |
+
{
|
123 |
+
if (a>b)
|
124 |
+
return(0);
|
125 |
+
Prop(i);
|
126 |
+
if ((a<=r1) && (r2<=b))
|
127 |
+
return(treeM[i] ? 0 : treeC[i]);
|
128 |
+
int m=(r1+r2)>>1,j=i<<1,k=j+1,v=0;
|
129 |
+
if (a<=m)
|
130 |
+
v+=Query(j,r1,m,a,b);
|
131 |
+
if (b>m)
|
132 |
+
v+=Query(k,m+1,r2,a,b);
|
133 |
+
return(v);
|
134 |
+
}
|
135 |
+
|
136 |
+
void Go(int i,int d)
|
137 |
+
{
|
138 |
+
int a=A[i],b=B[i],p;
|
139 |
+
if (a>b)
|
140 |
+
swap(a,b);
|
141 |
+
p=b-a;
|
142 |
+
if (p%2)
|
143 |
+
{
|
144 |
+
// odd difference, key point must be outside [a, b-1]
|
145 |
+
a/=2,b/=2;
|
146 |
+
Update(1,0,sz-1,a,b-1,d);
|
147 |
+
return;
|
148 |
+
}
|
149 |
+
// even difference, key point must be inside [a, b-1]
|
150 |
+
if (d<0)
|
151 |
+
{
|
152 |
+
SL.erase(a);
|
153 |
+
SR.erase(b-2);
|
154 |
+
}
|
155 |
+
else
|
156 |
+
{
|
157 |
+
SL.insert(a);
|
158 |
+
SR.insert(b-2);
|
159 |
+
}
|
160 |
+
}
|
161 |
+
|
162 |
+
int main()
|
163 |
+
{
|
164 |
+
// vars
|
165 |
+
int T,t;
|
166 |
+
int N,M;
|
167 |
+
int i,j,a,b;
|
168 |
+
int ans;
|
169 |
+
// testcase loop
|
170 |
+
Read(T);
|
171 |
+
Fox1(t,T)
|
172 |
+
{
|
173 |
+
// input, init pairs
|
174 |
+
Read(N),Read(M);
|
175 |
+
Fill(A,-1);
|
176 |
+
Fox(i,N<<1)
|
177 |
+
{
|
178 |
+
Read(j),j--;
|
179 |
+
L[i]=j;
|
180 |
+
if (A[j]<0)
|
181 |
+
A[j]=i;
|
182 |
+
else
|
183 |
+
B[j]=i;
|
184 |
+
}
|
185 |
+
// init data structures
|
186 |
+
for(sz=1;sz<=N;sz<<=1);
|
187 |
+
Fill(lazy,0);
|
188 |
+
Fill(treeM,0);
|
189 |
+
Fox(i,sz)
|
190 |
+
treeC[sz+i]=1;
|
191 |
+
FoxR1(i,sz-1)
|
192 |
+
treeC[i]=treeC[i*2]+treeC[i*2+1];
|
193 |
+
SL.clear(),SR.clear();
|
194 |
+
// insert pairs
|
195 |
+
Fox(i,N)
|
196 |
+
Go(i,1);
|
197 |
+
// process updates
|
198 |
+
ans=0;
|
199 |
+
while (M--)
|
200 |
+
{
|
201 |
+
// input
|
202 |
+
Read(i),Read(j),i--,j--;
|
203 |
+
a=L[i],b=L[j];
|
204 |
+
if (a==b)
|
205 |
+
goto Skip;
|
206 |
+
// remove pairs
|
207 |
+
Go(a,-1),Go(b,-1);
|
208 |
+
// update pairs
|
209 |
+
swap(L[i],L[j]);
|
210 |
+
if (A[a]==i)
|
211 |
+
A[a]=j;
|
212 |
+
else
|
213 |
+
B[a]=j;
|
214 |
+
if (A[b]==j)
|
215 |
+
A[b]=i;
|
216 |
+
else
|
217 |
+
B[b]=i;
|
218 |
+
// re-insert pairs
|
219 |
+
Go(a,1),Go(b,1);
|
220 |
+
Skip:;
|
221 |
+
// compute current answer
|
222 |
+
if (SL.empty())
|
223 |
+
a=0,b=N*2;
|
224 |
+
else
|
225 |
+
{
|
226 |
+
a=*SL.rbegin();
|
227 |
+
b=*SR.begin();
|
228 |
+
}
|
229 |
+
a/=2;
|
230 |
+
b/=2;
|
231 |
+
ans=(ans+Query(1,0,sz-1,a,b))%MOD;
|
232 |
+
}
|
233 |
+
// output
|
234 |
+
printf("Case #%d: %d\n",t,ans);
|
235 |
+
}
|
236 |
+
return(0);
|
237 |
+
}
|
2017/finals/holes.html
ADDED
@@ -0,0 +1,69 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
<p>
|
2 |
+
Kit is a young, obedient Fox who excels in his studies, diligently practices his hunting skills, and is always friendly. His parents couldn't be prouder! That is, except for one problem... he refuses to eat his vegetables!
|
3 |
+
</p>
|
4 |
+
|
5 |
+
<p>
|
6 |
+
In an effort to improve Kit's diet, his parents have set up a little game for him to play. They've dug a series of 2<strong>N</strong> + 2 small holes in a row on the ground, and numbered them from 0 to 2<strong>N</strong> + 1, from first to last. They've left the first and last holes empty, and filled the remaining 2<strong>N</strong> holes with one healthy vegetable each! There are <strong>N</strong> different types of vegetables, numbered from 1 to <strong>N</strong>, and a vegetable of type <strong>V<sub>i</sub></strong> has initially been placed into each hole <em>i</em>, such that exactly 2 vegetables of each type were used in total.
|
7 |
+
</p>
|
8 |
+
|
9 |
+
<p>
|
10 |
+
Kit must start inside hole 0, and jump forwards from hole to hole until he reaches hole 2<strong>N</strong> + 1. When he's inside any given hole <em>i</em>, he's agile enough to jump to either hole <em>i</em> + 1 or directly to hole <em>i</em> + 2, but he can't jump any further than that at once. Whenever he lands in a hole containing a vegetable, the rules of the game mandate that he must eat it!
|
11 |
+
</p>
|
12 |
+
|
13 |
+
<p>
|
14 |
+
Kit has agreed to play this game (not that he has much choice in the matter), but there's only so much he can take. The vegetables are tolerable as long as there's variety. He doesn't care how many he has to eat in total, but he absolutely refuses to eat multiple vegetables of any single type over the course of the game. In other words, for each vegetable type, he must only enter <i>at most</i> one of the two holes containing that vegetable on his way from hole 0 to hole 2<strong>N</strong> + 1.
|
15 |
+
</p>
|
16 |
+
|
17 |
+
<p>
|
18 |
+
Kit will play the game once per day for a period of <strong>M</strong> days. For some fun variety, at the start of each day <em>i</em>, his parents will swap the contents of two different holes <strong>A<sub>i</sub></strong> and <strong>B<sub>i</sub></strong>. Then, Kit will play the game using the current configuration of vegetables. Once he's done, his parents will replace any vegetables which he had eaten with new vegetables of the same types, thus resetting the game to the state it was in before Kit played it.
|
19 |
+
</p>
|
20 |
+
|
21 |
+
<p>
|
22 |
+
Each day, there might be no acceptable way for Kit to complete the game, or there might be many different ways for him to do so. Two ways are considered different if at least one hole is visited in one but not the other. In order to make the game more exciting for himself, Kit would like to count up the number of different ways he could potentially complete it each day. However, that's a lot of big numbers to keep track of, so he's only interested in the sum of these <strong>M</strong> values when taken modulo 1,000,000,007. Please help him compute this overall sum!
|
23 |
+
</p>
|
24 |
+
|
25 |
+
<h3>Input</h3>
|
26 |
+
|
27 |
+
<p>
|
28 |
+
Input begins with an integer <strong>T</strong>, the number of different rows of holes.
|
29 |
+
For each row of holes, there is first a line containing the space-separated integers <strong>N</strong> and <strong>M</strong>.
|
30 |
+
There is next a line containing 2<strong>N</strong> space-separated integers,
|
31 |
+
the <em>i</em>th of which is <strong>V<sub>i</sub></strong>.
|
32 |
+
Then <strong>M</strong> lines follow, the <em>i</em>th of which contains the space-separated integers
|
33 |
+
<strong>A<sub>i</sub></strong> and <strong>B<sub>i</sub></strong>
|
34 |
+
|
35 |
+
<h3>Output</h3>
|
36 |
+
|
37 |
+
<p>
|
38 |
+
For the <em>i</em>th row of holes, print a line containing "Case #<strong>i</strong>: "
|
39 |
+
followed by a single integer, the sum of the <strong>M</strong> days' answers modulo 1,000,000,007.
|
40 |
+
</p>
|
41 |
+
|
42 |
+
<h3>Constraints</h3>
|
43 |
+
|
44 |
+
<p>
|
45 |
+
1 ≤ <strong>T</strong> ≤ 30<br />
|
46 |
+
1 ≤ <strong>N, M</strong> ≤ 500,000 <br />
|
47 |
+
1 ≤ <strong>A<sub>i</sub></strong>, <strong>B<sub>i</sub></strong> ≤ 2<strong>N</strong> <br />
|
48 |
+
1 ≤ <strong>V<sub>i</sub></strong> ≤ <strong>N</strong> <br />
|
49 |
+
Both the sum of <strong>N</strong> values and the sum of <strong>M</strong> values across all <strong>T</strong> cases do not exceed 2,000,000.
|
50 |
+
</p>
|
51 |
+
|
52 |
+
<h3>Explanation of Sample</h3>
|
53 |
+
|
54 |
+
<p>
|
55 |
+
In the first case, both holes 1 and 2 will contain vegetables of type 1 even after their contents are swapped. There are then 2 different ways for Kit to validly reach hole 3 from hole 0, visiting either of these sequences of holes:
|
56 |
+
</p>
|
57 |
+
|
58 |
+
<pre>
|
59 |
+
0 -> 1 -> 3
|
60 |
+
0 -> 2 -> 3
|
61 |
+
</pre>
|
62 |
+
|
63 |
+
<p>
|
64 |
+
He can't quite jump far enough to reach hole 3 directly from hole 0, nor can he visit both holes 1 and 2, as that would require eating multiple vegetables of a single type.
|
65 |
+
</p>
|
66 |
+
|
67 |
+
<p>
|
68 |
+
In the second case, there are 2 different ways for Kit to validly complete the game on the first day, and only 1 way on the second day. This results in a final answer of (2 + 1) modulo 1,000,000,007 = 3.
|
69 |
+
</p>
|
2017/finals/holes.in
ADDED
@@ -0,0 +1,3 @@
|
|
|
|
|
|
|
|
|
1 |
+
version https://git-lfs.github.com/spec/v1
|
2 |
+
oid sha256:41bc8a8a4fb7549fe5a83923f79a4e404a416be097b2fc1d2cca3af56eaff32c
|
3 |
+
size 43256260
|
2017/finals/holes.md
ADDED
@@ -0,0 +1,82 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
Kit is a young, obedient Fox who excels in his studies, diligently practices
|
2 |
+
his hunting skills, and is always friendly. His parents couldn't be prouder!
|
3 |
+
That is, except for one problem... he refuses to eat his vegetables!
|
4 |
+
|
5 |
+
In an effort to improve Kit's diet, his parents have set up a little game for
|
6 |
+
him to play. They've dug a series of 2**N** \+ 2 small holes in a row on the
|
7 |
+
ground, and numbered them from 0 to 2**N** \+ 1, from first to last. They've
|
8 |
+
left the first and last holes empty, and filled the remaining 2**N** holes
|
9 |
+
with one healthy vegetable each! There are **N** different types of
|
10 |
+
vegetables, numbered from 1 to **N**, and a vegetable of type **Vi** has
|
11 |
+
initially been placed into each hole _i_, such that exactly 2 vegetables of
|
12 |
+
each type were used in total.
|
13 |
+
|
14 |
+
Kit must start inside hole 0, and jump forwards from hole to hole until he
|
15 |
+
reaches hole 2**N** \+ 1. When he's inside any given hole _i_, he's agile
|
16 |
+
enough to jump to either hole _i_ \+ 1 or directly to hole _i_ \+ 2, but he
|
17 |
+
can't jump any further than that at once. Whenever he lands in a hole
|
18 |
+
containing a vegetable, the rules of the game mandate that he must eat it!
|
19 |
+
|
20 |
+
Kit has agreed to play this game (not that he has much choice in the matter),
|
21 |
+
but there's only so much he can take. The vegetables are tolerable as long as
|
22 |
+
there's variety. He doesn't care how many he has to eat in total, but he
|
23 |
+
absolutely refuses to eat multiple vegetables of any single type over the
|
24 |
+
course of the game. In other words, for each vegetable type, he must only
|
25 |
+
enter _at most_ one of the two holes containing that vegetable on his way from
|
26 |
+
hole 0 to hole 2**N** \+ 1.
|
27 |
+
|
28 |
+
Kit will play the game once per day for a period of **M** days. For some fun
|
29 |
+
variety, at the start of each day _i_, his parents will swap the contents of
|
30 |
+
two different holes **Ai** and **Bi**. Then, Kit will play the game using the
|
31 |
+
current configuration of vegetables. Once he's done, his parents will replace
|
32 |
+
any vegetables which he had eaten with new vegetables of the same types, thus
|
33 |
+
resetting the game to the state it was in before Kit played it.
|
34 |
+
|
35 |
+
Each day, there might be no acceptable way for Kit to complete the game, or
|
36 |
+
there might be many different ways for him to do so. Two ways are considered
|
37 |
+
different if at least one hole is visited in one but not the other. In order
|
38 |
+
to make the game more exciting for himself, Kit would like to count up the
|
39 |
+
number of different ways he could potentially complete it each day. However,
|
40 |
+
that's a lot of big numbers to keep track of, so he's only interested in the
|
41 |
+
sum of these **M** values when taken modulo 1,000,000,007. Please help him
|
42 |
+
compute this overall sum!
|
43 |
+
|
44 |
+
### Input
|
45 |
+
|
46 |
+
Input begins with an integer **T**, the number of different rows of holes. For
|
47 |
+
each row of holes, there is first a line containing the space-separated
|
48 |
+
integers **N** and **M**. There is next a line containing 2**N** space-
|
49 |
+
separated integers, the _i_th of which is **Vi**. Then **M** lines follow, the
|
50 |
+
_i_th of which contains the space-separated integers **Ai** and **Bi**
|
51 |
+
|
52 |
+
### Output
|
53 |
+
|
54 |
+
For the _i_th row of holes, print a line containing "Case #**i**: " followed
|
55 |
+
by a single integer, the sum of the **M** days' answers modulo 1,000,000,007.
|
56 |
+
|
57 |
+
### Constraints
|
58 |
+
|
59 |
+
1 ≤ **T** ≤ 30
|
60 |
+
1 ≤ **N, M** ≤ 500,000
|
61 |
+
1 ≤ **Ai**, **Bi** ≤ 2**N**
|
62 |
+
1 ≤ **Vi** ≤ **N**
|
63 |
+
Both the sum of **N** values and the sum of **M** values across all **T**
|
64 |
+
cases do not exceed 2,000,000.
|
65 |
+
|
66 |
+
### Explanation of Sample
|
67 |
+
|
68 |
+
In the first case, both holes 1 and 2 will contain vegetables of type 1 even
|
69 |
+
after their contents are swapped. There are then 2 different ways for Kit to
|
70 |
+
validly reach hole 3 from hole 0, visiting either of these sequences of holes:
|
71 |
+
|
72 |
+
0 -> 1 -> 3
|
73 |
+
0 -> 2 -> 3
|
74 |
+
|
75 |
+
He can't quite jump far enough to reach hole 3 directly from hole 0, nor can
|
76 |
+
he visit both holes 1 and 2, as that would require eating multiple vegetables
|
77 |
+
of a single type.
|
78 |
+
|
79 |
+
In the second case, there are 2 different ways for Kit to validly complete the
|
80 |
+
game on the first day, and only 1 way on the second day. This results in a
|
81 |
+
final answer of (2 + 1) modulo 1,000,000,007 = 3.
|
82 |
+
|
2017/finals/holes.out
ADDED
@@ -0,0 +1,30 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
Case #1: 2
|
2 |
+
Case #2: 3
|
3 |
+
Case #3: 5
|
4 |
+
Case #4: 0
|
5 |
+
Case #5: 17
|
6 |
+
Case #6: 2
|
7 |
+
Case #7: 1
|
8 |
+
Case #8: 2
|
9 |
+
Case #9: 7
|
10 |
+
Case #10: 0
|
11 |
+
Case #11: 134098873
|
12 |
+
Case #12: 323526184
|
13 |
+
Case #13: 518272039
|
14 |
+
Case #14: 974275396
|
15 |
+
Case #15: 175772460
|
16 |
+
Case #16: 4477145
|
17 |
+
Case #17: 3216403
|
18 |
+
Case #18: 616019
|
19 |
+
Case #19: 4331672
|
20 |
+
Case #20: 3672991
|
21 |
+
Case #21: 3497579
|
22 |
+
Case #22: 4447010
|
23 |
+
Case #23: 2166340
|
24 |
+
Case #24: 1026410
|
25 |
+
Case #25: 861565
|
26 |
+
Case #26: 756
|
27 |
+
Case #27: 112
|
28 |
+
Case #28: 771
|
29 |
+
Case #29: 104
|
30 |
+
Case #30: 784
|
2017/finals/moles.cpp
ADDED
@@ -0,0 +1,247 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
|
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|
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|
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
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|
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|
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|
|
|
|
|
|
|
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|
|
|
|
|
|
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|
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|
|
|
|
|
|
|
|
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|
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|
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|
|
|
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|
|
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|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
// Hacker Cup 2017
|
2 |
+
// Final Round
|
3 |
+
// Fox Moles
|
4 |
+
// Jacob Plachta
|
5 |
+
|
6 |
+
#include <algorithm>
|
7 |
+
#include <functional>
|
8 |
+
#include <numeric>
|
9 |
+
#include <iostream>
|
10 |
+
#include <iomanip>
|
11 |
+
#include <cstdio>
|
12 |
+
#include <cmath>
|
13 |
+
#include <complex>
|
14 |
+
#include <cstdlib>
|
15 |
+
#include <ctime>
|
16 |
+
#include <cstring>
|
17 |
+
#include <cassert>
|
18 |
+
#include <string>
|
19 |
+
#include <vector>
|
20 |
+
#include <list>
|
21 |
+
#include <map>
|
22 |
+
#include <set>
|
23 |
+
#include <deque>
|
24 |
+
#include <queue>
|
25 |
+
#include <stack>
|
26 |
+
#include <bitset>
|
27 |
+
#include <sstream>
|
28 |
+
using namespace std;
|
29 |
+
|
30 |
+
#define LL long long
|
31 |
+
#define LD long double
|
32 |
+
#define PR pair<int,int>
|
33 |
+
|
34 |
+
#define Fox(i,n) for (i=0; i<n; i++)
|
35 |
+
#define Fox1(i,n) for (i=1; i<=n; i++)
|
36 |
+
#define FoxI(i,a,b) for (i=a; i<=b; i++)
|
37 |
+
#define FoxR(i,n) for (i=(n)-1; i>=0; i--)
|
38 |
+
#define FoxR1(i,n) for (i=n; i>0; i--)
|
39 |
+
#define FoxRI(i,a,b) for (i=b; i>=a; i--)
|
40 |
+
#define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++)
|
41 |
+
#define Min(a,b) a=min(a,b)
|
42 |
+
#define Max(a,b) a=max(a,b)
|
43 |
+
#define Sz(s) int((s).size())
|
44 |
+
#define All(s) (s).begin(),(s).end()
|
45 |
+
#define Fill(s,v) memset(s,v,sizeof(s))
|
46 |
+
#define pb push_back
|
47 |
+
#define mp make_pair
|
48 |
+
#define x first
|
49 |
+
#define y second
|
50 |
+
|
51 |
+
template<typename T> T Abs(T x) { return(x<0 ? -x : x); }
|
52 |
+
template<typename T> T Sqr(T x) { return(x*x); }
|
53 |
+
|
54 |
+
const int INF = (int)1e9;
|
55 |
+
const LD EPS = 1e-12;
|
56 |
+
const LD PI = acos(-1.0);
|
57 |
+
|
58 |
+
bool Read(int &x)
|
59 |
+
{
|
60 |
+
char c,r=0,n=0;
|
61 |
+
x=0;
|
62 |
+
for(;;)
|
63 |
+
{
|
64 |
+
c=getchar();
|
65 |
+
if ((c<0) && (!r))
|
66 |
+
return(0);
|
67 |
+
if ((c=='-') && (!r))
|
68 |
+
n=1;
|
69 |
+
else
|
70 |
+
if ((c>='0') && (c<='9'))
|
71 |
+
x=x*10+c-'0',r=1;
|
72 |
+
else
|
73 |
+
if (r)
|
74 |
+
break;
|
75 |
+
}
|
76 |
+
if (n)
|
77 |
+
x=-x;
|
78 |
+
return(1);
|
79 |
+
}
|
80 |
+
|
81 |
+
#define LIM 900009
|
82 |
+
|
83 |
+
int K,A,B;
|
84 |
+
map<int,int> M;
|
85 |
+
vector<int> con[LIM],con2[LIM];
|
86 |
+
bool col0[LIM];
|
87 |
+
int col[LIM],ind[LIM],P1[LIM],P2[LIM],dist[LIM];
|
88 |
+
|
89 |
+
int Make(int i)
|
90 |
+
{
|
91 |
+
if (M.count(i))
|
92 |
+
return(M[i]);
|
93 |
+
return(M[i]=K++);
|
94 |
+
}
|
95 |
+
|
96 |
+
bool DFS(int i,int c)
|
97 |
+
{
|
98 |
+
// conflict?
|
99 |
+
if ((col0[i]) && (c) || (col[i]>=0) && (c!=col[i]))
|
100 |
+
return(0);
|
101 |
+
// already coloured?
|
102 |
+
if (col[i]>=0)
|
103 |
+
return(1);
|
104 |
+
// colour node and its neighbours
|
105 |
+
col[i]=c;
|
106 |
+
int j;
|
107 |
+
Fox(j,Sz(con[i]))
|
108 |
+
if (!DFS(con[i][j],1-c))
|
109 |
+
return(0);
|
110 |
+
return(1);
|
111 |
+
}
|
112 |
+
|
113 |
+
bool MatchBFS()
|
114 |
+
{
|
115 |
+
int i,a,b,a2;
|
116 |
+
queue<int> Q;
|
117 |
+
Fill(dist,60);
|
118 |
+
Fox(a,A)
|
119 |
+
if (P1[a]<0)
|
120 |
+
dist[a]=0,Q.push(a);
|
121 |
+
while (!Q.empty())
|
122 |
+
{
|
123 |
+
a=Q.front(),Q.pop();
|
124 |
+
if (dist[a]<dist[A])
|
125 |
+
Fox(i,Sz(con2[a]))
|
126 |
+
{
|
127 |
+
b=con2[a][i];
|
128 |
+
a2=P2[b];
|
129 |
+
if (dist[a2]>=INF)
|
130 |
+
dist[a2]=dist[a]+1,Q.push(a2);
|
131 |
+
}
|
132 |
+
}
|
133 |
+
return(dist[A]<INF);
|
134 |
+
}
|
135 |
+
|
136 |
+
bool MatchDFS(int a)
|
137 |
+
{
|
138 |
+
if (a==A)
|
139 |
+
return(1);
|
140 |
+
int i,b,a2;
|
141 |
+
Fox(i,Sz(con2[a]))
|
142 |
+
{
|
143 |
+
b=con2[a][i];
|
144 |
+
a2=P2[b];
|
145 |
+
if ((dist[a2]==dist[a]+1) && (MatchDFS(a2)))
|
146 |
+
{
|
147 |
+
P1[a]=b;
|
148 |
+
P2[b]=a;
|
149 |
+
return(1);
|
150 |
+
}
|
151 |
+
}
|
152 |
+
dist[a]=INF;
|
153 |
+
return(0);
|
154 |
+
}
|
155 |
+
|
156 |
+
int MaxMatching()
|
157 |
+
{
|
158 |
+
int a,b,m=0;
|
159 |
+
Fill(P1,-1);
|
160 |
+
Fox(b,B)
|
161 |
+
P2[b]=A;
|
162 |
+
while (MatchBFS())
|
163 |
+
{
|
164 |
+
Fox(a,A)
|
165 |
+
if ((P1[a]<0) && (MatchDFS(a)))
|
166 |
+
m++;
|
167 |
+
}
|
168 |
+
return(m);
|
169 |
+
}
|
170 |
+
|
171 |
+
int main()
|
172 |
+
{
|
173 |
+
// vars
|
174 |
+
int T,t;
|
175 |
+
int N=0;
|
176 |
+
int i,j,a,b,p,r,z,ans;
|
177 |
+
// testcase loop
|
178 |
+
Read(T);
|
179 |
+
Fox1(t,T)
|
180 |
+
{
|
181 |
+
// init
|
182 |
+
Fox(i,K)
|
183 |
+
con[i].clear();
|
184 |
+
Fox(i,A)
|
185 |
+
con2[i].clear();
|
186 |
+
K=0;
|
187 |
+
M.clear();
|
188 |
+
Fill(col0,0);
|
189 |
+
Fill(col,-1);
|
190 |
+
Fill(ind,-1);
|
191 |
+
Fill(P1,-1);
|
192 |
+
Fill(P2,-1);
|
193 |
+
// input
|
194 |
+
Read(N);
|
195 |
+
while (N--)
|
196 |
+
{
|
197 |
+
Read(p),Read(r);
|
198 |
+
i=Make(p),a=Make(p-r),b=Make(p+r);
|
199 |
+
col0[i]=1;
|
200 |
+
con[a].pb(b);
|
201 |
+
con[b].pb(a);
|
202 |
+
}
|
203 |
+
// attempt to 2-color the graph (initially just from forced nodes)
|
204 |
+
Fox(z,2)
|
205 |
+
Fox(i,K)
|
206 |
+
if (col[i]<0)
|
207 |
+
{
|
208 |
+
if ((!z) && (!col0[i]))
|
209 |
+
continue;
|
210 |
+
if (!DFS(i,0))
|
211 |
+
{
|
212 |
+
ans=-1;
|
213 |
+
goto Done;
|
214 |
+
}
|
215 |
+
}
|
216 |
+
// construct the bipartite graph, ignoring forced nodes
|
217 |
+
A=B=0;
|
218 |
+
Fox(i,K)
|
219 |
+
if (!col0[i])
|
220 |
+
{
|
221 |
+
Fox(j,Sz(con[i]))
|
222 |
+
if (col0[con[i][j]])
|
223 |
+
goto Skip;
|
224 |
+
if (!col[i])
|
225 |
+
ind[i]=A++;
|
226 |
+
else
|
227 |
+
ind[i]=B++;
|
228 |
+
Skip:;
|
229 |
+
}
|
230 |
+
Fox(i,K)
|
231 |
+
if ((ind[i]>=0) && (!col[i]))
|
232 |
+
{
|
233 |
+
a=ind[i];
|
234 |
+
Fox(j,Sz(con[i]))
|
235 |
+
{
|
236 |
+
b=ind[con[i][j]];
|
237 |
+
if (b>=0)
|
238 |
+
con2[a].pb(b);
|
239 |
+
}
|
240 |
+
}
|
241 |
+
ans=A+B-MaxMatching()+1;
|
242 |
+
// output
|
243 |
+
Done:;
|
244 |
+
printf("Case #%d: %d\n",t,ans);
|
245 |
+
}
|
246 |
+
return(0);
|
247 |
+
}
|
2017/finals/moles.html
ADDED
@@ -0,0 +1,88 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
<p>
|
2 |
+
It's dinner time!
|
3 |
+
A group of <strong>N</strong> Foxen are standing silently in a field, which can be represented as an infinite number line,
|
4 |
+
patiently waiting for their meals to make an appearance.
|
5 |
+
The <em>i</em>th Fox is standing at position <strong>P<sub>i</sub></strong>,
|
6 |
+
with no two Foxen standing at the same position.
|
7 |
+
There's also one hole in the ground at each each integral position on the number line.
|
8 |
+
Each of these holes is the entrance to a mole's den, and the Foxen know that some of these delicious critters are
|
9 |
+
bound to show up sooner or later!
|
10 |
+
</p>
|
11 |
+
|
12 |
+
<p>
|
13 |
+
A little-known fact about Foxen is that, in addition to having an acute array of regular senses,
|
14 |
+
they possess a SONAR-like ability to emit imperceptible sound waves and use them to discern objects at great distances.
|
15 |
+
The <em>i</em>th Fox has tuned their wavelength to a distance of <strong>R<sub>i</sub></strong>,
|
16 |
+
allowing them to only detect moles which emerge from holes at a distance of exactly <strong>R<sub>i</sub></strong> away from them
|
17 |
+
(that is, at either position <strong>P<sub>i</sub></strong> - <strong>R<sub>i</sub></strong> or
|
18 |
+
<strong>P<sub>i</sub></strong> + <strong>R<sub>i</sub></strong>).
|
19 |
+
|
20 |
+
</p>
|
21 |
+
|
22 |
+
<p>
|
23 |
+
All of a sudden, some number of moles have just popped up from various holes all at once!
|
24 |
+
No mole popped up at any Fox's position, no two moles popped up from the same hole, and
|
25 |
+
every mole was detected by at least one Fox.
|
26 |
+
Furthermore, each Fox <em>i</em> has determined that there's <i>exactly</i> 1 mole
|
27 |
+
at a distance of <strong>R<sub>i</sub></strong> away from it (as opposed to there being either 0 or 2 such moles).
|
28 |
+
</p>
|
29 |
+
|
30 |
+
<p>
|
31 |
+
Following this initial event, there's been quite some commotion.
|
32 |
+
Some moles may have retreated back underground, and some new moles may have emerged, all in any order.
|
33 |
+
At every point in time, the set of moles on the surface is subject to all of the same restrictions as before, with one difference:
|
34 |
+
Each Fox <em>i</em> continues to be sure that <i>at least</i> 1 mole is still present
|
35 |
+
at a distance of <strong>R<sub>i</sub></strong> away from it, but can no longer determine whether or not
|
36 |
+
there are perhaps now 2 such moles instead.
|
37 |
+
</p>
|
38 |
+
|
39 |
+
<p>
|
40 |
+
After some time of this, the Foxen have decided that they're ready to pounce and "invite" some of the moles
|
41 |
+
currently on the surface over for dinner.
|
42 |
+
Unfortunately, they've started to become rather overwhelmed with trying to keep track of which moles
|
43 |
+
may be on the surface, or even roughly how many of them there might be.
|
44 |
+
Assuming that the Foxen's initial observations were correct, and that some unknown amount of time has since gone by
|
45 |
+
with moles surfacing or departing, please help the Foxen determine the number of different quantities of moles which
|
46 |
+
could possibly have ended up on the surface.
|
47 |
+
</p>
|
48 |
+
|
49 |
+
<p>
|
50 |
+
If it's impossible for their set of initial observations to have been accurate in the first place, output -1 instead.
|
51 |
+
</p>
|
52 |
+
|
53 |
+
|
54 |
+
<h3>Input</h3>
|
55 |
+
|
56 |
+
<p>
|
57 |
+
Input begins with an integer <strong>T</strong>, the number of different fields.
|
58 |
+
For each field, there is first a line containing the integer <strong>N</strong>.
|
59 |
+
Then <strong>N</strong> lines follow, the <em>i</em>th of which contains the space-separated integers
|
60 |
+
<strong>P<sub>i</sub></strong> and <strong>R<sub>i</sub></strong>.
|
61 |
+
</p>
|
62 |
+
|
63 |
+
<h3>Output</h3>
|
64 |
+
|
65 |
+
<p>
|
66 |
+
For the <em>i</em>th field, print a line containing "Case #<strong>i</strong>: "
|
67 |
+
followed by a single integer, the number of different quantities of moles which could possibly end up on the surface at any point,
|
68 |
+
or -1 if the Foxen's initial observations must have been inaccurate.
|
69 |
+
</p>
|
70 |
+
|
71 |
+
<h3>Constraints</h3>
|
72 |
+
|
73 |
+
<p>
|
74 |
+
1 ≤ <strong>T</strong> ≤ 30<br />
|
75 |
+
1 ≤ <strong>N</strong> ≤ 5,000 <br />
|
76 |
+
0 ≤ <strong>P<sub>i</sub></strong> ≤ 1,000,000,000 <br />
|
77 |
+
1 ≤ <strong>R<sub>i</sub></strong> ≤ 1,000,000,000 <br />
|
78 |
+
</p>
|
79 |
+
|
80 |
+
<h3>Explanation of Sample</h3>
|
81 |
+
|
82 |
+
<p>
|
83 |
+
In the first case, it's possible for there to eventually be 1 mole (at either position -1 or 1), or 2 moles (at both positions -1 and 1). There can't be 0 moles due to the restriction that the Fox must detect at least 1 of them, and there can't be more than 2 moles as they'd have to be at positions which the Fox is unable to detect.
|
84 |
+
</p>
|
85 |
+
|
86 |
+
<p>
|
87 |
+
In the third case, it's impossible for a set of moles to have initially popped up such that each Fox would have detected <em>exactly</em> one of them.
|
88 |
+
</p>
|
2017/finals/moles.in
ADDED
The diff for this file is too large to render.
See raw diff
|
|
2017/finals/moles.md
ADDED
@@ -0,0 +1,75 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
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|
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|
|
|
|
|
|
|
|
|
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|
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|
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|
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|
|
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|
|
|
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|
|
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|
|
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|
|
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|
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|
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|
|
|
|
|
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|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
It's dinner time! A group of **N** Foxen are standing silently in a field,
|
2 |
+
which can be represented as an infinite number line, patiently waiting for
|
3 |
+
their meals to make an appearance. The _i_th Fox is standing at position
|
4 |
+
**Pi**, with no two Foxen standing at the same position. There's also one hole
|
5 |
+
in the ground at each each integral position on the number line. Each of these
|
6 |
+
holes is the entrance to a mole's den, and the Foxen know that some of these
|
7 |
+
delicious critters are bound to show up sooner or later!
|
8 |
+
|
9 |
+
A little-known fact about Foxen is that, in addition to having an acute array
|
10 |
+
of regular senses, they possess a SONAR-like ability to emit imperceptible
|
11 |
+
sound waves and use them to discern objects at great distances. The _i_th Fox
|
12 |
+
has tuned their wavelength to a distance of **Ri**, allowing them to only
|
13 |
+
detect moles which emerge from holes at a distance of exactly **Ri** away from
|
14 |
+
them (that is, at either position **Pi** \- **Ri** or **Pi** \+ **Ri**).
|
15 |
+
|
16 |
+
All of a sudden, some number of moles have just popped up from various holes
|
17 |
+
all at once! No mole popped up at any Fox's position, no two moles popped up
|
18 |
+
from the same hole, and every mole was detected by at least one Fox.
|
19 |
+
Furthermore, each Fox _i_ has determined that there's _exactly_ 1 mole at a
|
20 |
+
distance of **Ri** away from it (as opposed to there being either 0 or 2 such
|
21 |
+
moles).
|
22 |
+
|
23 |
+
Following this initial event, there's been quite some commotion. Some moles
|
24 |
+
may have retreated back underground, and some new moles may have emerged, all
|
25 |
+
in any order. At every point in time, the set of moles on the surface is
|
26 |
+
subject to all of the same restrictions as before, with one difference: Each
|
27 |
+
Fox _i_ continues to be sure that _at least_ 1 mole is still present at a
|
28 |
+
distance of **Ri** away from it, but can no longer determine whether or not
|
29 |
+
there are perhaps now 2 such moles instead.
|
30 |
+
|
31 |
+
After some time of this, the Foxen have decided that they're ready to pounce
|
32 |
+
and "invite" some of the moles currently on the surface over for dinner.
|
33 |
+
Unfortunately, they've started to become rather overwhelmed with trying to
|
34 |
+
keep track of which moles may be on the surface, or even roughly how many of
|
35 |
+
them there might be. Assuming that the Foxen's initial observations were
|
36 |
+
correct, and that some unknown amount of time has since gone by with moles
|
37 |
+
surfacing or departing, please help the Foxen determine the number of
|
38 |
+
different quantities of moles which could possibly have ended up on the
|
39 |
+
surface.
|
40 |
+
|
41 |
+
If it's impossible for their set of initial observations to have been accurate
|
42 |
+
in the first place, output -1 instead.
|
43 |
+
|
44 |
+
### Input
|
45 |
+
|
46 |
+
Input begins with an integer **T**, the number of different fields. For each
|
47 |
+
field, there is first a line containing the integer **N**. Then **N** lines
|
48 |
+
follow, the _i_th of which contains the space-separated integers **Pi** and
|
49 |
+
**Ri**.
|
50 |
+
|
51 |
+
### Output
|
52 |
+
|
53 |
+
For the _i_th field, print a line containing "Case #**i**: " followed by a
|
54 |
+
single integer, the number of different quantities of moles which could
|
55 |
+
possibly end up on the surface at any point, or -1 if the Foxen's initial
|
56 |
+
observations must have been inaccurate.
|
57 |
+
|
58 |
+
### Constraints
|
59 |
+
|
60 |
+
1 ≤ **T** ≤ 30
|
61 |
+
1 ≤ **N** ≤ 5,000
|
62 |
+
0 ≤ **Pi** ≤ 1,000,000,000
|
63 |
+
1 ≤ **Ri** ≤ 1,000,000,000
|
64 |
+
|
65 |
+
### Explanation of Sample
|
66 |
+
|
67 |
+
In the first case, it's possible for there to eventually be 1 mole (at either
|
68 |
+
position -1 or 1), or 2 moles (at both positions -1 and 1). There can't be 0
|
69 |
+
moles due to the restriction that the Fox must detect at least 1 of them, and
|
70 |
+
there can't be more than 2 moles as they'd have to be at positions which the
|
71 |
+
Fox is unable to detect.
|
72 |
+
|
73 |
+
In the third case, it's impossible for a set of moles to have initially popped
|
74 |
+
up such that each Fox would have detected _exactly_ one of them.
|
75 |
+
|
2017/finals/moles.out
ADDED
@@ -0,0 +1,30 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
Case #1: 2
|
2 |
+
Case #2: 3
|
3 |
+
Case #3: -1
|
4 |
+
Case #4: 4
|
5 |
+
Case #5: 10
|
6 |
+
Case #6: 2
|
7 |
+
Case #7: 5
|
8 |
+
Case #8: 1
|
9 |
+
Case #9: 2
|
10 |
+
Case #10: -1
|
11 |
+
Case #11: 1363
|
12 |
+
Case #12: 1493
|
13 |
+
Case #13: 68
|
14 |
+
Case #14: 123
|
15 |
+
Case #15: 417
|
16 |
+
Case #16: 1215
|
17 |
+
Case #17: -1
|
18 |
+
Case #18: 79
|
19 |
+
Case #19: 133
|
20 |
+
Case #20: 134
|
21 |
+
Case #21: 964
|
22 |
+
Case #22: 38
|
23 |
+
Case #23: 1489
|
24 |
+
Case #24: -1
|
25 |
+
Case #25: 345
|
26 |
+
Case #26: 113
|
27 |
+
Case #27: 1026
|
28 |
+
Case #28: 795
|
29 |
+
Case #29: 659
|
30 |
+
Case #30: 1076
|
2017/finals/patrols.cpp
ADDED
@@ -0,0 +1,123 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
// Hacker Cup 2017
|
2 |
+
// Final Round
|
3 |
+
// Fox Patrols
|
4 |
+
// Jacob Plachta
|
5 |
+
|
6 |
+
#include <algorithm>
|
7 |
+
#include <functional>
|
8 |
+
#include <numeric>
|
9 |
+
#include <iostream>
|
10 |
+
#include <iomanip>
|
11 |
+
#include <cstdio>
|
12 |
+
#include <cmath>
|
13 |
+
#include <complex>
|
14 |
+
#include <cstdlib>
|
15 |
+
#include <ctime>
|
16 |
+
#include <cstring>
|
17 |
+
#include <cassert>
|
18 |
+
#include <string>
|
19 |
+
#include <vector>
|
20 |
+
#include <list>
|
21 |
+
#include <map>
|
22 |
+
#include <set>
|
23 |
+
#include <deque>
|
24 |
+
#include <queue>
|
25 |
+
#include <stack>
|
26 |
+
#include <bitset>
|
27 |
+
#include <sstream>
|
28 |
+
using namespace std;
|
29 |
+
|
30 |
+
#define LL long long
|
31 |
+
#define LD long double
|
32 |
+
#define PR pair<int,int>
|
33 |
+
|
34 |
+
#define Fox(i,n) for (i=0; i<n; i++)
|
35 |
+
#define Fox1(i,n) for (i=1; i<=n; i++)
|
36 |
+
#define FoxI(i,a,b) for (i=a; i<=b; i++)
|
37 |
+
#define FoxR(i,n) for (i=(n)-1; i>=0; i--)
|
38 |
+
#define FoxR1(i,n) for (i=n; i>0; i--)
|
39 |
+
#define FoxRI(i,a,b) for (i=b; i>=a; i--)
|
40 |
+
#define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++)
|
41 |
+
#define Min(a,b) a=min(a,b)
|
42 |
+
#define Max(a,b) a=max(a,b)
|
43 |
+
#define Sz(s) int((s).size())
|
44 |
+
#define All(s) (s).begin(),(s).end()
|
45 |
+
#define Fill(s,v) memset(s,v,sizeof(s))
|
46 |
+
#define pb push_back
|
47 |
+
#define mp make_pair
|
48 |
+
#define x first
|
49 |
+
#define y second
|
50 |
+
|
51 |
+
template<typename T> T Abs(T x) { return(x<0 ? -x : x); }
|
52 |
+
template<typename T> T Sqr(T x) { return(x*x); }
|
53 |
+
|
54 |
+
const int INF = (int)1e9;
|
55 |
+
const LD EPS = 1e-12;
|
56 |
+
const LD PI = acos(-1.0);
|
57 |
+
|
58 |
+
bool Read(int &x)
|
59 |
+
{
|
60 |
+
char c,r=0,n=0;
|
61 |
+
x=0;
|
62 |
+
for(;;)
|
63 |
+
{
|
64 |
+
c=getchar();
|
65 |
+
if ((c<0) && (!r))
|
66 |
+
return(0);
|
67 |
+
if ((c=='-') && (!r))
|
68 |
+
n=1;
|
69 |
+
else
|
70 |
+
if ((c>='0') && (c<='9'))
|
71 |
+
x=x*10+c-'0',r=1;
|
72 |
+
else
|
73 |
+
if (r)
|
74 |
+
break;
|
75 |
+
}
|
76 |
+
if (n)
|
77 |
+
x=-x;
|
78 |
+
return(1);
|
79 |
+
}
|
80 |
+
|
81 |
+
#define MOD 1000000007
|
82 |
+
|
83 |
+
// a^b
|
84 |
+
int pw(LL a,int b)
|
85 |
+
{
|
86 |
+
int v=1;
|
87 |
+
while (b)
|
88 |
+
{
|
89 |
+
if (b&1)
|
90 |
+
v=v*a%MOD;
|
91 |
+
a=a*a%MOD;
|
92 |
+
b>>=1;
|
93 |
+
}
|
94 |
+
return(v);
|
95 |
+
}
|
96 |
+
|
97 |
+
// number of skylines of length n with max height h
|
98 |
+
LL cnt(int n,int h)
|
99 |
+
{
|
100 |
+
return((pw(h,n)-pw(h-1,n)+MOD)%MOD);
|
101 |
+
}
|
102 |
+
|
103 |
+
int main()
|
104 |
+
{
|
105 |
+
// vars
|
106 |
+
int T,t;
|
107 |
+
int N,M,H;
|
108 |
+
int h,ans;
|
109 |
+
// testcase loop
|
110 |
+
Read(T);
|
111 |
+
Fox1(t,T)
|
112 |
+
{
|
113 |
+
// input
|
114 |
+
Read(N),Read(M),Read(H);
|
115 |
+
// consider all possible max tree heights
|
116 |
+
ans=0;
|
117 |
+
Fox1(h,H)
|
118 |
+
ans=(ans+cnt(N,h)*cnt(M,h))%MOD;
|
119 |
+
// output
|
120 |
+
printf("Case #%d: %d\n",t,ans);
|
121 |
+
}
|
122 |
+
return(0);
|
123 |
+
}
|
2017/finals/patrols.html
ADDED
@@ -0,0 +1,82 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
<p>
|
2 |
+
A certain well-hidden valley is home to a thriving population of mysterious creatures — Foxen!
|
3 |
+
However, keeping the valley safe from outsiders (such as humans) is a necessity.
|
4 |
+
To that end, a group of Foxen have been sent out to patrol the border.
|
5 |
+
</p>
|
6 |
+
|
7 |
+
<p>
|
8 |
+
On their patrol route, the Foxen know that they're going to pass by an interesting, rectangular forest.
|
9 |
+
When viewed from above, the forest can be modeled as a grid of cells
|
10 |
+
with <strong>R</strong> rows and <strong>C</strong> columns. The rows are numbered from 1 to <strong>R</strong> from North to South, while the column are numbered from 1 to <strong>C</strong> from West to East.
|
11 |
+
One tree is growing in the center of each cell, and each tree's height (in metres) is some positive integer no larger than <strong>H</strong>.
|
12 |
+
</p>
|
13 |
+
|
14 |
+
<p>
|
15 |
+
If the Foxen were to look at the forest from the North side, all of the trees in any given column of cells would obscure each other and blend together.
|
16 |
+
In fact, the Foxen would really only be able make out the overall shape of the forest's "skyline" when viewed from that direction.
|
17 |
+
This Northern skyline can be expressed as a sequence of <strong>C</strong> positive integers,
|
18 |
+
with the <em>i</em>th one being the largest of the <strong>R</strong> tree heights in the <em>i</em>th column.
|
19 |
+
</p>
|
20 |
+
|
21 |
+
<p>
|
22 |
+
Similarly, if they were to look at the forest from the West side, they would only be able to make out the shape of its skyline from that direction.
|
23 |
+
This Western skyline is a sequence of <strong>R</strong> positive integers,
|
24 |
+
with the <em>i</em>th one being the largest of the <strong>C</strong> tree heights in the <em>i</em>th row.
|
25 |
+
</p>
|
26 |
+
|
27 |
+
<p>
|
28 |
+
On their way to the forest, the Foxen find themselves wondering about what it might look like.
|
29 |
+
They've done their research and are aware of its dimensions <strong>R</strong> and <strong>C</strong>,
|
30 |
+
as well as the maximum possible height of its trees <strong>H</strong>, but they don't know the actual heights of any of its trees.
|
31 |
+
They'd like to determine how many different, distinct-looking forests they might end up finding.
|
32 |
+
A forest is a set of heights for all <strong>R</strong>x<strong>C</strong> trees,
|
33 |
+
and two forests are considered to be distinct-looking from one another if their Northern skyline sequences differ and/or their Western skyline sequences differ.
|
34 |
+
</p>
|
35 |
+
|
36 |
+
<p>
|
37 |
+
Please help the Foxen determine the number of possible different, distinct-looking forests! As this quantity may be quite large, they're only interested in its value when taken modulo 1,000,000,007.
|
38 |
+
</p>
|
39 |
+
|
40 |
+
|
41 |
+
<h3>Input</h3>
|
42 |
+
|
43 |
+
<p>
|
44 |
+
Input begins with an integer <strong>T</strong>, the number of different forests visited by the Foxen.
|
45 |
+
For each forest, there is a single line containing the three space-separated integers
|
46 |
+
<strong>R</strong>, <strong>C</strong>, and <strong>H</strong>.
|
47 |
+
</p>
|
48 |
+
|
49 |
+
<h3>Output</h3>
|
50 |
+
|
51 |
+
<p>
|
52 |
+
For the <em>i</em>th forest, print a line containing "Case #<strong>i</strong>: "
|
53 |
+
followed by the number of possible different, distinct-looking forests modulo 1,000,000,007.
|
54 |
+
</p>
|
55 |
+
|
56 |
+
<h3>Constraints</h3>
|
57 |
+
|
58 |
+
<p>
|
59 |
+
1 ≤ <strong>T</strong> ≤ 30 <br />
|
60 |
+
1 ≤ <strong>R</strong>, <strong>C</strong>, <strong>H</strong> ≤ 500,000 <br />
|
61 |
+
</p>
|
62 |
+
|
63 |
+
<h3>Explanation of Sample</h3>
|
64 |
+
|
65 |
+
<p>
|
66 |
+
In the first case, there are 10 possible different, distinct-looking forests which consist of a 2x2 grid of trees, with each tree being either 1m or 2m tall.
|
67 |
+
For example, the following 2 forests look different (even though their Western skylines are equal, their Northern skylines differ), so both should be counted:
|
68 |
+
</p>
|
69 |
+
|
70 |
+
<pre>
|
71 |
+
1 2 2 1
|
72 |
+
1 1 1 1
|
73 |
+
</pre>
|
74 |
+
|
75 |
+
<p>
|
76 |
+
On the other hand, the following 2 forests look identical to one another from both the North and the West, so only one of them should be counted:
|
77 |
+
</p>
|
78 |
+
|
79 |
+
<pre>
|
80 |
+
1 2 2 2
|
81 |
+
2 1 2 1
|
82 |
+
</pre>
|
2017/finals/patrols.in
ADDED
@@ -0,0 +1,31 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
30
|
2 |
+
2 2 2
|
3 |
+
3 3 3
|
4 |
+
10 10 2
|
5 |
+
10 10 3
|
6 |
+
12345 54321 98765
|
7 |
+
1 1 1
|
8 |
+
1 1 500000
|
9 |
+
1 500000 1
|
10 |
+
1 500000 500000
|
11 |
+
500000 1 1
|
12 |
+
500000 1 500000
|
13 |
+
500000 500000 1
|
14 |
+
500000 500000 500000
|
15 |
+
320813 317748 38903
|
16 |
+
138867 351451 187697
|
17 |
+
364035 163376 439098
|
18 |
+
122612 43988 469986
|
19 |
+
226214 129928 408977
|
20 |
+
284198 155331 429992
|
21 |
+
484558 292080 226183
|
22 |
+
171737 440766 26340
|
23 |
+
254725 193588 492598
|
24 |
+
36190 169975 160386
|
25 |
+
41408 136075 416935
|
26 |
+
192383 203512 259578
|
27 |
+
60442 2849 34087
|
28 |
+
373965 441433 45490
|
29 |
+
138969 349737 156766
|
30 |
+
216121 452029 88461
|
31 |
+
162647 304920 107641
|
2017/finals/patrols.md
ADDED
@@ -0,0 +1,72 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
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|
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|
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|
|
|
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|
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|
|
|
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|
|
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|
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|
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|
|
|
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|
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|
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|
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|
|
|
|
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|
|
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|
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|
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|
|
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|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
A certain well-hidden valley is home to a thriving population of mysterious
|
2 |
+
creatures — Foxen! However, keeping the valley safe from outsiders (such as
|
3 |
+
humans) is a necessity. To that end, a group of Foxen have been sent out to
|
4 |
+
patrol the border.
|
5 |
+
|
6 |
+
On their patrol route, the Foxen know that they're going to pass by an
|
7 |
+
interesting, rectangular forest. When viewed from above, the forest can be
|
8 |
+
modeled as a grid of cells with **R** rows and **C** columns. The rows are
|
9 |
+
numbered from 1 to **R** from North to South, while the column are numbered
|
10 |
+
from 1 to **C** from West to East. One tree is growing in the center of each
|
11 |
+
cell, and each tree's height (in metres) is some positive integer no larger
|
12 |
+
than **H**.
|
13 |
+
|
14 |
+
If the Foxen were to look at the forest from the North side, all of the trees
|
15 |
+
in any given column of cells would obscure each other and blend together. In
|
16 |
+
fact, the Foxen would really only be able make out the overall shape of the
|
17 |
+
forest's "skyline" when viewed from that direction. This Northern skyline can
|
18 |
+
be expressed as a sequence of **C** positive integers, with the _i_th one
|
19 |
+
being the largest of the **R** tree heights in the _i_th column.
|
20 |
+
|
21 |
+
Similarly, if they were to look at the forest from the West side, they would
|
22 |
+
only be able to make out the shape of its skyline from that direction. This
|
23 |
+
Western skyline is a sequence of **R** positive integers, with the _i_th one
|
24 |
+
being the largest of the **C** tree heights in the _i_th row.
|
25 |
+
|
26 |
+
On their way to the forest, the Foxen find themselves wondering about what it
|
27 |
+
might look like. They've done their research and are aware of its dimensions
|
28 |
+
**R** and **C**, as well as the maximum possible height of its trees **H**,
|
29 |
+
but they don't know the actual heights of any of its trees. They'd like to
|
30 |
+
determine how many different, distinct-looking forests they might end up
|
31 |
+
finding. A forest is a set of heights for all **R**x**C** trees, and two
|
32 |
+
forests are considered to be distinct-looking from one another if their
|
33 |
+
Northern skyline sequences differ and/or their Western skyline sequences
|
34 |
+
differ.
|
35 |
+
|
36 |
+
Please help the Foxen determine the number of possible different, distinct-
|
37 |
+
looking forests! As this quantity may be quite large, they're only interested
|
38 |
+
in its value when taken modulo 1,000,000,007.
|
39 |
+
|
40 |
+
### Input
|
41 |
+
|
42 |
+
Input begins with an integer **T**, the number of different forests visited by
|
43 |
+
the Foxen. For each forest, there is a single line containing the three space-
|
44 |
+
separated integers **R**, **C**, and **H**.
|
45 |
+
|
46 |
+
### Output
|
47 |
+
|
48 |
+
For the _i_th forest, print a line containing "Case #**i**: " followed by the
|
49 |
+
number of possible different, distinct-looking forests modulo 1,000,000,007.
|
50 |
+
|
51 |
+
### Constraints
|
52 |
+
|
53 |
+
1 ≤ **T** ≤ 30
|
54 |
+
1 ≤ **R**, **C**, **H** ≤ 500,000
|
55 |
+
|
56 |
+
### Explanation of Sample
|
57 |
+
|
58 |
+
In the first case, there are 10 possible different, distinct-looking forests
|
59 |
+
which consist of a 2x2 grid of trees, with each tree being either 1m or 2m
|
60 |
+
tall. For example, the following 2 forests look different (even though their
|
61 |
+
Western skylines are equal, their Northern skylines differ), so both should be
|
62 |
+
counted:
|
63 |
+
|
64 |
+
1 2 2 1
|
65 |
+
1 1 1 1
|
66 |
+
|
67 |
+
On the other hand, the following 2 forests look identical to one another from
|
68 |
+
both the North and the West, so only one of them should be counted:
|
69 |
+
|
70 |
+
1 2 2 2
|
71 |
+
2 1 2 1
|
72 |
+
|
2017/finals/patrols.out
ADDED
@@ -0,0 +1,30 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
Case #1: 10
|
2 |
+
Case #2: 411
|
3 |
+
Case #3: 1046530
|
4 |
+
Case #4: 367947134
|
5 |
+
Case #5: 151251795
|
6 |
+
Case #6: 1
|
7 |
+
Case #7: 500000
|
8 |
+
Case #8: 1
|
9 |
+
Case #9: 640164345
|
10 |
+
Case #10: 1
|
11 |
+
Case #11: 640164345
|
12 |
+
Case #12: 1
|
13 |
+
Case #13: 784439684
|
14 |
+
Case #14: 628695691
|
15 |
+
Case #15: 473900163
|
16 |
+
Case #16: 161834349
|
17 |
+
Case #17: 102371132
|
18 |
+
Case #18: 83427294
|
19 |
+
Case #19: 535513662
|
20 |
+
Case #20: 527573954
|
21 |
+
Case #21: 223154519
|
22 |
+
Case #22: 426038426
|
23 |
+
Case #23: 86290130
|
24 |
+
Case #24: 856759245
|
25 |
+
Case #25: 511958015
|
26 |
+
Case #26: 672344838
|
27 |
+
Case #27: 350475264
|
28 |
+
Case #28: 968392936
|
29 |
+
Case #29: 340846376
|
30 |
+
Case #30: 730119034
|
2017/finals/poles.cpp
ADDED
@@ -0,0 +1,202 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
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|
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|
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|
|
|
|
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
// Hacker Cup 2017
|
2 |
+
// Final Round
|
3 |
+
// Fox Poles
|
4 |
+
// Jacob Plachta
|
5 |
+
|
6 |
+
#include <algorithm>
|
7 |
+
#include <functional>
|
8 |
+
#include <numeric>
|
9 |
+
#include <iostream>
|
10 |
+
#include <iomanip>
|
11 |
+
#include <cstdio>
|
12 |
+
#include <cmath>
|
13 |
+
#include <complex>
|
14 |
+
#include <cstdlib>
|
15 |
+
#include <ctime>
|
16 |
+
#include <cstring>
|
17 |
+
#include <cassert>
|
18 |
+
#include <string>
|
19 |
+
#include <vector>
|
20 |
+
#include <list>
|
21 |
+
#include <map>
|
22 |
+
#include <set>
|
23 |
+
#include <deque>
|
24 |
+
#include <queue>
|
25 |
+
#include <stack>
|
26 |
+
#include <bitset>
|
27 |
+
#include <sstream>
|
28 |
+
using namespace std;
|
29 |
+
|
30 |
+
#define LL long long
|
31 |
+
#define LD long double
|
32 |
+
#define PR pair<int,int>
|
33 |
+
|
34 |
+
#define Fox(i,n) for (i=0; i<n; i++)
|
35 |
+
#define Fox1(i,n) for (i=1; i<=n; i++)
|
36 |
+
#define FoxI(i,a,b) for (i=a; i<=b; i++)
|
37 |
+
#define FoxR(i,n) for (i=(n)-1; i>=0; i--)
|
38 |
+
#define FoxR1(i,n) for (i=n; i>0; i--)
|
39 |
+
#define FoxRI(i,a,b) for (i=b; i>=a; i--)
|
40 |
+
#define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++)
|
41 |
+
#define Min(a,b) a=min(a,b)
|
42 |
+
#define Max(a,b) a=max(a,b)
|
43 |
+
#define Sz(s) int((s).size())
|
44 |
+
#define All(s) (s).begin(),(s).end()
|
45 |
+
#define Fill(s,v) memset(s,v,sizeof(s))
|
46 |
+
#define pb push_back
|
47 |
+
#define mp make_pair
|
48 |
+
#define x first
|
49 |
+
#define y second
|
50 |
+
|
51 |
+
template<typename T> T Abs(T x) { return(x<0 ? -x : x); }
|
52 |
+
template<typename T> T Sqr(T x) { return(x*x); }
|
53 |
+
|
54 |
+
const int INF = (int)1e9;
|
55 |
+
const LD EPS = 1e-10;
|
56 |
+
const LD PI = acos(-1.0);
|
57 |
+
|
58 |
+
bool Read(int &x)
|
59 |
+
{
|
60 |
+
char c,r=0,n=0;
|
61 |
+
x=0;
|
62 |
+
for(;;)
|
63 |
+
{
|
64 |
+
c=getchar();
|
65 |
+
if ((c<0) && (!r))
|
66 |
+
return(0);
|
67 |
+
if ((c=='-') && (!r))
|
68 |
+
n=1;
|
69 |
+
else
|
70 |
+
if ((c>='0') && (c<='9'))
|
71 |
+
x=x*10+c-'0',r=1;
|
72 |
+
else
|
73 |
+
if (r)
|
74 |
+
break;
|
75 |
+
}
|
76 |
+
if (n)
|
77 |
+
x=-x;
|
78 |
+
return(1);
|
79 |
+
}
|
80 |
+
|
81 |
+
#define LIM 800001
|
82 |
+
|
83 |
+
pair<int,LD> O[LIM],P[LIM];
|
84 |
+
LD mx[LIM];
|
85 |
+
vector<int> ch[LIM];
|
86 |
+
int ss,st[LIM];
|
87 |
+
|
88 |
+
LD Calc(LD h,LD a1,LD a2)
|
89 |
+
{
|
90 |
+
return(-h*(log(cos(a2))-log(cos(a1))));
|
91 |
+
}
|
92 |
+
|
93 |
+
int main()
|
94 |
+
{
|
95 |
+
// vars
|
96 |
+
int T,t;
|
97 |
+
int N;
|
98 |
+
bool D[2];
|
99 |
+
LD A[2];
|
100 |
+
int i,j,c,p,w,z;
|
101 |
+
double v;
|
102 |
+
LD a,a1,a2,tot,ans[2];
|
103 |
+
// testcase loop
|
104 |
+
Read(T);
|
105 |
+
Fox1(t,T)
|
106 |
+
{
|
107 |
+
// input
|
108 |
+
Read(N);
|
109 |
+
Fox(i,2)
|
110 |
+
{
|
111 |
+
Read(j);
|
112 |
+
D[i]=int(j<0);
|
113 |
+
A[i]=j*PI/180;
|
114 |
+
}
|
115 |
+
Fox(i,N)
|
116 |
+
{
|
117 |
+
scanf("%d%lf",&O[i].x,&v);
|
118 |
+
O[i].y=v;
|
119 |
+
}
|
120 |
+
// consider each direction
|
121 |
+
Fox(z,2)
|
122 |
+
{
|
123 |
+
// reverse poles horizontally if necessary
|
124 |
+
memcpy(P,O,sizeof(P));
|
125 |
+
if (D[z])
|
126 |
+
Fox(i,N)
|
127 |
+
P[i].x=-P[i].x;
|
128 |
+
// sort poles left-to-right
|
129 |
+
sort(P,P+N);
|
130 |
+
// init
|
131 |
+
Fox(i,N+1)
|
132 |
+
ch[i].clear();
|
133 |
+
// consider poles left-to-right, constructing convex hull of coverage
|
134 |
+
ss=0;
|
135 |
+
Fox(i,N)
|
136 |
+
{
|
137 |
+
// find this pole's parent, if any
|
138 |
+
while (ss>0)
|
139 |
+
{
|
140 |
+
j=st[ss-1];
|
141 |
+
a=atan2(P[i].x-P[j].x,P[j].y-P[i].y);
|
142 |
+
if (a<mx[j]-EPS)
|
143 |
+
break;
|
144 |
+
ss--;
|
145 |
+
}
|
146 |
+
if (!ss)
|
147 |
+
{
|
148 |
+
mx[i]=Abs(A[z]);
|
149 |
+
ch[N].pb(i);
|
150 |
+
}
|
151 |
+
else
|
152 |
+
{
|
153 |
+
mx[i]=a;
|
154 |
+
ch[j].pb(i);
|
155 |
+
}
|
156 |
+
st[ss++]=i;
|
157 |
+
}
|
158 |
+
// augment tree
|
159 |
+
Fox(j,Sz(ch[N])-1)
|
160 |
+
ch[ch[N][j]].pb(ch[N][j+1]);
|
161 |
+
Fox(i,N)
|
162 |
+
Fox(j,Sz(ch[i])-1)
|
163 |
+
ch[ch[i][j]].pb(ch[i][j+1]);
|
164 |
+
// consider each left pole
|
165 |
+
ans[z]=0;
|
166 |
+
Fox(i,N)
|
167 |
+
{
|
168 |
+
a=0;
|
169 |
+
Fox(c,Sz(ch[i]))
|
170 |
+
{
|
171 |
+
j=ch[i][c];
|
172 |
+
a1=max(a,atan2(P[j].x-P[i].x,P[i].y));
|
173 |
+
a2=atan2(P[j].x-P[i].x,P[i].y-P[j].y);
|
174 |
+
// partial shadow until bottom
|
175 |
+
if (a1>mx[i]-EPS)
|
176 |
+
break;
|
177 |
+
ans[z]+=Calc(P[i].y,a,a1);
|
178 |
+
// full shadow until top
|
179 |
+
w=P[j].x-P[i].x;
|
180 |
+
if (a2>mx[i]-EPS)
|
181 |
+
{
|
182 |
+
ans[z]+=w*(mx[i]-a1);
|
183 |
+
goto Done;
|
184 |
+
}
|
185 |
+
ans[z]+=w*(a2-a1);
|
186 |
+
a=a2;
|
187 |
+
}
|
188 |
+
ans[z]+=Calc(P[i].y,a,mx[i]);
|
189 |
+
Done:;
|
190 |
+
}
|
191 |
+
}
|
192 |
+
// compute answer
|
193 |
+
if (D[0]==D[1])
|
194 |
+
tot=Abs(ans[0]-ans[1]);
|
195 |
+
else
|
196 |
+
tot=ans[0]+ans[1];
|
197 |
+
tot/=A[1]-A[0];
|
198 |
+
// output
|
199 |
+
printf("Case #%d: %.9lf\n",t,(double)tot);
|
200 |
+
}
|
201 |
+
return(0);
|
202 |
+
}
|
2017/finals/poles.html
ADDED
@@ -0,0 +1,55 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
<p>
|
2 |
+
While taking a walk through the woods, a group of Foxen have come upon a curious sight — a row of <strong>N</strong> wooden poles sticking straight up out of the ground! Who placed them there, and why? The Foxen have no clue.
|
3 |
+
</p>
|
4 |
+
|
5 |
+
<p>
|
6 |
+
Looking at the poles from the side, they can be modeled as vertical line segments rising upwards from a number line (which represents the ground), with the <em>i</em>th pole at distinct integral position <strong>P<sub>i</sub></strong> and having a real-valued height of <strong>H<sub>i</sub></strong>.
|
7 |
+
</p>
|
8 |
+
|
9 |
+
<p>
|
10 |
+
One of the Foxen, Ozy, is fascinated by the shadows being cast on the ground by the poles. The sun is shining down on the poles from some point very high up in the sky, resulting in infinitely many rays of light descending towards the number line at all possible positions along it, but all travelling in some uniform direction. Each ray of light stops travelling as soon as it comes into contact with either a pole or the ground. Any point on the ground which is incapable of being reached by rays of light (because they would get blocked by at least one pole before reaching that point) is considered to be covered in shadows.
|
11 |
+
</p>
|
12 |
+
|
13 |
+
<p>
|
14 |
+
The sunlight's direction can be described by a real value <em>a</em>, with absolute value no larger than 80, where <em>a</em> is the signed angle difference (in degrees) between the rays' direction and a vector pointing directly downwards. As an example, let's imagine that there's a single pole at position 50 and with a height of 100. If <em>a</em> = 45, then sunlight is shining diagonally down and to the right, meaning that the pole obstructs rays of light from being able to reach any points on the ground in the interval [50, 150], effectively casting a shadow with length 100 to the right. If <em>a</em> = -45, then sunlight is shining diagonally down and to the left, causing the pole to cast a shadow with 100 to the left instead (over the interval [-50, 50]). If <em>a</em> = 0, then sunlight is shining directly downwards onto the ground, resulting in the pole not casting any shadow.
|
15 |
+
</p>
|
16 |
+
|
17 |
+
<p>
|
18 |
+
Ozy is planning on returning by himself tomorrow in order to observe the poles again, but he doesn't know at what time of day he'll be able to make the trip. He does at least have it narrowed down to being within some interval of time, during which he knows that the sunlight's direction <em>a</em> will range from <strong>A</strong> and <strong>B</strong>, inclusive. Given that the sunlight's direction <em>a</em> will be a real number drawn uniformly at random from the interval [<strong>A</strong>, <strong>B</strong>] when Ozy visits the poles tomorrow, please help him predict the expected total length of ground which will be covered in shadows at that time.
|
19 |
+
</p>
|
20 |
+
|
21 |
+
|
22 |
+
<h3>Input</h3>
|
23 |
+
|
24 |
+
<p>
|
25 |
+
Input begins with an integer <strong>T</strong>, the number of different sets of poles.
|
26 |
+
For each set of poles, there is first a line containing the space-separated integers <strong>N</strong>, <strong>A</strong>, and <strong>B</strong>.
|
27 |
+
Then <strong>N</strong> lines follow, the <em>i</em>th of which contains the integer
|
28 |
+
<strong>P<sub>i</sub></strong> and the real number <strong>H<sub>i</sub></strong> separated by a space.
|
29 |
+
The poles' heights are given with at most 4 digits after the decimal point.
|
30 |
+
</p>
|
31 |
+
|
32 |
+
<h3>Output</h3>
|
33 |
+
|
34 |
+
<p>
|
35 |
+
For the <em>i</em>th set of poles, print a line containing "Case #<strong>i</strong>: "
|
36 |
+
followed by a single real number, the expected length of ground which will be covered in shadows.
|
37 |
+
Your output should have at most 10<sup>-6</sup> absolute or relative error.
|
38 |
+
</p>
|
39 |
+
|
40 |
+
<h3>Constraints</h3>
|
41 |
+
|
42 |
+
<p>
|
43 |
+
1 ≤ <strong>T</strong> ≤ 30<br />
|
44 |
+
1 ≤ <strong>N</strong> ≤ 500,000 <br />
|
45 |
+
-80 ≤ <strong>A</strong> < <strong>B</strong> ≤ 80 <br />
|
46 |
+
0 ≤ <strong>P<sub>i</sub></strong> ≤ 1,000,000,000 <br />
|
47 |
+
1 ≤ <strong>H<sub>i</sub></strong> ≤ 1,000,000 <br />
|
48 |
+
The sum of <strong>N</strong> values across all <strong>T</strong> cases does not exceed 2,000,000.
|
49 |
+
</p>
|
50 |
+
|
51 |
+
<h3>Explanation of Sample</h3>
|
52 |
+
|
53 |
+
<p>
|
54 |
+
In the first case, the sunlight's direction <em>a</em> is drawn uniformly at random from the interval [44, 46]. As described above, the length of ground covered in shadows when <em>a</em> = 45 is exactly 100. When <em>a</em> = 44, the shadow's length is ~96.57, and when <em>a</em> = 46, its length is ~103.55. However, note that its expected length for this distribution of possible <em>a</em> values is not equal to the average of those sample lengths.
|
55 |
+
</p>
|
2017/finals/poles.in
ADDED
@@ -0,0 +1,3 @@
|
|
|
|
|
|
|
|
|
1 |
+
version https://git-lfs.github.com/spec/v1
|
2 |
+
oid sha256:e9d2250a37156e0f4c67b9967987999557727d4f04df2b1b75f0eb6addfecd5d
|
3 |
+
size 23912078
|
2017/finals/poles.md
ADDED
@@ -0,0 +1,73 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
While taking a walk through the woods, a group of Foxen have come upon a
|
2 |
+
curious sight — a row of **N** wooden poles sticking straight up out of the
|
3 |
+
ground! Who placed them there, and why? The Foxen have no clue.
|
4 |
+
|
5 |
+
Looking at the poles from the side, they can be modeled as vertical line
|
6 |
+
segments rising upwards from a number line (which represents the ground), with
|
7 |
+
the _i_th pole at distinct integral position **Pi** and having a real-valued
|
8 |
+
height of **Hi**.
|
9 |
+
|
10 |
+
One of the Foxen, Ozy, is fascinated by the shadows being cast on the ground
|
11 |
+
by the poles. The sun is shining down on the poles from some point very high
|
12 |
+
up in the sky, resulting in infinitely many rays of light descending towards
|
13 |
+
the number line at all possible positions along it, but all travelling in some
|
14 |
+
uniform direction. Each ray of light stops travelling as soon as it comes into
|
15 |
+
contact with either a pole or the ground. Any point on the ground which is
|
16 |
+
incapable of being reached by rays of light (because they would get blocked by
|
17 |
+
at least one pole before reaching that point) is considered to be covered in
|
18 |
+
shadows.
|
19 |
+
|
20 |
+
The sunlight's direction can be described by a real value _a_, with absolute
|
21 |
+
value no larger than 80, where _a_ is the signed angle difference (in degrees)
|
22 |
+
between the rays' direction and a vector pointing directly downwards. As an
|
23 |
+
example, let's imagine that there's a single pole at position 50 and with a
|
24 |
+
height of 100. If _a_ = 45, then sunlight is shining diagonally down and to
|
25 |
+
the right, meaning that the pole obstructs rays of light from being able to
|
26 |
+
reach any points on the ground in the interval [50, 150], effectively casting
|
27 |
+
a shadow with length 100 to the right. If _a_ = -45, then sunlight is shining
|
28 |
+
diagonally down and to the left, causing the pole to cast a shadow with 100 to
|
29 |
+
the left instead (over the interval [-50, 50]). If _a_ = 0, then sunlight is
|
30 |
+
shining directly downwards onto the ground, resulting in the pole not casting
|
31 |
+
any shadow.
|
32 |
+
|
33 |
+
Ozy is planning on returning by himself tomorrow in order to observe the poles
|
34 |
+
again, but he doesn't know at what time of day he'll be able to make the trip.
|
35 |
+
He does at least have it narrowed down to being within some interval of time,
|
36 |
+
during which he knows that the sunlight's direction _a_ will range from **A**
|
37 |
+
and **B**, inclusive. Given that the sunlight's direction _a_ will be a real
|
38 |
+
number drawn uniformly at random from the interval [**A**, **B**] when Ozy
|
39 |
+
visits the poles tomorrow, please help him predict the expected total length
|
40 |
+
of ground which will be covered in shadows at that time.
|
41 |
+
|
42 |
+
### Input
|
43 |
+
|
44 |
+
Input begins with an integer **T**, the number of different sets of poles. For
|
45 |
+
each set of poles, there is first a line containing the space-separated
|
46 |
+
integers **N**, **A**, and **B**. Then **N** lines follow, the _i_th of which
|
47 |
+
contains the integer **Pi** and the real number **Hi** separated by a space.
|
48 |
+
The poles' heights are given with at most 4 digits after the decimal point.
|
49 |
+
|
50 |
+
### Output
|
51 |
+
|
52 |
+
For the _i_th set of poles, print a line containing "Case #**i**: " followed
|
53 |
+
by a single real number, the expected length of ground which will be covered
|
54 |
+
in shadows. Your output should have at most 10-6 absolute or relative error.
|
55 |
+
|
56 |
+
### Constraints
|
57 |
+
|
58 |
+
1 ≤ **T** ≤ 30
|
59 |
+
1 ≤ **N** ≤ 500,000
|
60 |
+
-80 ≤ **A** < **B** ≤ 80
|
61 |
+
0 ≤ **Pi** ≤ 1,000,000,000
|
62 |
+
1 ≤ **Hi** ≤ 1,000,000
|
63 |
+
The sum of **N** values across all **T** cases does not exceed 2,000,000.
|
64 |
+
|
65 |
+
### Explanation of Sample
|
66 |
+
|
67 |
+
In the first case, the sunlight's direction _a_ is drawn uniformly at random
|
68 |
+
from the interval [44, 46]. As described above, the length of ground covered
|
69 |
+
in shadows when _a_ = 45 is exactly 100. When _a_ = 44, the shadow's length is
|
70 |
+
~96.57, and when _a_ = 46, its length is ~103.55. However, note that its
|
71 |
+
expected length for this distribution of possible _a_ values is not equal to
|
72 |
+
the average of those sample lengths.
|
73 |
+
|
2017/finals/poles.out
ADDED
@@ -0,0 +1,30 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
Case #1: 100.020314016
|
2 |
+
Case #2: 199.157629122
|
3 |
+
Case #3: 5.731972014
|
4 |
+
Case #4: 29594.141688738
|
5 |
+
Case #5: 155.088749996
|
6 |
+
Case #6: 540.964700227
|
7 |
+
Case #7: 126.382789019
|
8 |
+
Case #8: 1.586234811
|
9 |
+
Case #9: 540.964700227
|
10 |
+
Case #10: 127.947392195
|
11 |
+
Case #11: 1461095.416128784
|
12 |
+
Case #12: 1432754.504519450
|
13 |
+
Case #13: 1628184.422929403
|
14 |
+
Case #14: 177448.735071926
|
15 |
+
Case #15: 2504.658183565
|
16 |
+
Case #16: 6199.844885954
|
17 |
+
Case #17: 1661.458735424
|
18 |
+
Case #18: 1342.586511189
|
19 |
+
Case #19: 2426.361263193
|
20 |
+
Case #20: 3884.965136178
|
21 |
+
Case #21: 1772.164136795
|
22 |
+
Case #22: 992.500138918
|
23 |
+
Case #23: 5276.187756659
|
24 |
+
Case #24: 142985.603513587
|
25 |
+
Case #25: 1024.929137712
|
26 |
+
Case #26: 1114.779671090
|
27 |
+
Case #27: 4987.183650761
|
28 |
+
Case #28: 34755.461192247
|
29 |
+
Case #29: 2071.210292890
|
30 |
+
Case #30: 1734.160405930
|
2017/finals/strolls.cpp
ADDED
@@ -0,0 +1,170 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
// Hacker Cup 2017
|
2 |
+
// Final Round
|
3 |
+
// Fox Strolls
|
4 |
+
// Jacob Plachta
|
5 |
+
|
6 |
+
#include <algorithm>
|
7 |
+
#include <functional>
|
8 |
+
#include <numeric>
|
9 |
+
#include <iostream>
|
10 |
+
#include <iomanip>
|
11 |
+
#include <cstdio>
|
12 |
+
#include <cmath>
|
13 |
+
#include <complex>
|
14 |
+
#include <cstdlib>
|
15 |
+
#include <ctime>
|
16 |
+
#include <cstring>
|
17 |
+
#include <cassert>
|
18 |
+
#include <string>
|
19 |
+
#include <vector>
|
20 |
+
#include <list>
|
21 |
+
#include <map>
|
22 |
+
#include <set>
|
23 |
+
#include <deque>
|
24 |
+
#include <queue>
|
25 |
+
#include <stack>
|
26 |
+
#include <bitset>
|
27 |
+
#include <sstream>
|
28 |
+
using namespace std;
|
29 |
+
|
30 |
+
#define LL long long
|
31 |
+
#define LD long double
|
32 |
+
#define PR pair<int,int>
|
33 |
+
|
34 |
+
#define Fox(i,n) for (i=0; i<n; i++)
|
35 |
+
#define Fox1(i,n) for (i=1; i<=n; i++)
|
36 |
+
#define FoxI(i,a,b) for (i=a; i<=b; i++)
|
37 |
+
#define FoxR(i,n) for (i=(n)-1; i>=0; i--)
|
38 |
+
#define FoxR1(i,n) for (i=n; i>0; i--)
|
39 |
+
#define FoxRI(i,a,b) for (i=b; i>=a; i--)
|
40 |
+
#define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++)
|
41 |
+
#define Min(a,b) a=min(a,b)
|
42 |
+
#define Max(a,b) a=max(a,b)
|
43 |
+
#define Sz(s) int((s).size())
|
44 |
+
#define All(s) (s).begin(),(s).end()
|
45 |
+
#define Fill(s,v) memset(s,v,sizeof(s))
|
46 |
+
#define pb push_back
|
47 |
+
#define mp make_pair
|
48 |
+
#define x first
|
49 |
+
#define y second
|
50 |
+
|
51 |
+
template<typename T> T Abs(T x) { return(x<0 ? -x : x); }
|
52 |
+
template<typename T> T Sqr(T x) { return(x*x); }
|
53 |
+
|
54 |
+
const int INF = (int)1e9;
|
55 |
+
const LD EPS = 1e-12;
|
56 |
+
const LD PI = acos(-1.0);
|
57 |
+
|
58 |
+
bool Read(int &x)
|
59 |
+
{
|
60 |
+
char c,r=0,n=0;
|
61 |
+
x=0;
|
62 |
+
for(;;)
|
63 |
+
{
|
64 |
+
c=GETCHAR();
|
65 |
+
if ((c<0) && (!r))
|
66 |
+
return(0);
|
67 |
+
if ((c=='-') && (!r))
|
68 |
+
n=1;
|
69 |
+
else
|
70 |
+
if ((c>='0') && (c<='9'))
|
71 |
+
x=x*10+c-'0',r=1;
|
72 |
+
else
|
73 |
+
if (r)
|
74 |
+
break;
|
75 |
+
}
|
76 |
+
if (n)
|
77 |
+
x=-x;
|
78 |
+
return(1);
|
79 |
+
}
|
80 |
+
|
81 |
+
#define LIM 500000
|
82 |
+
|
83 |
+
int main()
|
84 |
+
{
|
85 |
+
// vars
|
86 |
+
int T,t;
|
87 |
+
int N;
|
88 |
+
int i,j,a,b,h;
|
89 |
+
LL ans1,ans2;
|
90 |
+
static int H[LIM];
|
91 |
+
static PR P[LIM];
|
92 |
+
stack<PR> ST;
|
93 |
+
set<int> S;
|
94 |
+
set<int>::iterator I;
|
95 |
+
// testcase loop
|
96 |
+
Read(T);
|
97 |
+
Fox1(t,T)
|
98 |
+
{
|
99 |
+
// input
|
100 |
+
Read(N);
|
101 |
+
Fox(i,N)
|
102 |
+
{
|
103 |
+
Read(H[i]);
|
104 |
+
P[i]=mp(H[i],i);
|
105 |
+
}
|
106 |
+
// PART 1: sum of distances
|
107 |
+
ans1=0;
|
108 |
+
// get base sum assuming no bridges
|
109 |
+
Fox(i,N)
|
110 |
+
ans1+=H[i];
|
111 |
+
ans1*=N-1;
|
112 |
+
Fox1(i,N-1)
|
113 |
+
ans1+=(LL)i*(N-i);
|
114 |
+
// process trees bottom-to-top
|
115 |
+
sort(P,P+N);
|
116 |
+
S.clear();
|
117 |
+
S.insert(-1);
|
118 |
+
S.insert(N);
|
119 |
+
Fox(i,N)
|
120 |
+
{
|
121 |
+
h=P[i].x;
|
122 |
+
j=P[i].y;
|
123 |
+
// find interval of paths having this as their shortest tree
|
124 |
+
I=S.lower_bound(j);
|
125 |
+
b=*I - 1;
|
126 |
+
I--;
|
127 |
+
a=*I + 1;
|
128 |
+
// subtract total distance saved by bridges
|
129 |
+
ans1-=(LL)h*2*((LL)(j-a)*(b-j) + (b-a));
|
130 |
+
// insert this tree
|
131 |
+
S.insert(j);
|
132 |
+
}
|
133 |
+
// PART 2: number of bridges
|
134 |
+
ans2=0;
|
135 |
+
// process trees left-to-right
|
136 |
+
while (!ST.empty())
|
137 |
+
ST.pop();
|
138 |
+
ST.push(mp(0,0));
|
139 |
+
Fox1(i,N-1)
|
140 |
+
if (H[i]>=H[i-1])
|
141 |
+
{
|
142 |
+
// at least as high as envelope's highest point, so build single bridge backwards
|
143 |
+
if (ST.top().x==H[i-1])
|
144 |
+
ST.pop();
|
145 |
+
ST.push(mp(H[i-1],i));
|
146 |
+
ans2++;
|
147 |
+
}
|
148 |
+
else
|
149 |
+
{
|
150 |
+
// clip off higher parts of envelope
|
151 |
+
j=i-1;
|
152 |
+
while (ST.top().x>H[i])
|
153 |
+
j=ST.top().y,ST.pop();
|
154 |
+
// build row of bridges back across, and update envelope
|
155 |
+
if (!ST.top().x)
|
156 |
+
ans2+=i-j;
|
157 |
+
else
|
158 |
+
{
|
159 |
+
ans2+=i-ST.top().y;
|
160 |
+
h=ST.top().x,ST.pop();
|
161 |
+
if (H[i]!=h)
|
162 |
+
ST.push(mp(h,j));
|
163 |
+
}
|
164 |
+
ST.push(mp(H[i],i));
|
165 |
+
}
|
166 |
+
// output
|
167 |
+
printf("Case #%d: %lld %lld\n",t,ans1,ans2);
|
168 |
+
}
|
169 |
+
return(0);
|
170 |
+
}
|
2017/finals/strolls.html
ADDED
@@ -0,0 +1,73 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
<p>
|
2 |
+
A certain forest has <strong>N</strong> trees growing in it, and there just so happens to be a Fox living at the top of each one!
|
3 |
+
The trees are numbered from 1 to <strong>N</strong>, and their bases are all arranged in a straight line on the ground,
|
4 |
+
with 1 metre between the bases of each pair of adjacent trees <em>i</em> and <em>i</em> + 1.
|
5 |
+
Each tree <em>i</em> is <strong>H<sub>i</sub></strong> metres tall.
|
6 |
+
</p>
|
7 |
+
|
8 |
+
</p>
|
9 |
+
The Foxen are all good friends with one another, and frequently like to go out for strolls to visit each other's homes.
|
10 |
+
Rather than jumping directly between the trees, they prefer to always keep their paws firmly planted on some surface, such as tree trunks or the ground.
|
11 |
+
As such, the shortest possible trip from the top of tree <em>i</em> to the top of another tree <em>j</em> requires
|
12 |
+
climbing down tree <em>i</em> to the ground, walking along the ground to the base of tree <em>j</em>,
|
13 |
+
and then climbing up to its top, resulting in a total distance of
|
14 |
+
<strong>H<sub>i</sub></strong> + |<em>j</em> - <em>i</em>| + <strong>H<sub>j</sub></strong> metres traveled.
|
15 |
+
</p>
|
16 |
+
|
17 |
+
<p>
|
18 |
+
However, the Foxen aren't terribly satisfied with how long their trips currently take.
|
19 |
+
Given the frequency of their strolls, they've decided to invest in reducing their travel distance
|
20 |
+
by constructing some bridges between the tree trunks.
|
21 |
+
They're only interested in building metre-long, horizontal bridges.
|
22 |
+
Each bridge may be constructed to connect any pair of adjacent trees <em>i</em> and <em>i</em> + 1,
|
23 |
+
and it may be placed at any height above the ground, as long as it touches both of those tree trunks
|
24 |
+
(in other words, its height must be no larger than the minimum of <strong>H<sub>i</sub></strong>
|
25 |
+
and <strong>H<sub>i+1</sub></strong>).
|
26 |
+
Multiple bridges may be constructed at different heights between any given pair of adjacent trees.
|
27 |
+
Once bridges are installed, the Foxen will be willing to walk across them, potentially saving them the time
|
28 |
+
of descending all the way to the ground during their strolls.
|
29 |
+
</p>
|
30 |
+
|
31 |
+
<p>
|
32 |
+
There are <strong>N</strong> * <strong>(N - 1)</strong> / 2 different pairs of Foxen <em>i</em> and <em>j</em>
|
33 |
+
who might want to meet up, requiring a trip to be taken from the top of tree <em>i</em> to the top of tree
|
34 |
+
<em>j</em> (or vice versa).
|
35 |
+
The Foxen's top priority is minimizing the sum of the <strong>N</strong> * <strong>(N - 1)</strong> / 2
|
36 |
+
pairwise shortest distances of strolls which would be required for these visits to take place.
|
37 |
+
Please help them determine the minimum possible value of this sum, assuming that they construct
|
38 |
+
as many bridges as it takes.
|
39 |
+
That being said, they don't want to spend all day constructing bridges either...
|
40 |
+
as such, they're also interested in the minimum number of bridges which they can construct to achieve that minimum
|
41 |
+
possible sum of pairwise distances.
|
42 |
+
</p>
|
43 |
+
|
44 |
+
<h3>Input</h3>
|
45 |
+
|
46 |
+
<p>
|
47 |
+
Input begins with an integer <strong>T</strong>, the number of forests.
|
48 |
+
For each forest, there are two lines. The first line contains the integer <strong>N</strong>.
|
49 |
+
The second line contains <strong>N</strong> space-separated integers,
|
50 |
+
the <em>i</em>th of which is the integer <strong>H<sub>i</sub></strong>.
|
51 |
+
</p>
|
52 |
+
|
53 |
+
<h3>Output</h3>
|
54 |
+
|
55 |
+
<p>
|
56 |
+
For the <em>i</em>th forest, print a line containing "Case #<strong>i</strong>: "
|
57 |
+
followed by two integers: the minimum possible sum of pairwise shortest distances after any number of bridges are built (in metres), and the minimum number of bridges required to achieve that minimum sum.
|
58 |
+
</p>
|
59 |
+
|
60 |
+
<h3>Constraints</h3>
|
61 |
+
|
62 |
+
<p>
|
63 |
+
1 ≤ <strong>T</strong> ≤ 30 <br />
|
64 |
+
2 ≤ <strong>N</strong> ≤ 500,000 <br />
|
65 |
+
1 ≤ <strong>H<sub>i</sub></strong> ≤ 10,000,000 <br />
|
66 |
+
The sum of <strong>N</strong> values across all <strong>T</strong> cases does not exceed 6,000,000.
|
67 |
+
</p>
|
68 |
+
|
69 |
+
<h3>Explanation of Sample</h3>
|
70 |
+
|
71 |
+
<p>
|
72 |
+
In the first case, the Foxen's optimal strategy is to construct one bridge between the first 2 trees at a height of 20m, and another bridge between the last 2 trees at a height of 10m. With this configuration, the shortest path between the first 2 trees' tops is 11m long, as is the shortest path for the last 2 trees. The shortest path between the first tree's top and the last tree's top is 22m long. The resulting sum of pairwise shortest distances is then 44m, which is the minimum possible sum.
|
73 |
+
</p>
|
2017/finals/strolls.in
ADDED
@@ -0,0 +1,3 @@
|
|
|
|
|
|
|
|
|
1 |
+
version https://git-lfs.github.com/spec/v1
|
2 |
+
oid sha256:9fe710744cbf537741bbb7ebe8a9888e7917244eb30214ea02d0b11e65453b84
|
3 |
+
size 36557012
|
2017/finals/strolls.md
ADDED
@@ -0,0 +1,70 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
A certain forest has **N** trees growing in it, and there just so happens to
|
2 |
+
be a Fox living at the top of each one! The trees are numbered from 1 to
|
3 |
+
**N**, and their bases are all arranged in a straight line on the ground, with
|
4 |
+
1 metre between the bases of each pair of adjacent trees _i_ and _i_ \+ 1.
|
5 |
+
Each tree _i_ is **Hi** metres tall.
|
6 |
+
|
7 |
+
The Foxen are all good friends with one another, and frequently like to go out
|
8 |
+
for strolls to visit each other's homes. Rather than jumping directly between
|
9 |
+
the trees, they prefer to always keep their paws firmly planted on some
|
10 |
+
surface, such as tree trunks or the ground. As such, the shortest possible
|
11 |
+
trip from the top of tree _i_ to the top of another tree _j_ requires climbing
|
12 |
+
down tree _i_ to the ground, walking along the ground to the base of tree _j_,
|
13 |
+
and then climbing up to its top, resulting in a total distance of **Hi** \+
|
14 |
+
|_j_ \- _i_| + **Hj** metres traveled.
|
15 |
+
|
16 |
+
However, the Foxen aren't terribly satisfied with how long their trips
|
17 |
+
currently take. Given the frequency of their strolls, they've decided to
|
18 |
+
invest in reducing their travel distance by constructing some bridges between
|
19 |
+
the tree trunks. They're only interested in building metre-long, horizontal
|
20 |
+
bridges. Each bridge may be constructed to connect any pair of adjacent trees
|
21 |
+
_i_ and _i_ \+ 1, and it may be placed at any height above the ground, as long
|
22 |
+
as it touches both of those tree trunks (in other words, its height must be no
|
23 |
+
larger than the minimum of **Hi** and **Hi+1**). Multiple bridges may be
|
24 |
+
constructed at different heights between any given pair of adjacent trees.
|
25 |
+
Once bridges are installed, the Foxen will be willing to walk across them,
|
26 |
+
potentially saving them the time of descending all the way to the ground
|
27 |
+
during their strolls.
|
28 |
+
|
29 |
+
There are **N** * **(N - 1)** / 2 different pairs of Foxen _i_ and _j_ who
|
30 |
+
might want to meet up, requiring a trip to be taken from the top of tree _i_
|
31 |
+
to the top of tree _j_ (or vice versa). The Foxen's top priority is minimizing
|
32 |
+
the sum of the **N** * **(N - 1)** / 2 pairwise shortest distances of strolls
|
33 |
+
which would be required for these visits to take place. Please help them
|
34 |
+
determine the minimum possible value of this sum, assuming that they construct
|
35 |
+
as many bridges as it takes. That being said, they don't want to spend all day
|
36 |
+
constructing bridges either... as such, they're also interested in the minimum
|
37 |
+
number of bridges which they can construct to achieve that minimum possible
|
38 |
+
sum of pairwise distances.
|
39 |
+
|
40 |
+
### Input
|
41 |
+
|
42 |
+
Input begins with an integer **T**, the number of forests. For each forest,
|
43 |
+
there are two lines. The first line contains the integer **N**. The second
|
44 |
+
line contains **N** space-separated integers, the _i_th of which is the
|
45 |
+
integer **Hi**.
|
46 |
+
|
47 |
+
### Output
|
48 |
+
|
49 |
+
For the _i_th forest, print a line containing "Case #**i**: " followed by two
|
50 |
+
integers: the minimum possible sum of pairwise shortest distances after any
|
51 |
+
number of bridges are built (in metres), and the minimum number of bridges
|
52 |
+
required to achieve that minimum sum.
|
53 |
+
|
54 |
+
### Constraints
|
55 |
+
|
56 |
+
1 ≤ **T** ≤ 30
|
57 |
+
2 ≤ **N** ≤ 500,000
|
58 |
+
1 ≤ **Hi** ≤ 10,000,000
|
59 |
+
The sum of **N** values across all **T** cases does not exceed 6,000,000.
|
60 |
+
|
61 |
+
### Explanation of Sample
|
62 |
+
|
63 |
+
In the first case, the Foxen's optimal strategy is to construct one bridge
|
64 |
+
between the first 2 trees at a height of 20m, and another bridge between the
|
65 |
+
last 2 trees at a height of 10m. With this configuration, the shortest path
|
66 |
+
between the first 2 trees' tops is 11m long, as is the shortest path for the
|
67 |
+
last 2 trees. The shortest path between the first tree's top and the last
|
68 |
+
tree's top is 22m long. The resulting sum of pairwise shortest distances is
|
69 |
+
then 44m, which is the minimum possible sum.
|
70 |
+
|
2017/finals/strolls.out
ADDED
@@ -0,0 +1,30 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
Case #1: 44 2
|
2 |
+
Case #2: 30 6
|
3 |
+
Case #3: 158 9
|
4 |
+
Case #4: 213877494 8
|
5 |
+
Case #5: 67828 26
|
6 |
+
Case #6: 40 4
|
7 |
+
Case #7: 62 5
|
8 |
+
Case #8: 64 4
|
9 |
+
Case #9: 66 5
|
10 |
+
Case #10: 1 1
|
11 |
+
Case #11: 1270694467527793009 3991523
|
12 |
+
Case #12: 1270089310668572086 4233586
|
13 |
+
Case #13: 1271258716744254028 4187288
|
14 |
+
Case #14: 1076962260755154 368610
|
15 |
+
Case #15: 154085676243923872 2143830
|
16 |
+
Case #16: 34868734969578 192125
|
17 |
+
Case #17: 42013086963397592 1016400
|
18 |
+
Case #18: 14227652173282473 970767
|
19 |
+
Case #19: 63891446396267944 1209911
|
20 |
+
Case #20: 54609935358656 317923
|
21 |
+
Case #21: 46568399312743270 3842959
|
22 |
+
Case #22: 42359072334166 267152801
|
23 |
+
Case #23: 23050539441884206 17994431223
|
24 |
+
Case #24: 20169516401652989 16526262785
|
25 |
+
Case #25: 9348834263019686 9893968817
|
26 |
+
Case #26: 32509644062872258 22648141115
|
27 |
+
Case #27: 54275273806021 317817471
|
28 |
+
Case #28: 3562914776995796 5205231509
|
29 |
+
Case #29: 46421890364145250 28680626068
|
30 |
+
Case #30: 2608880906805372 4195541434
|
2017/finals/tolls.cpp
ADDED
@@ -0,0 +1,210 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
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|
|
|
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|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
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|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
// Hacker Cup 2017
|
2 |
+
// Final Round
|
3 |
+
// Fox Tolls
|
4 |
+
// Jacob Plachta
|
5 |
+
|
6 |
+
#include <algorithm>
|
7 |
+
#include <functional>
|
8 |
+
#include <numeric>
|
9 |
+
#include <iostream>
|
10 |
+
#include <iomanip>
|
11 |
+
#include <cstdio>
|
12 |
+
#include <cmath>
|
13 |
+
#include <complex>
|
14 |
+
#include <cstdlib>
|
15 |
+
#include <ctime>
|
16 |
+
#include <cstring>
|
17 |
+
#include <cassert>
|
18 |
+
#include <string>
|
19 |
+
#include <vector>
|
20 |
+
#include <list>
|
21 |
+
#include <map>
|
22 |
+
#include <set>
|
23 |
+
#include <deque>
|
24 |
+
#include <queue>
|
25 |
+
#include <stack>
|
26 |
+
#include <bitset>
|
27 |
+
#include <sstream>
|
28 |
+
using namespace std;
|
29 |
+
|
30 |
+
#define LL long long
|
31 |
+
#define LD long double
|
32 |
+
#define PR pair<int,int>
|
33 |
+
|
34 |
+
#define Fox(i,n) for (i=0; i<n; i++)
|
35 |
+
#define Fox1(i,n) for (i=1; i<=n; i++)
|
36 |
+
#define FoxI(i,a,b) for (i=a; i<=b; i++)
|
37 |
+
#define FoxR(i,n) for (i=(n)-1; i>=0; i--)
|
38 |
+
#define FoxR1(i,n) for (i=n; i>0; i--)
|
39 |
+
#define FoxRI(i,a,b) for (i=b; i>=a; i--)
|
40 |
+
#define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++)
|
41 |
+
#define Min(a,b) a=min(a,b)
|
42 |
+
#define Max(a,b) a=max(a,b)
|
43 |
+
#define Sz(s) int((s).size())
|
44 |
+
#define All(s) (s).begin(),(s).end()
|
45 |
+
#define Fill(s,v) memset(s,v,sizeof(s))
|
46 |
+
#define pb push_back
|
47 |
+
#define mp make_pair
|
48 |
+
#define x first
|
49 |
+
#define y second
|
50 |
+
|
51 |
+
template<typename T> T Abs(T x) { return(x<0 ? -x : x); }
|
52 |
+
template<typename T> T Sqr(T x) { return(x*x); }
|
53 |
+
|
54 |
+
const int INF = (int)1e9;
|
55 |
+
const LD EPS = 1e-12;
|
56 |
+
const LD PI = acos(-1.0);
|
57 |
+
|
58 |
+
bool Read(int &x)
|
59 |
+
{
|
60 |
+
char c,r=0,n=0;
|
61 |
+
x=0;
|
62 |
+
for(;;)
|
63 |
+
{
|
64 |
+
c=getchar();
|
65 |
+
if ((c<0) && (!r))
|
66 |
+
return(0);
|
67 |
+
if ((c=='-') && (!r))
|
68 |
+
n=1;
|
69 |
+
else
|
70 |
+
if ((c>='0') && (c<='9'))
|
71 |
+
x=x*10+c-'0',r=1;
|
72 |
+
else
|
73 |
+
if (r)
|
74 |
+
break;
|
75 |
+
}
|
76 |
+
if (n)
|
77 |
+
x=-x;
|
78 |
+
return(1);
|
79 |
+
}
|
80 |
+
|
81 |
+
#define LIM 500000
|
82 |
+
#define LIM2 2100000
|
83 |
+
#define MOD 1000000007
|
84 |
+
|
85 |
+
int ind;
|
86 |
+
int L[2][LIM],R[2][LIM],S[2][LIM];
|
87 |
+
vector<int> con[LIM];
|
88 |
+
|
89 |
+
int sz;
|
90 |
+
vector<int> lst[LIM2];
|
91 |
+
|
92 |
+
void rec(int z,int i,int p)
|
93 |
+
{
|
94 |
+
int j,k;
|
95 |
+
L[z][i]=ind++;
|
96 |
+
Fox(j,Sz(con[i]))
|
97 |
+
if ((k=con[i][j])!=p)
|
98 |
+
rec(z,k,i);
|
99 |
+
R[z][i]=ind-1;
|
100 |
+
S[z][i]=R[z][i]-L[z][i];
|
101 |
+
}
|
102 |
+
|
103 |
+
int Query(int i,int r1,int r2,int a,int b,int v1,int v2)
|
104 |
+
{
|
105 |
+
if ((a<=r1) && (r2<=b))
|
106 |
+
{
|
107 |
+
return(
|
108 |
+
lower_bound(lst[i].begin(),lst[i].end(),v2+1) -
|
109 |
+
lower_bound(lst[i].begin(),lst[i].end(),v1)
|
110 |
+
);
|
111 |
+
}
|
112 |
+
i<<=1;
|
113 |
+
int m=(r1+r2)>>1,s=0;
|
114 |
+
if (a<=m)
|
115 |
+
s+=Query(i,r1,m,a,b,v1,v2);
|
116 |
+
if (b>m)
|
117 |
+
s+=Query(i+1,m+1,r2,a,b,v1,v2);
|
118 |
+
return(s);
|
119 |
+
}
|
120 |
+
|
121 |
+
int main()
|
122 |
+
{
|
123 |
+
// vars
|
124 |
+
int T,t;
|
125 |
+
int N,M;
|
126 |
+
int i,j,k,a,b,z,z1,z2;
|
127 |
+
int ans,tot;
|
128 |
+
char c1,c2;
|
129 |
+
// testcase loop
|
130 |
+
Read(T);
|
131 |
+
Fox1(t,T)
|
132 |
+
{
|
133 |
+
// input
|
134 |
+
Read(N),Read(M);
|
135 |
+
Fox(z,2)
|
136 |
+
{
|
137 |
+
// clear graph
|
138 |
+
Fox(i,N)
|
139 |
+
con[i].clear();
|
140 |
+
// input tree
|
141 |
+
Fox(i,N-1)
|
142 |
+
{
|
143 |
+
Read(j),Read(k),j--,k--;
|
144 |
+
con[j].pb(k);
|
145 |
+
con[k].pb(j);
|
146 |
+
}
|
147 |
+
// calculate DFS ordering for this tree
|
148 |
+
ind=0;
|
149 |
+
rec(z,0,0);
|
150 |
+
}
|
151 |
+
// construct segment tree
|
152 |
+
for(sz=1;sz<N;sz<<=1);
|
153 |
+
Fox(i,sz<<1)
|
154 |
+
lst[i].clear();
|
155 |
+
Fox(i,N)
|
156 |
+
lst[sz+L[0][i]].pb(L[1][i]);
|
157 |
+
FoxR1(i,sz-1)
|
158 |
+
{
|
159 |
+
j=i<<1,k=j+1;
|
160 |
+
a=b=0;
|
161 |
+
while ((a<Sz(lst[j])) && (b<Sz(lst[k])))
|
162 |
+
if (lst[j][a]<lst[k][b])
|
163 |
+
lst[i].pb(lst[j][a++]);
|
164 |
+
else
|
165 |
+
lst[i].pb(lst[k][b++]);
|
166 |
+
while (a<Sz(lst[j]))
|
167 |
+
lst[i].pb(lst[j][a++]);
|
168 |
+
while (b<Sz(lst[k]))
|
169 |
+
lst[i].pb(lst[k][b++]);
|
170 |
+
}
|
171 |
+
// handle days
|
172 |
+
tot=0;
|
173 |
+
while (M--)
|
174 |
+
{
|
175 |
+
// input
|
176 |
+
Read(i),Read(j),i--,j--;
|
177 |
+
scanf("%c %c",&c1,&c2);
|
178 |
+
z1=c1=='T' ? 0 : 1;
|
179 |
+
z2=c2=='T' ? 0 : 1;
|
180 |
+
// same tree?
|
181 |
+
if (z1==z2)
|
182 |
+
{
|
183 |
+
z=z1;
|
184 |
+
if (L[z][i]>L[z][j])
|
185 |
+
swap(i,j);
|
186 |
+
// nested subtrees?
|
187 |
+
if (L[z][j]<=R[z][i])
|
188 |
+
{
|
189 |
+
ans=2*S[z][i] + S[z][j];
|
190 |
+
goto Done;
|
191 |
+
}
|
192 |
+
// disjoint subtrees
|
193 |
+
ans=2*(S[z][i] + S[z][j]);
|
194 |
+
goto Done;
|
195 |
+
}
|
196 |
+
// different trees
|
197 |
+
if (z1)
|
198 |
+
swap(i,j);
|
199 |
+
ans=2*(S[0][i] + S[1][j]);
|
200 |
+
if ((S[0][i]) && (S[1][j]))
|
201 |
+
ans-=Query(1,0,sz-1,L[0][i]+1,R[0][i],L[1][j]+1,R[1][j]);
|
202 |
+
// update total
|
203 |
+
Done:;
|
204 |
+
tot=((LL)tot*12345+ans)%MOD;
|
205 |
+
}
|
206 |
+
//output
|
207 |
+
printf("Case #%d: %d\n",t,tot);
|
208 |
+
}
|
209 |
+
return(0);
|
210 |
+
}
|
2017/finals/tolls.html
ADDED
@@ -0,0 +1,100 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
<p>
|
2 |
+
A group of <strong>N</strong> Foxen reside in a peaceful forest community.
|
3 |
+
Each Fox's property consists of a tree stump as well as an underground den.
|
4 |
+
There are <strong>N</strong> - 1 two-way paths on the ground running amongst the tree stumps,
|
5 |
+
with the <em>i</em>th path connecting the stumps belonging to two different Foxen
|
6 |
+
<strong>A<sub>i</sub></strong> and <strong>B<sub>i</sub></strong>,
|
7 |
+
such that all <strong>N</strong> stumps can be reached from one another by following a sequence of paths.
|
8 |
+
Similarly, there are <strong>N</strong> - 1 underground tunnels running amongst the dens,
|
9 |
+
with the <em>i</em>th tunnel connecting the dens belonging to Foxen
|
10 |
+
<strong>C<sub>i</sub></strong> and <strong>D<sub>i</sub></strong>,
|
11 |
+
such that all <strong>N</strong> dens can be reached from one another.
|
12 |
+
There's additionally a passageway connecting the tree stump and den belonging to the 1st Fox,
|
13 |
+
which is the only way in the whole forest to get underground from the surface and vice versa.
|
14 |
+
</p>
|
15 |
+
|
16 |
+
<p>
|
17 |
+
At night the Foxen sleep in their dens, but during the daytime, they like to emerge and relax lazily on their tree stumps.
|
18 |
+
Each day, every Fox takes a trip from their den to their tree stump,
|
19 |
+
taking the unique shortest path through the system of tunnels and paths to get there.
|
20 |
+
However, this often requires passing through other Foxen's properties, which they don't appreciate a whole lot.
|
21 |
+
To compensate, the Foxen have started charging each other tolls for said passage.
|
22 |
+
They don't have much of a currency, but Foxen do love crackers, so those will do.
|
23 |
+
Over a given period of <strong>M</strong> days, on the <em>i</em>th day, two different Foxen
|
24 |
+
<strong>W<sub>i</sub></strong> and <strong>X<sub>i</sub></strong> will each charge tolls for one of their pieces of property.
|
25 |
+
If <strong>Y<sub>i</sub></strong> = "T",
|
26 |
+
then Fox <strong>W<sub>i</sub></strong> will be charging tolls for passage through their tree stump.
|
27 |
+
Otherwise, if <strong>Y<sub>i</sub></strong> = "D",
|
28 |
+
then Fox <strong>W<sub>i</sub></strong> will instead be charging tolls for passage through their den.
|
29 |
+
Similarly, Fox <strong>X<sub>i</sub></strong> will be charging tolls for passage through either their tree stump
|
30 |
+
(if <strong>Z<sub>i</sub></strong>= "T") or their den (if <strong>Z<sub>i</sub></strong>= "D").
|
31 |
+
</p>
|
32 |
+
|
33 |
+
<p>
|
34 |
+
Each day, whenever a Fox passes through another Fox's den or stump which is subject to tolls on that day,
|
35 |
+
they'll normally need to pay up with 2 crackers. However, if they've already paid a toll earlier on that same trip,
|
36 |
+
then the property-owning Fox will take pity and only charge them 1 cracker instead of 2.
|
37 |
+
As such, a Fox's daily trip may end up costing them at most 3 crackers. A Fox will never charge themselves a toll, of course. If a pair of Foxen both owe each other crackers, they'll still both pay up as normal, rather than attempting to minimize the number of cracker transactions performed.
|
38 |
+
</p>
|
39 |
+
|
40 |
+
<p>
|
41 |
+
The Foxen are having some trouble keeping track of how many crackers they owe one another. On each of the <strong>M</strong> days, they'd like to count up the total number of crackers which will be charged as part of the tolls for the <strong>N</strong> trips taken on that day. To avoid dealing with too many large numbers, they'd like to combine these <strong>M</strong> cracker counts into a single value as follows (where <strong>V<sub>i</sub></strong> is the <em>i</em>th day's count):
|
42 |
+
</p>
|
43 |
+
|
44 |
+
<p>
|
45 |
+
|
46 |
+
( ... (((<strong>V<sub>1</sub></strong> * 12,345) + <strong>V<sub>2</sub></strong>) * 12,345 + <strong>V<sub>3</sub></strong>) ... * 12,345 + <strong>V<sub>M</sub></strong>) modulo 1,000,000,007
|
47 |
+
</p>
|
48 |
+
|
49 |
+
<p>
|
50 |
+
Please help the Foxen compute this combined value!
|
51 |
+
</p>
|
52 |
+
|
53 |
+
<h3>Input</h3>
|
54 |
+
|
55 |
+
<p>
|
56 |
+
Input begins with an integer <strong>T</strong>, the number of different communities of Foxen.
|
57 |
+
For each community of Foxen, there is first a line containing the space-separated integers <strong>N</strong> and <strong>M</strong>.
|
58 |
+
Then <strong>N - 1</strong> lines follow, the <em>i</em>th of which contains the space-separated integers
|
59 |
+
<strong>A<sub>i</sub></strong> and <strong>B<sub>i</sub></strong>.
|
60 |
+
Then <strong>N - 1</strong> lines follow, the <em>i</em>th of which contains the space-separated integers
|
61 |
+
<strong>C<sub>i</sub></strong> and <strong>D<sub>i</sub></strong>.
|
62 |
+
Then <strong>M</strong> lines follow, the <em>i</em>th of which contains the integers
|
63 |
+
<strong>W<sub>i</sub></strong> and <strong>X<sub>i</sub></strong> and the characters
|
64 |
+
<strong>Y<sub>i</sub></strong> and <strong>Z<sub>i</sub></strong>, all separated by spaces.
|
65 |
+
</p>
|
66 |
+
|
67 |
+
<h3>Output</h3>
|
68 |
+
|
69 |
+
<p>
|
70 |
+
For the <em>i</em>th community of Foxen, print a line containing "Case #<strong>i</strong>: "
|
71 |
+
followed by a single integer, the requested combined value based on the <strong>M</strong> days' cracker counts, modulo 1,000,000,007.
|
72 |
+
</p>
|
73 |
+
|
74 |
+
<h3>Constraints</h3>
|
75 |
+
|
76 |
+
<p>
|
77 |
+
1 ≤ <strong>T</strong> ≤ 30<br />
|
78 |
+
2 ≤ <strong>N</strong> ≤ 500,000 <br />
|
79 |
+
1 ≤ <strong>M</strong> ≤ 500,000 <br />
|
80 |
+
1 ≤
|
81 |
+
<strong>A<sub>i</sub></strong>, <strong>B<sub>i</sub></strong>,
|
82 |
+
<strong>C<sub>i</sub></strong>, <strong>D<sub>i</sub></strong>,
|
83 |
+
<strong>W<sub>i</sub></strong>, <strong>X<sub>i</sub></strong>
|
84 |
+
≤ <strong>N</strong> <br />
|
85 |
+
Both the sum of <strong>N</strong> values and the sum of <strong>M</strong> values across all <strong>T</strong> cases do not exceed 1,500,000.
|
86 |
+
</p>
|
87 |
+
|
88 |
+
<h3>Explanation of Sample</h3>
|
89 |
+
|
90 |
+
<p>
|
91 |
+
In the first case, Fox 1 doesn't need to pay any tolls to get from its den to its tree stump, while Fox 2 must pay 2 crackers to complete its trip due to passing through Fox's 1 tree stump.
|
92 |
+
</p>
|
93 |
+
|
94 |
+
<p>
|
95 |
+
In the second case, 5 crackers will be charged on the first day (the 3 Foxen must pay 0, 2, and 3 crackers, respectively), 2 crackers will be charged on the second day, and none will be charged on the third day. This results in a final answer of (((5 * 12,345) + 2) * 12,345) + 0) modulo 1,000,000,007 = 762,019,815.
|
96 |
+
</p>
|
97 |
+
|
98 |
+
<p>
|
99 |
+
In the third case, 7, 6, and 4 crackers will be charged on each of the three days, respectively.
|
100 |
+
</p>
|
2017/finals/tolls.in
ADDED
@@ -0,0 +1,3 @@
|
|
|
|
|
|
|
|
|
1 |
+
version https://git-lfs.github.com/spec/v1
|
2 |
+
oid sha256:1df58783e9feba1244b6e94657061d40c81c94a6d4c1df75f194e0f854256cb0
|
3 |
+
size 47154883
|
2017/finals/tolls.md
ADDED
@@ -0,0 +1,89 @@
|
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|
|
|
|
1 |
+
A group of **N** Foxen reside in a peaceful forest community. Each Fox's
|
2 |
+
property consists of a tree stump as well as an underground den. There are
|
3 |
+
**N** \- 1 two-way paths on the ground running amongst the tree stumps, with
|
4 |
+
the _i_th path connecting the stumps belonging to two different Foxen **Ai**
|
5 |
+
and **Bi**, such that all **N** stumps can be reached from one another by
|
6 |
+
following a sequence of paths. Similarly, there are **N** \- 1 underground
|
7 |
+
tunnels running amongst the dens, with the _i_th tunnel connecting the dens
|
8 |
+
belonging to Foxen **Ci** and **Di**, such that all **N** dens can be reached
|
9 |
+
from one another. There's additionally a passageway connecting the tree stump
|
10 |
+
and den belonging to the 1st Fox, which is the only way in the whole forest to
|
11 |
+
get underground from the surface and vice versa.
|
12 |
+
|
13 |
+
At night the Foxen sleep in their dens, but during the daytime, they like to
|
14 |
+
emerge and relax lazily on their tree stumps. Each day, every Fox takes a trip
|
15 |
+
from their den to their tree stump, taking the unique shortest path through
|
16 |
+
the system of tunnels and paths to get there. However, this often requires
|
17 |
+
passing through other Foxen's properties, which they don't appreciate a whole
|
18 |
+
lot. To compensate, the Foxen have started charging each other tolls for said
|
19 |
+
passage. They don't have much of a currency, but Foxen do love crackers, so
|
20 |
+
those will do. Over a given period of **M** days, on the _i_th day, two
|
21 |
+
different Foxen **Wi** and **Xi** will each charge tolls for one of their
|
22 |
+
pieces of property. If **Yi** = "T", then Fox **Wi** will be charging tolls
|
23 |
+
for passage through their tree stump. Otherwise, if **Yi** = "D", then Fox
|
24 |
+
**Wi** will instead be charging tolls for passage through their den.
|
25 |
+
Similarly, Fox **Xi** will be charging tolls for passage through either their
|
26 |
+
tree stump (if **Zi**= "T") or their den (if **Zi**= "D").
|
27 |
+
|
28 |
+
Each day, whenever a Fox passes through another Fox's den or stump which is
|
29 |
+
subject to tolls on that day, they'll normally need to pay up with 2 crackers.
|
30 |
+
However, if they've already paid a toll earlier on that same trip, then the
|
31 |
+
property-owning Fox will take pity and only charge them 1 cracker instead of
|
32 |
+
2. As such, a Fox's daily trip may end up costing them at most 3 crackers. A
|
33 |
+
Fox will never charge themselves a toll, of course. If a pair of Foxen both
|
34 |
+
owe each other crackers, they'll still both pay up as normal, rather than
|
35 |
+
attempting to minimize the number of cracker transactions performed.
|
36 |
+
|
37 |
+
The Foxen are having some trouble keeping track of how many crackers they owe
|
38 |
+
one another. On each of the **M** days, they'd like to count up the total
|
39 |
+
number of crackers which will be charged as part of the tolls for the **N**
|
40 |
+
trips taken on that day. To avoid dealing with too many large numbers, they'd
|
41 |
+
like to combine these **M** cracker counts into a single value as follows
|
42 |
+
(where **Vi** is the _i_th day's count):
|
43 |
+
|
44 |
+
( ... (((**V1** * 12,345) + **V2**) * 12,345 + **V3**) ... * 12,345 + **VM**)
|
45 |
+
modulo 1,000,000,007
|
46 |
+
|
47 |
+
Please help the Foxen compute this combined value!
|
48 |
+
|
49 |
+
### Input
|
50 |
+
|
51 |
+
Input begins with an integer **T**, the number of different communities of
|
52 |
+
Foxen. For each community of Foxen, there is first a line containing the
|
53 |
+
space-separated integers **N** and **M**. Then **N - 1** lines follow, the
|
54 |
+
_i_th of which contains the space-separated integers **Ai** and **Bi**. Then
|
55 |
+
**N - 1** lines follow, the _i_th of which contains the space-separated
|
56 |
+
integers **Ci** and **Di**. Then **M** lines follow, the _i_th of which
|
57 |
+
contains the integers **Wi** and **Xi** and the characters **Yi** and **Zi**,
|
58 |
+
all separated by spaces.
|
59 |
+
|
60 |
+
### Output
|
61 |
+
|
62 |
+
For the _i_th community of Foxen, print a line containing "Case #**i**: "
|
63 |
+
followed by a single integer, the requested combined value based on the **M**
|
64 |
+
days' cracker counts, modulo 1,000,000,007.
|
65 |
+
|
66 |
+
### Constraints
|
67 |
+
|
68 |
+
1 ≤ **T** ≤ 30
|
69 |
+
2 ≤ **N** ≤ 500,000
|
70 |
+
1 ≤ **M** ≤ 500,000
|
71 |
+
1 ≤ **Ai**, **Bi**, **Ci**, **Di**, **Wi**, **Xi** ≤ **N**
|
72 |
+
Both the sum of **N** values and the sum of **M** values across all **T**
|
73 |
+
cases do not exceed 1,500,000.
|
74 |
+
|
75 |
+
### Explanation of Sample
|
76 |
+
|
77 |
+
In the first case, Fox 1 doesn't need to pay any tolls to get from its den to
|
78 |
+
its tree stump, while Fox 2 must pay 2 crackers to complete its trip due to
|
79 |
+
passing through Fox's 1 tree stump.
|
80 |
+
|
81 |
+
In the second case, 5 crackers will be charged on the first day (the 3 Foxen
|
82 |
+
must pay 0, 2, and 3 crackers, respectively), 2 crackers will be charged on
|
83 |
+
the second day, and none will be charged on the third day. This results in a
|
84 |
+
final answer of (((5 * 12,345) + 2) * 12,345) + 0) modulo 1,000,000,007 =
|
85 |
+
762,019,815.
|
86 |
+
|
87 |
+
In the third case, 7, 6, and 4 crackers will be charged on each of the three
|
88 |
+
days, respectively.
|
89 |
+
|
2017/finals/tolls.out
ADDED
@@ -0,0 +1,30 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
Case #1: 2
|
2 |
+
Case #2: 762019815
|
3 |
+
Case #3: 66867242
|
4 |
+
Case #4: 537335547
|
5 |
+
Case #5: 874784450
|
6 |
+
Case #6: 2
|
7 |
+
Case #7: 24692
|
8 |
+
Case #8: 73447302
|
9 |
+
Case #9: 669976683
|
10 |
+
Case #10: 219446081
|
11 |
+
Case #11: 904947270
|
12 |
+
Case #12: 409408195
|
13 |
+
Case #13: 865385861
|
14 |
+
Case #14: 389014816
|
15 |
+
Case #15: 181175579
|
16 |
+
Case #16: 908478491
|
17 |
+
Case #17: 794274425
|
18 |
+
Case #18: 546269402
|
19 |
+
Case #19: 475069547
|
20 |
+
Case #20: 702572093
|
21 |
+
Case #21: 326135465
|
22 |
+
Case #22: 35062074
|
23 |
+
Case #23: 726328157
|
24 |
+
Case #24: 388616994
|
25 |
+
Case #25: 186249620
|
26 |
+
Case #26: 395528232
|
27 |
+
Case #27: 912756755
|
28 |
+
Case #28: 708169574
|
29 |
+
Case #29: 158143482
|
30 |
+
Case #30: 67974385
|
2017/quals/592385891651966.jpg
ADDED
Git LFS Details
|
2017/quals/fightingthezombie.html
ADDED
@@ -0,0 +1,86 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
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|
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|
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|
|
|
|
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|
|
|
|
|
|
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|
|
|
|
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
<p>
|
2 |
+
<strong>"Okay, Wizard, cast your spell!"</strong>
|
3 |
+
</p>
|
4 |
+
|
5 |
+
<p>
|
6 |
+
But which of your many spells to cast? In the ever-popular role-playing game
|
7 |
+
<em>Dungeons & Dragons</em>, or <em>D&D</em>, you determine a spell's damage
|
8 |
+
by rolling polyhedral
|
9 |
+
dice with 4, 6, 8, 10, 12, or 20 sides. Since there's a lot of dice-rolling
|
10 |
+
involved, players use shorthand to denote which dice should be rolled.
|
11 |
+
<strong>X</strong>d<strong>Y</strong> means
|
12 |
+
"roll a <strong>Y</strong>-sided die <strong>X</strong> times, and sum the rolls''.
|
13 |
+
Sometimes, you must add or subtract a value <strong>Z</strong> after
|
14 |
+
you finish rolling, in which case the notation is
|
15 |
+
<strong>X</strong>d<strong>Y</strong>+<strong>Z</strong> or
|
16 |
+
<strong>X</strong>d<strong>Y</strong>-<strong>Z</strong> respectively.
|
17 |
+
</p>
|
18 |
+
|
19 |
+
<p>
|
20 |
+
For example, if you roll 2d4+1, you'll end up with a result between 3 and 9
|
21 |
+
inclusive. If you roll 1d6-3, your result will be between -2 and 3 inclusive.
|
22 |
+
</p>
|
23 |
+
|
24 |
+
<p>
|
25 |
+
In <em>D&D</em>, wizards are powerful but flimsy spellcasters. As a wizard
|
26 |
+
fighting a zombie, your best strategy is to maximize the chance that you can
|
27 |
+
kill the zombie with a single spell before it has a chance to retaliate. What
|
28 |
+
spell should you cast?
|
29 |
+
</p>
|
30 |
+
|
31 |
+
|
32 |
+
|
33 |
+
<h3>Input</h3>
|
34 |
+
|
35 |
+
<p>
|
36 |
+
Input begins with an integer <strong>T</strong>, the number of zombies
|
37 |
+
you'll fight. For each zombie, there are two lines. The first contains two
|
38 |
+
integers, <strong>H</strong> and <strong>S</strong>, the minimum amount of
|
39 |
+
damage it takes to defeat the zombie, and the number of spells you have prepared,
|
40 |
+
respectively. The second line contains <strong>S</strong> spell descriptions separated by
|
41 |
+
single spaces. A spell description is simply the amount of damage a spell does
|
42 |
+
in the notation described above.
|
43 |
+
</p>
|
44 |
+
|
45 |
+
|
46 |
+
|
47 |
+
<h3>Output</h3>
|
48 |
+
|
49 |
+
<p>
|
50 |
+
For each zombie, print a line containing the probability of defeating the zombie if you select your spell optimally.
|
51 |
+
</p>
|
52 |
+
|
53 |
+
<p>
|
54 |
+
Absolute and relative errors of up to 1e-6 will be ignored.
|
55 |
+
</p>
|
56 |
+
|
57 |
+
<h3>Constraints</h3>
|
58 |
+
|
59 |
+
<p>
|
60 |
+
1 ≤ <strong>T</strong> ≤ 1,000 <br />
|
61 |
+
1 ≤ <strong>H</strong> ≤ 10,000 <br />
|
62 |
+
2 ≤ <strong>S</strong> ≤ 10 <br />
|
63 |
+
</p>
|
64 |
+
|
65 |
+
<p>
|
66 |
+
Additionally, the following constraints will hold for each spell:
|
67 |
+
</p>
|
68 |
+
|
69 |
+
<p>
|
70 |
+
1 ≤ <strong>X</strong> ≤ 20 <br />
|
71 |
+
<strong>Y</strong> ∈ {4, 6, 8, 10, 12, 20} <br />
|
72 |
+
1 ≤ <strong>Z</strong> ≤ 10,000, if <strong>Z</strong> is specified. <br />
|
73 |
+
<strong>X</strong>, <strong>Y</strong>, and <strong>Z</strong>
|
74 |
+
will be integers with no leading zeros. <br />
|
75 |
+
</p>
|
76 |
+
|
77 |
+
<h3>Explanation of Sample</h3>
|
78 |
+
|
79 |
+
<p>
|
80 |
+
In the first case, you can guarantee a kill with the first spell, which must always do at least 2 damage.
|
81 |
+
</p>
|
82 |
+
|
83 |
+
<p>
|
84 |
+
In the third case, your first spell is the best. If you roll a 4, you'll do the requisite 8 damage. The second spell requires
|
85 |
+
rolling a 4 on two dice rather than just one, and the third spell requires rolling a 4 on all three dice.
|
86 |
+
</p>
|
2017/quals/fightingthezombie.in
ADDED
The diff for this file is too large to render.
See raw diff
|
|
2017/quals/fightingthezombie.java
ADDED
@@ -0,0 +1,107 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
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|
|
|
|
|
|
|
1 |
+
// Hacker Cup 2017
|
2 |
+
// Qualification Round
|
3 |
+
// Fighting The Zombie
|
4 |
+
// Wesley May
|
5 |
+
|
6 |
+
import java.util.*;
|
7 |
+
|
8 |
+
public class FightingTheZombie
|
9 |
+
{
|
10 |
+
static int H;
|
11 |
+
|
12 |
+
public static void main(String args[])
|
13 |
+
{
|
14 |
+
double EPS = 0.000001;
|
15 |
+
Scanner scan = new Scanner(System.in);
|
16 |
+
int T = scan.nextInt();
|
17 |
+
|
18 |
+
for(int ca=1;ca <= T;ca++)
|
19 |
+
{
|
20 |
+
H = scan.nextInt();
|
21 |
+
int S = scan.nextInt();
|
22 |
+
Spell[] a = new Spell[S];
|
23 |
+
|
24 |
+
for(int i=0;i < S;i++)
|
25 |
+
{
|
26 |
+
String str = scan.next();
|
27 |
+
int X, Y, Z;
|
28 |
+
|
29 |
+
int didx = str.indexOf("d");
|
30 |
+
int plusidx = str.indexOf("+");
|
31 |
+
int minusidx = str.indexOf("-");
|
32 |
+
|
33 |
+
X = Integer.parseInt(str.substring(0, didx));
|
34 |
+
|
35 |
+
if(plusidx >= 0)
|
36 |
+
{
|
37 |
+
Y = Integer.parseInt(str.substring(didx+1, plusidx));
|
38 |
+
Z = Integer.parseInt(str.substring(plusidx));
|
39 |
+
}
|
40 |
+
else if(minusidx >= 0)
|
41 |
+
{
|
42 |
+
Y = Integer.parseInt(str.substring(didx+1,minusidx));
|
43 |
+
Z = Integer.parseInt(str.substring(minusidx));
|
44 |
+
}
|
45 |
+
else
|
46 |
+
{
|
47 |
+
Y = Integer.parseInt(str.substring(didx+1));
|
48 |
+
Z = 0;
|
49 |
+
}
|
50 |
+
|
51 |
+
a[i] = new Spell(X, Y, Z);
|
52 |
+
}
|
53 |
+
|
54 |
+
Arrays.sort(a);
|
55 |
+
|
56 |
+
System.out.printf("Case #%d: %.6f\n", ca, a[0].p);
|
57 |
+
}
|
58 |
+
}
|
59 |
+
|
60 |
+
static class Spell implements Comparable<Spell>
|
61 |
+
{
|
62 |
+
int x, y, z;
|
63 |
+
public double p;
|
64 |
+
double[][] dp;
|
65 |
+
|
66 |
+
public Spell(int ix, int iy, int iz)
|
67 |
+
{
|
68 |
+
x = ix;
|
69 |
+
y = iy;
|
70 |
+
z = iz;
|
71 |
+
int newH = H - z;
|
72 |
+
|
73 |
+
if(newH <= 0)
|
74 |
+
{
|
75 |
+
p = 1;
|
76 |
+
return;
|
77 |
+
}
|
78 |
+
|
79 |
+
dp = new double[newH+1][x+1];
|
80 |
+
for(int i=0;i < dp.length;i++)
|
81 |
+
Arrays.fill(dp[i], -1);
|
82 |
+
|
83 |
+
p = f(newH, x);
|
84 |
+
}
|
85 |
+
|
86 |
+
private double f(int h, int d)
|
87 |
+
{
|
88 |
+
if(h <= 0) return 1;
|
89 |
+
if(d == 0) return 0;
|
90 |
+
if(dp[h][d] >= 0) return dp[h][d];
|
91 |
+
|
92 |
+
double ans = 0;
|
93 |
+
for(int i=1;i <= y;i++)
|
94 |
+
{
|
95 |
+
ans += (1.0 / y) * f(h-i, d-1);
|
96 |
+
}
|
97 |
+
|
98 |
+
return dp[h][d] = ans;
|
99 |
+
}
|
100 |
+
|
101 |
+
public int compareTo(Spell that) {
|
102 |
+
if(this.p > that.p) return -1;
|
103 |
+
if(this.p < that.p) return 1;
|
104 |
+
return 0;
|
105 |
+
}
|
106 |
+
}
|
107 |
+
}
|
2017/quals/fightingthezombie.md
ADDED
@@ -0,0 +1,58 @@
|
|
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|
|
|
1 |
+
**"Okay, Wizard, cast your spell!"**
|
2 |
+
|
3 |
+
But which of your many spells to cast? In the ever-popular role-playing game
|
4 |
+
_Dungeons & Dragons_, or _D&D_, you determine a spell's damage by rolling
|
5 |
+
polyhedral dice with 4, 6, 8, 10, 12, or 20 sides. Since there's a lot of
|
6 |
+
dice-rolling involved, players use shorthand to denote which dice should be
|
7 |
+
rolled. **X**d**Y** means "roll a **Y**-sided die **X** times, and sum the
|
8 |
+
rolls''. Sometimes, you must add or subtract a value **Z** after you finish
|
9 |
+
rolling, in which case the notation is **X**d**Y**+**Z** or **X**d**Y**-**Z**
|
10 |
+
respectively.
|
11 |
+
|
12 |
+
For example, if you roll 2d4+1, you'll end up with a result between 3 and 9
|
13 |
+
inclusive. If you roll 1d6-3, your result will be between -2 and 3 inclusive.
|
14 |
+
|
15 |
+
In _D&D_, wizards are powerful but flimsy spellcasters. As a wizard fighting a
|
16 |
+
zombie, your best strategy is to maximize the chance that you can kill the
|
17 |
+
zombie with a single spell before it has a chance to retaliate. What spell
|
18 |
+
should you cast?
|
19 |
+
|
20 |
+
### Input
|
21 |
+
|
22 |
+
Input begins with an integer **T**, the number of zombies you'll fight. For
|
23 |
+
each zombie, there are two lines. The first contains two integers, **H** and
|
24 |
+
**S**, the minimum amount of damage it takes to defeat the zombie, and the
|
25 |
+
number of spells you have prepared, respectively. The second line contains
|
26 |
+
**S** spell descriptions separated by single spaces. A spell description is
|
27 |
+
simply the amount of damage a spell does in the notation described above.
|
28 |
+
|
29 |
+
### Output
|
30 |
+
|
31 |
+
For each zombie, print a line containing the probability of defeating the
|
32 |
+
zombie if you select your spell optimally.
|
33 |
+
|
34 |
+
Absolute and relative errors of up to 1e-6 will be ignored.
|
35 |
+
|
36 |
+
### Constraints
|
37 |
+
|
38 |
+
1 ≤ **T** ≤ 1,000
|
39 |
+
1 ≤ **H** ≤ 10,000
|
40 |
+
2 ≤ **S** ≤ 10
|
41 |
+
|
42 |
+
Additionally, the following constraints will hold for each spell:
|
43 |
+
|
44 |
+
1 ≤ **X** ≤ 20
|
45 |
+
**Y** ∈ {4, 6, 8, 10, 12, 20}
|
46 |
+
1 ≤ **Z** ≤ 10,000, if **Z** is specified.
|
47 |
+
**X**, **Y**, and **Z** will be integers with no leading zeros.
|
48 |
+
|
49 |
+
### Explanation of Sample
|
50 |
+
|
51 |
+
In the first case, you can guarantee a kill with the first spell, which must
|
52 |
+
always do at least 2 damage.
|
53 |
+
|
54 |
+
In the third case, your first spell is the best. If you roll a 4, you'll do
|
55 |
+
the requisite 8 damage. The second spell requires rolling a 4 on two dice
|
56 |
+
rather than just one, and the third spell requires rolling a 4 on all three
|
57 |
+
dice.
|
58 |
+
|
2017/quals/fightingthezombie.out
ADDED
@@ -0,0 +1,2005 @@
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|
1 |
+
Case #1: 1.000000
|
2 |
+
Case #2: 0.998520
|
3 |
+
Case #3: 0.250000
|
4 |
+
Case #4: 0.002500
|
5 |
+
Case #5: 0.400000
|
6 |
+
Case #6: 1.000000
|
7 |
+
Case #7: 0.000233
|
8 |
+
Case #8: 1.000000
|
9 |
+
Case #9: 1.000000
|
10 |
+
Case #10: 1.000000
|
11 |
+
Case #11: 0.999959
|
12 |
+
Case #12: 1.000000
|
13 |
+
Case #13: 0.999869
|
14 |
+
Case #14: 0.999892
|
15 |
+
Case #15: 1.000000
|
16 |
+
Case #16: 1.000000
|
17 |
+
Case #17: 1.000000
|
18 |
+
Case #18: 1.000000
|
19 |
+
Case #19: 1.000000
|
20 |
+
Case #20: 0.964421
|
21 |
+
Case #21: 0.000000
|
22 |
+
Case #22: 0.983925
|
23 |
+
Case #23: 0.999995
|
24 |
+
Case #24: 1.000000
|
25 |
+
Case #25: 0.999973
|
26 |
+
Case #26: 1.000000
|
27 |
+
Case #27: 0.981068
|
28 |
+
Case #28: 0.998299
|
29 |
+
Case #29: 0.954580
|
30 |
+
Case #30: 1.000000
|
31 |
+
Case #31: 1.000000
|
32 |
+
Case #32: 0.999939
|
33 |
+
Case #33: 1.000000
|
34 |
+
Case #34: 0.864193
|
35 |
+
Case #35: 1.000000
|
36 |
+
Case #36: 0.999976
|
37 |
+
Case #37: 1.000000
|
38 |
+
Case #38: 1.000000
|
39 |
+
Case #39: 1.000000
|
40 |
+
Case #40: 0.546425
|
41 |
+
Case #41: 1.000000
|
42 |
+
Case #42: 0.999659
|
43 |
+
Case #43: 1.000000
|
44 |
+
Case #44: 1.000000
|
45 |
+
Case #45: 0.999892
|
46 |
+
Case #46: 1.000000
|
47 |
+
Case #47: 1.000000
|
48 |
+
Case #48: 0.336282
|
49 |
+
Case #49: 1.000000
|
50 |
+
Case #50: 0.999982
|
51 |
+
Case #51: 0.998978
|
52 |
+
Case #52: 1.000000
|
53 |
+
Case #53: 0.977943
|
54 |
+
Case #54: 1.000000
|
55 |
+
Case #55: 1.000000
|
56 |
+
Case #56: 1.000000
|
57 |
+
Case #57: 1.000000
|
58 |
+
Case #58: 0.999991
|
59 |
+
Case #59: 0.000016
|
60 |
+
Case #60: 0.999366
|
61 |
+
Case #61: 1.000000
|
62 |
+
Case #62: 0.968075
|
63 |
+
Case #63: 0.086123
|
64 |
+
Case #64: 1.000000
|
65 |
+
Case #65: 0.999976
|
66 |
+
Case #66: 1.000000
|
67 |
+
Case #67: 1.000000
|
68 |
+
Case #68: 0.982362
|
69 |
+
Case #69: 0.996338
|
70 |
+
Case #70: 1.000000
|
71 |
+
Case #71: 0.987706
|
72 |
+
Case #72: 1.000000
|
73 |
+
Case #73: 0.971707
|
74 |
+
Case #74: 0.000018
|
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2017/quals/lazyloading.html
ADDED
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1 |
+
<p>
|
2 |
+
Wilson works for a moving company. His primary duty is to load household items into a moving truck.
|
3 |
+
Wilson has a bag that he uses to move these items. He puts a bunch of items in the bag, moves them
|
4 |
+
to the truck, and then drops the items off.
|
5 |
+
</p>
|
6 |
+
|
7 |
+
<p>
|
8 |
+
Wilson has a bit of a reputation as a lazy worker. Julie is Wilson's supervisor, and she's keen to make sure
|
9 |
+
that he doesn't slack off. She wants Wilson to carry at least 50 pounds of items in his bag every time he
|
10 |
+
goes to the truck.
|
11 |
+
</p>
|
12 |
+
|
13 |
+
<p>
|
14 |
+
Luckily for Wilson, his bag is opaque. When he carries a bagful of items, Julie can tell how many items are
|
15 |
+
in the bag (based on the height of the stack in the bag), and she can tell the weight of the top item. She can't, however,
|
16 |
+
tell how much the other items in the bag weigh.
|
17 |
+
She assumes that every item in the bag weighs at least as much as this top item, because surely Wilson,
|
18 |
+
as lazy as he is, would at least not be so dense as to put heavier items on top of lighter ones. Alas, Julie is woefully
|
19 |
+
ignorant of the extent of Wilson's lack of dedication to his duty, and this assumption is frequently incorrect.
|
20 |
+
</p>
|
21 |
+
|
22 |
+
<p>
|
23 |
+
Today there are <strong>N</strong> items to be moved, and Wilson, paid by the hour as he is, wants to maximize the
|
24 |
+
number of trips he makes to move all of them to the truck. What is the maximum number of trips Wilson can make
|
25 |
+
without getting berated by Julie?
|
26 |
+
</p>
|
27 |
+
|
28 |
+
<p>
|
29 |
+
Note that Julie is not aware of what items are to be moved today, and she is not keeping track of what Wilson has already
|
30 |
+
moved when she examines each bag of items. She simply assumes that each bagful contains a total weight of at least
|
31 |
+
<strong>k</strong> * <strong>w</strong>
|
32 |
+
where <strong>k</strong> is the number of items in the bag, and <strong>w</strong> is the weight of the top item.
|
33 |
+
</p>
|
34 |
+
|
35 |
+
<h3>Input</h3>
|
36 |
+
|
37 |
+
<p>
|
38 |
+
Input begins with an integer <strong>T</strong>, the number of days Wilson "works" at his job.
|
39 |
+
For each day, there is first a line containing the integer <strong>N</strong>.
|
40 |
+
Then there are <strong>N</strong> lines, the <strong>i</strong>th of which contains a single integer,
|
41 |
+
the weight of the <strong>i</strong>th item, <strong>W<sub>i</sub></strong>.
|
42 |
+
</p>
|
43 |
+
|
44 |
+
<h3>Output</h3>
|
45 |
+
|
46 |
+
<p>
|
47 |
+
For the <strong>i</strong>th day, print a line containing "Case #<strong>i</strong>: " followed by the maximum number
|
48 |
+
of trips Wilson can take that day.
|
49 |
+
</p>
|
50 |
+
|
51 |
+
<h3>Constraints</h3>
|
52 |
+
|
53 |
+
<p>
|
54 |
+
1 ≤ <strong>T</strong> ≤ 500 <br />
|
55 |
+
1 ≤ <strong>N</strong> ≤ 100 <br />
|
56 |
+
1 ≤ <strong>W<sub>i</sub></strong> ≤ 100 <br />
|
57 |
+
</p>
|
58 |
+
|
59 |
+
<p>
|
60 |
+
On every day, it is guaranteed that the total weight of all of the items is at least 50 pounds.
|
61 |
+
</p>
|
62 |
+
|
63 |
+
|
64 |
+
<h3>Explanation of Sample</h3>
|
65 |
+
|
66 |
+
<p>
|
67 |
+
In the first case, Wilson can make two trips by stacking a 30-pound item on top of a 1-pound item,
|
68 |
+
making the bag appear to contain 60 pounds.
|
69 |
+
</p>
|
70 |
+
|
71 |
+
<p>
|
72 |
+
In the second case, Wilson needs to put all the items in the bag at once and can only make one trip.
|
73 |
+
</p>
|
74 |
+
|
75 |
+
<p>
|
76 |
+
In the third case, one possible solution is to put the items with odd weight in the bag for the first trip,
|
77 |
+
and then the items with even weight in the bag for the second trip, making sure to put the heaviest item on top.
|
78 |
+
</p>
|
2017/quals/lazyloading.in
ADDED
The diff for this file is too large to render.
See raw diff
|
|
2017/quals/lazyloading.java
ADDED
@@ -0,0 +1,46 @@
|
|
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|
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|
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|
|
|
1 |
+
// Hacker Cup 2017
|
2 |
+
// Qualification Round
|
3 |
+
// Lazy Loading
|
4 |
+
// Wesley May
|
5 |
+
|
6 |
+
import java.util.*;
|
7 |
+
|
8 |
+
public class LazyLoading
|
9 |
+
{
|
10 |
+
public static void main(String args[])
|
11 |
+
{
|
12 |
+
Scanner scan = new Scanner(System.in);
|
13 |
+
int T = scan.nextInt();
|
14 |
+
|
15 |
+
for(int ca=1;ca <= T;ca++)
|
16 |
+
{
|
17 |
+
int N = scan.nextInt();
|
18 |
+
|
19 |
+
int[] W = new int[N];
|
20 |
+
|
21 |
+
for(int i=0;i < N;i++)
|
22 |
+
W[i] = scan.nextInt();
|
23 |
+
|
24 |
+
Arrays.sort(W);
|
25 |
+
|
26 |
+
int ans = 0;
|
27 |
+
int L = 0;
|
28 |
+
int R = N-1;
|
29 |
+
|
30 |
+
while(L <= R)
|
31 |
+
{
|
32 |
+
int count = 1;
|
33 |
+
while(count * W[R] < 50)
|
34 |
+
count++;
|
35 |
+
|
36 |
+
if(count <= R - L + 1)
|
37 |
+
ans++;
|
38 |
+
|
39 |
+
L += count - 1;
|
40 |
+
R--;
|
41 |
+
}
|
42 |
+
|
43 |
+
System.out.println("Case #" + ca + ": " + ans);
|
44 |
+
}
|
45 |
+
}
|
46 |
+
}
|
2017/quals/lazyloading.md
ADDED
@@ -0,0 +1,63 @@
|
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|
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|
|
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|
|
|
|
|
|
|
1 |
+
Wilson works for a moving company. His primary duty is to load household items
|
2 |
+
into a moving truck. Wilson has a bag that he uses to move these items. He
|
3 |
+
puts a bunch of items in the bag, moves them to the truck, and then drops the
|
4 |
+
items off.
|
5 |
+
|
6 |
+
Wilson has a bit of a reputation as a lazy worker. Julie is Wilson's
|
7 |
+
supervisor, and she's keen to make sure that he doesn't slack off. She wants
|
8 |
+
Wilson to carry at least 50 pounds of items in his bag every time he goes to
|
9 |
+
the truck.
|
10 |
+
|
11 |
+
Luckily for Wilson, his bag is opaque. When he carries a bagful of items,
|
12 |
+
Julie can tell how many items are in the bag (based on the height of the stack
|
13 |
+
in the bag), and she can tell the weight of the top item. She can't, however,
|
14 |
+
tell how much the other items in the bag weigh. She assumes that every item in
|
15 |
+
the bag weighs at least as much as this top item, because surely Wilson, as
|
16 |
+
lazy as he is, would at least not be so dense as to put heavier items on top
|
17 |
+
of lighter ones. Alas, Julie is woefully ignorant of the extent of Wilson's
|
18 |
+
lack of dedication to his duty, and this assumption is frequently incorrect.
|
19 |
+
|
20 |
+
Today there are **N** items to be moved, and Wilson, paid by the hour as he
|
21 |
+
is, wants to maximize the number of trips he makes to move all of them to the
|
22 |
+
truck. What is the maximum number of trips Wilson can make without getting
|
23 |
+
berated by Julie?
|
24 |
+
|
25 |
+
Note that Julie is not aware of what items are to be moved today, and she is
|
26 |
+
not keeping track of what Wilson has already moved when she examines each bag
|
27 |
+
of items. She simply assumes that each bagful contains a total weight of at
|
28 |
+
least **k** * **w** where **k** is the number of items in the bag, and **w**
|
29 |
+
is the weight of the top item.
|
30 |
+
|
31 |
+
### Input
|
32 |
+
|
33 |
+
Input begins with an integer **T**, the number of days Wilson "works" at his
|
34 |
+
job. For each day, there is first a line containing the integer **N**. Then
|
35 |
+
there are **N** lines, the **i**th of which contains a single integer, the
|
36 |
+
weight of the **i**th item, **Wi**.
|
37 |
+
|
38 |
+
### Output
|
39 |
+
|
40 |
+
For the **i**th day, print a line containing "Case #**i**: " followed by the
|
41 |
+
maximum number of trips Wilson can take that day.
|
42 |
+
|
43 |
+
### Constraints
|
44 |
+
|
45 |
+
1 ≤ **T** ≤ 500
|
46 |
+
1 ≤ **N** ≤ 100
|
47 |
+
1 ≤ **Wi** ≤ 100
|
48 |
+
|
49 |
+
On every day, it is guaranteed that the total weight of all of the items is at
|
50 |
+
least 50 pounds.
|
51 |
+
|
52 |
+
### Explanation of Sample
|
53 |
+
|
54 |
+
In the first case, Wilson can make two trips by stacking a 30-pound item on
|
55 |
+
top of a 1-pound item, making the bag appear to contain 60 pounds.
|
56 |
+
|
57 |
+
In the second case, Wilson needs to put all the items in the bag at once and
|
58 |
+
can only make one trip.
|
59 |
+
|
60 |
+
In the third case, one possible solution is to put the items with odd weight
|
61 |
+
in the bag for the first trip, and then the items with even weight in the bag
|
62 |
+
for the second trip, making sure to put the heaviest item on top.
|
63 |
+
|
2017/quals/lazyloading.out
ADDED
@@ -0,0 +1,1005 @@
|
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1 |
+
Case #1: 2
|
2 |
+
Case #2: 1
|
3 |
+
Case #3: 2
|
4 |
+
Case #4: 3
|
5 |
+
Case #5: 8
|
6 |
+
Case #6: 30
|
7 |
+
Case #7: 39
|
8 |
+
Case #8: 42
|
9 |
+
Case #9: 23
|
10 |
+
Case #10: 66
|
11 |
+
Case #11: 16
|
12 |
+
Case #12: 32
|
13 |
+
Case #13: 3
|
14 |
+
Case #14: 64
|
15 |
+
Case #15: 6
|
16 |
+
Case #16: 24
|
17 |
+
Case #17: 36
|
18 |
+
Case #18: 22
|
19 |
+
Case #19: 41
|
20 |
+
Case #20: 50
|
21 |
+
Case #21: 2
|
22 |
+
Case #22: 69
|
23 |
+
Case #23: 63
|
24 |
+
Case #24: 57
|
25 |
+
Case #25: 36
|
26 |
+
Case #26: 16
|
27 |
+
Case #27: 26
|
28 |
+
Case #28: 56
|
29 |
+
Case #29: 3
|
30 |
+
Case #30: 76
|
31 |
+
Case #31: 12
|
32 |
+
Case #32: 25
|
33 |
+
Case #33: 71
|
34 |
+
Case #34: 16
|
35 |
+
Case #35: 5
|
36 |
+
Case #36: 21
|
37 |
+
Case #37: 25
|
38 |
+
Case #38: 4
|
39 |
+
Case #39: 22
|
40 |
+
Case #40: 35
|
41 |
+
Case #41: 6
|
42 |
+
Case #42: 10
|
43 |
+
Case #43: 26
|
44 |
+
Case #44: 4
|
45 |
+
Case #45: 61
|
46 |
+
Case #46: 15
|
47 |
+
Case #47: 6
|
48 |
+
Case #48: 12
|
49 |
+
Case #49: 26
|
50 |
+
Case #50: 67
|
51 |
+
Case #51: 23
|
52 |
+
Case #52: 58
|
53 |
+
Case #53: 41
|
54 |
+
Case #54: 67
|
55 |
+
Case #55: 20
|
56 |
+
Case #56: 11
|
57 |
+
Case #57: 74
|
58 |
+
Case #58: 53
|
59 |
+
Case #59: 43
|
60 |
+
Case #60: 2
|
61 |
+
Case #61: 72
|
62 |
+
Case #62: 14
|
63 |
+
Case #63: 34
|
64 |
+
Case #64: 63
|
65 |
+
Case #65: 4
|
66 |
+
Case #66: 54
|
67 |
+
Case #67: 53
|
68 |
+
Case #68: 60
|
69 |
+
Case #69: 56
|
70 |
+
Case #70: 16
|
71 |
+
Case #71: 9
|
72 |
+
Case #72: 45
|
73 |
+
Case #73: 37
|
74 |
+
Case #74: 69
|
75 |
+
Case #75: 23
|
76 |
+
Case #76: 8
|
77 |
+
Case #77: 12
|
78 |
+
Case #78: 57
|
79 |
+
Case #79: 62
|
80 |
+
Case #80: 77
|
81 |
+
Case #81: 54
|
82 |
+
Case #82: 69
|
83 |
+
Case #83: 48
|
84 |
+
Case #84: 18
|
85 |
+
Case #85: 1
|
86 |
+
Case #86: 72
|
87 |
+
Case #87: 46
|
88 |
+
Case #88: 34
|
89 |
+
Case #89: 36
|
90 |
+
Case #90: 43
|
91 |
+
Case #91: 21
|
92 |
+
Case #92: 14
|
93 |
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Case #93: 13
|
94 |
+
Case #94: 8
|
95 |
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Case #95: 43
|
96 |
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Case #96: 21
|
97 |
+
Case #97: 67
|
98 |
+
Case #98: 5
|
99 |
+
Case #99: 24
|
100 |
+
Case #100: 21
|
101 |
+
Case #101: 60
|
102 |
+
Case #102: 43
|
103 |
+
Case #103: 1
|
104 |
+
Case #104: 29
|
105 |
+
Case #105: 9
|
106 |
+
Case #106: 65
|
107 |
+
Case #107: 57
|
108 |
+
Case #108: 24
|
109 |
+
Case #109: 1
|
110 |
+
Case #110: 25
|
111 |
+
Case #111: 30
|
112 |
+
Case #112: 65
|
113 |
+
Case #113: 23
|
114 |
+
Case #114: 75
|
115 |
+
Case #115: 40
|
116 |
+
Case #116: 16
|
117 |
+
Case #117: 22
|
118 |
+
Case #118: 52
|
119 |
+
Case #119: 11
|
120 |
+
Case #120: 60
|
121 |
+
Case #121: 13
|
122 |
+
Case #122: 1
|
123 |
+
Case #123: 55
|
124 |
+
Case #124: 47
|
125 |
+
Case #125: 20
|
126 |
+
Case #126: 46
|
127 |
+
Case #127: 45
|
128 |
+
Case #128: 28
|
129 |
+
Case #129: 15
|
130 |
+
Case #130: 57
|
131 |
+
Case #131: 50
|
132 |
+
Case #132: 5
|
133 |
+
Case #133: 63
|
134 |
+
Case #134: 28
|
135 |
+
Case #135: 33
|
136 |
+
Case #136: 32
|
137 |
+
Case #137: 69
|
138 |
+
Case #138: 60
|
139 |
+
Case #139: 11
|
140 |
+
Case #140: 54
|
141 |
+
Case #141: 71
|
142 |
+
Case #142: 44
|
143 |
+
Case #143: 72
|
144 |
+
Case #144: 60
|
145 |
+
Case #145: 3
|
146 |
+
Case #146: 13
|
147 |
+
Case #147: 16
|
148 |
+
Case #148: 8
|
149 |
+
Case #149: 20
|
150 |
+
Case #150: 41
|
151 |
+
Case #151: 60
|
152 |
+
Case #152: 65
|
153 |
+
Case #153: 76
|
154 |
+
Case #154: 53
|
155 |
+
Case #155: 51
|
156 |
+
Case #156: 71
|
157 |
+
Case #157: 48
|
158 |
+
Case #158: 33
|
159 |
+
Case #159: 41
|
160 |
+
Case #160: 68
|
161 |
+
Case #161: 41
|
162 |
+
Case #162: 25
|
163 |
+
Case #163: 60
|
164 |
+
Case #164: 23
|
165 |
+
Case #165: 13
|
166 |
+
Case #166: 2
|
167 |
+
Case #167: 35
|
168 |
+
Case #168: 45
|
169 |
+
Case #169: 56
|
170 |
+
Case #170: 54
|
171 |
+
Case #171: 48
|
172 |
+
Case #172: 46
|
173 |
+
Case #173: 3
|
174 |
+
Case #174: 57
|
175 |
+
Case #175: 4
|
176 |
+
Case #176: 9
|
177 |
+
Case #177: 61
|
178 |
+
Case #178: 40
|
179 |
+
Case #179: 9
|
180 |
+
Case #180: 15
|
181 |
+
Case #181: 65
|
182 |
+
Case #182: 1
|
183 |
+
Case #183: 16
|
184 |
+
Case #184: 72
|
185 |
+
Case #185: 70
|
186 |
+
Case #186: 52
|
187 |
+
Case #187: 1
|
188 |
+
Case #188: 72
|
189 |
+
Case #189: 34
|
190 |
+
Case #190: 49
|
191 |
+
Case #191: 54
|
192 |
+
Case #192: 2
|
193 |
+
Case #193: 11
|
194 |
+
Case #194: 36
|
195 |
+
Case #195: 51
|
196 |
+
Case #196: 54
|
197 |
+
Case #197: 54
|
198 |
+
Case #198: 76
|
199 |
+
Case #199: 30
|
200 |
+
Case #200: 69
|
201 |
+
Case #201: 28
|
202 |
+
Case #202: 17
|
203 |
+
Case #203: 58
|
204 |
+
Case #204: 19
|
205 |
+
Case #205: 20
|
206 |
+
Case #206: 44
|
207 |
+
Case #207: 68
|
208 |
+
Case #208: 69
|
209 |
+
Case #209: 36
|
210 |
+
Case #210: 29
|
211 |
+
Case #211: 29
|
212 |
+
Case #212: 6
|
213 |
+
Case #213: 16
|
214 |
+
Case #214: 60
|
215 |
+
Case #215: 66
|
216 |
+
Case #216: 72
|
217 |
+
Case #217: 38
|
218 |
+
Case #218: 30
|
219 |
+
Case #219: 54
|
220 |
+
Case #220: 39
|
221 |
+
Case #221: 13
|
222 |
+
Case #222: 5
|
223 |
+
Case #223: 11
|
224 |
+
Case #224: 19
|
225 |
+
Case #225: 7
|
226 |
+
Case #226: 23
|
227 |
+
Case #227: 7
|
228 |
+
Case #228: 68
|
229 |
+
Case #229: 53
|
230 |
+
Case #230: 33
|
231 |
+
Case #231: 65
|
232 |
+
Case #232: 7
|
233 |
+
Case #233: 14
|
234 |
+
Case #234: 15
|
235 |
+
Case #235: 69
|
236 |
+
Case #236: 7
|
237 |
+
Case #237: 48
|
238 |
+
Case #238: 24
|
239 |
+
Case #239: 23
|
240 |
+
Case #240: 64
|
241 |
+
Case #241: 70
|
242 |
+
Case #242: 16
|
243 |
+
Case #243: 27
|
244 |
+
Case #244: 54
|
245 |
+
Case #245: 54
|
246 |
+
Case #246: 39
|
247 |
+
Case #247: 22
|
248 |
+
Case #248: 42
|
249 |
+
Case #249: 37
|
250 |
+
Case #250: 68
|
251 |
+
Case #251: 23
|
252 |
+
Case #252: 40
|
253 |
+
Case #253: 24
|
254 |
+
Case #254: 57
|
255 |
+
Case #255: 55
|
256 |
+
Case #256: 77
|
257 |
+
Case #257: 25
|
258 |
+
Case #258: 26
|
259 |
+
Case #259: 38
|
260 |
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Case #260: 24
|
261 |
+
Case #261: 59
|
262 |
+
Case #262: 48
|
263 |
+
Case #263: 14
|
264 |
+
Case #264: 21
|
265 |
+
Case #265: 56
|
266 |
+
Case #266: 61
|
267 |
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Case #267: 8
|
268 |
+
Case #268: 57
|
269 |
+
Case #269: 7
|
270 |
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Case #270: 32
|
271 |
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Case #271: 29
|
272 |
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Case #272: 13
|
273 |
+
Case #273: 37
|
274 |
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Case #274: 27
|
275 |
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Case #275: 47
|
276 |
+
Case #276: 75
|
277 |
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Case #277: 47
|
278 |
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Case #278: 41
|
279 |
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Case #279: 12
|
280 |
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Case #280: 1
|
281 |
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Case #281: 16
|
282 |
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Case #282: 27
|
283 |
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Case #283: 73
|
284 |
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Case #284: 1
|
285 |
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Case #285: 52
|
286 |
+
Case #286: 60
|
287 |
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Case #287: 65
|
288 |
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Case #288: 14
|
289 |
+
Case #289: 26
|
290 |
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Case #290: 27
|
291 |
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Case #291: 52
|
292 |
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Case #292: 56
|
293 |
+
Case #293: 64
|
294 |
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Case #294: 19
|
295 |
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Case #295: 54
|
296 |
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Case #296: 21
|
297 |
+
Case #297: 41
|
298 |
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Case #298: 39
|
299 |
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Case #299: 10
|
300 |
+
Case #300: 59
|
301 |
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Case #301: 20
|
302 |
+
Case #302: 75
|
303 |
+
Case #303: 74
|
304 |
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Case #304: 21
|
305 |
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Case #305: 40
|
306 |
+
Case #306: 64
|
307 |
+
Case #307: 74
|
308 |
+
Case #308: 63
|
309 |
+
Case #309: 52
|
310 |
+
Case #310: 35
|
311 |
+
Case #311: 61
|
312 |
+
Case #312: 73
|
313 |
+
Case #313: 69
|
314 |
+
Case #314: 28
|
315 |
+
Case #315: 24
|
316 |
+
Case #316: 52
|
317 |
+
Case #317: 74
|
318 |
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Case #318: 57
|
319 |
+
Case #319: 53
|
320 |
+
Case #320: 66
|
321 |
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Case #321: 14
|
322 |
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Case #322: 20
|
323 |
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Case #323: 63
|
324 |
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Case #324: 58
|
325 |
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Case #325: 42
|
326 |
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Case #326: 13
|
327 |
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Case #327: 25
|
328 |
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Case #328: 7
|
329 |
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Case #329: 25
|
330 |
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Case #330: 42
|
331 |
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Case #331: 17
|
332 |
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Case #332: 16
|
333 |
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Case #333: 29
|
334 |
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Case #334: 30
|
335 |
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Case #335: 25
|
336 |
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Case #336: 53
|
337 |
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Case #337: 54
|
338 |
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Case #338: 12
|
339 |
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Case #339: 57
|
340 |
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Case #340: 72
|
341 |
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Case #341: 10
|
342 |
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Case #342: 70
|
343 |
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Case #343: 3
|
344 |
+
Case #344: 56
|
345 |
+
Case #345: 1
|
346 |
+
Case #346: 33
|
347 |
+
Case #347: 38
|
348 |
+
Case #348: 68
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824 |
+
Case #824: 21
|
825 |
+
Case #825: 13
|
826 |
+
Case #826: 73
|
827 |
+
Case #827: 56
|
828 |
+
Case #828: 27
|
829 |
+
Case #829: 8
|
830 |
+
Case #830: 36
|
831 |
+
Case #831: 24
|
832 |
+
Case #832: 9
|
833 |
+
Case #833: 38
|
834 |
+
Case #834: 34
|
835 |
+
Case #835: 43
|
836 |
+
Case #836: 73
|
837 |
+
Case #837: 38
|
838 |
+
Case #838: 56
|
839 |
+
Case #839: 29
|
840 |
+
Case #840: 34
|
841 |
+
Case #841: 69
|
842 |
+
Case #842: 5
|
843 |
+
Case #843: 21
|
844 |
+
Case #844: 11
|
845 |
+
Case #845: 40
|
846 |
+
Case #846: 13
|
847 |
+
Case #847: 66
|
848 |
+
Case #848: 10
|
849 |
+
Case #849: 23
|
850 |
+
Case #850: 29
|
851 |
+
Case #851: 51
|
852 |
+
Case #852: 26
|
853 |
+
Case #853: 33
|
854 |
+
Case #854: 10
|
855 |
+
Case #855: 32
|
856 |
+
Case #856: 17
|
857 |
+
Case #857: 71
|
858 |
+
Case #858: 1
|
859 |
+
Case #859: 35
|
860 |
+
Case #860: 48
|
861 |
+
Case #861: 50
|
862 |
+
Case #862: 44
|
863 |
+
Case #863: 63
|
864 |
+
Case #864: 55
|
865 |
+
Case #865: 34
|
866 |
+
Case #866: 62
|
867 |
+
Case #867: 61
|
868 |
+
Case #868: 28
|
869 |
+
Case #869: 15
|
870 |
+
Case #870: 55
|
871 |
+
Case #871: 55
|
872 |
+
Case #872: 44
|
873 |
+
Case #873: 6
|
874 |
+
Case #874: 30
|
875 |
+
Case #875: 61
|
876 |
+
Case #876: 45
|
877 |
+
Case #877: 15
|
878 |
+
Case #878: 66
|
879 |
+
Case #879: 19
|
880 |
+
Case #880: 32
|
881 |
+
Case #881: 23
|
882 |
+
Case #882: 21
|
883 |
+
Case #883: 69
|
884 |
+
Case #884: 67
|
885 |
+
Case #885: 39
|
886 |
+
Case #886: 2
|
887 |
+
Case #887: 48
|
888 |
+
Case #888: 45
|
889 |
+
Case #889: 66
|
890 |
+
Case #890: 10
|
891 |
+
Case #891: 40
|
892 |
+
Case #892: 23
|
893 |
+
Case #893: 65
|
894 |
+
Case #894: 34
|
895 |
+
Case #895: 26
|
896 |
+
Case #896: 30
|
897 |
+
Case #897: 26
|
898 |
+
Case #898: 59
|
899 |
+
Case #899: 67
|
900 |
+
Case #900: 41
|
901 |
+
Case #901: 10
|
902 |
+
Case #902: 56
|
903 |
+
Case #903: 19
|
904 |
+
Case #904: 43
|
905 |
+
Case #905: 42
|
906 |
+
Case #906: 6
|
907 |
+
Case #907: 9
|
908 |
+
Case #908: 54
|
909 |
+
Case #909: 10
|
910 |
+
Case #910: 29
|
911 |
+
Case #911: 32
|
912 |
+
Case #912: 46
|
913 |
+
Case #913: 39
|
914 |
+
Case #914: 52
|
915 |
+
Case #915: 33
|
916 |
+
Case #916: 37
|
917 |
+
Case #917: 4
|
918 |
+
Case #918: 6
|
919 |
+
Case #919: 5
|
920 |
+
Case #920: 72
|
921 |
+
Case #921: 68
|
922 |
+
Case #922: 37
|
923 |
+
Case #923: 74
|
924 |
+
Case #924: 60
|
925 |
+
Case #925: 62
|
926 |
+
Case #926: 74
|
927 |
+
Case #927: 14
|
928 |
+
Case #928: 44
|
929 |
+
Case #929: 51
|
930 |
+
Case #930: 50
|
931 |
+
Case #931: 42
|
932 |
+
Case #932: 9
|
933 |
+
Case #933: 24
|
934 |
+
Case #934: 51
|
935 |
+
Case #935: 26
|
936 |
+
Case #936: 65
|
937 |
+
Case #937: 54
|
938 |
+
Case #938: 21
|
939 |
+
Case #939: 25
|
940 |
+
Case #940: 32
|
941 |
+
Case #941: 59
|
942 |
+
Case #942: 45
|
943 |
+
Case #943: 23
|
944 |
+
Case #944: 27
|
945 |
+
Case #945: 22
|
946 |
+
Case #946: 65
|
947 |
+
Case #947: 20
|
948 |
+
Case #948: 12
|
949 |
+
Case #949: 8
|
950 |
+
Case #950: 24
|
951 |
+
Case #951: 34
|
952 |
+
Case #952: 19
|
953 |
+
Case #953: 7
|
954 |
+
Case #954: 42
|
955 |
+
Case #955: 26
|
956 |
+
Case #956: 18
|
957 |
+
Case #957: 22
|
958 |
+
Case #958: 2
|
959 |
+
Case #959: 46
|
960 |
+
Case #960: 54
|
961 |
+
Case #961: 2
|
962 |
+
Case #962: 62
|
963 |
+
Case #963: 45
|
964 |
+
Case #964: 10
|
965 |
+
Case #965: 21
|
966 |
+
Case #966: 61
|
967 |
+
Case #967: 71
|
968 |
+
Case #968: 25
|
969 |
+
Case #969: 33
|
970 |
+
Case #970: 36
|
971 |
+
Case #971: 55
|
972 |
+
Case #972: 29
|
973 |
+
Case #973: 77
|
974 |
+
Case #974: 71
|
975 |
+
Case #975: 60
|
976 |
+
Case #976: 71
|
977 |
+
Case #977: 67
|
978 |
+
Case #978: 13
|
979 |
+
Case #979: 5
|
980 |
+
Case #980: 24
|
981 |
+
Case #981: 47
|
982 |
+
Case #982: 50
|
983 |
+
Case #983: 17
|
984 |
+
Case #984: 26
|
985 |
+
Case #985: 23
|
986 |
+
Case #986: 54
|
987 |
+
Case #987: 38
|
988 |
+
Case #988: 73
|
989 |
+
Case #989: 51
|
990 |
+
Case #990: 15
|
991 |
+
Case #991: 50
|
992 |
+
Case #992: 52
|
993 |
+
Case #993: 5
|
994 |
+
Case #994: 35
|
995 |
+
Case #995: 55
|
996 |
+
Case #996: 56
|
997 |
+
Case #997: 49
|
998 |
+
Case #998: 72
|
999 |
+
Case #999: 31
|
1000 |
+
Case #1000: 13
|
1001 |
+
Case #1001: 32
|
1002 |
+
Case #1002: 12
|
1003 |
+
Case #1003: 67
|
1004 |
+
Case #1004: 21
|
1005 |
+
Case #1005: 31
|
2017/quals/progresspie.html
ADDED
@@ -0,0 +1,72 @@
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|
1 |
+
<p>
|
2 |
+
Some progress bars fill you with anticipation.
|
3 |
+
Some are finished before you know it and make you wonder why there was a progress bar at all.
|
4 |
+
</p>
|
5 |
+
|
6 |
+
<p>
|
7 |
+
Some progress bars progress at a pleasant, steady rate.
|
8 |
+
Some are chaotic, lurching forward and then pausing for long periods.
|
9 |
+
Some seem to slow down as they go, never quite reaching 100%.
|
10 |
+
</p>
|
11 |
+
|
12 |
+
<p>
|
13 |
+
Some progress bars are in fact not bars at all, but circles.
|
14 |
+
</p>
|
15 |
+
|
16 |
+
<p>
|
17 |
+
On your screen is a progress pie, a sort of progress bar that shows its progress as a sector of a circle.
|
18 |
+
Envision your screen as a square on the plane with its bottom-left corner at (0, 0), and its upper-right corner at (100, 100).
|
19 |
+
Every point on the screen is either white or black.
|
20 |
+
Initially, the progress is 0%, and all points on the screen are white. When the progress percentage, <strong>P</strong>, is greater than 0%, a sector of angle (<strong>P</strong>% * 360) degrees
|
21 |
+
is colored black, anchored by the line segment from the center of the square to the center of the top side, and proceeding clockwise.
|
22 |
+
</p>
|
23 |
+
|
24 |
+
<img src="{{PHOTO_ID:592385891651966}}" />
|
25 |
+
|
26 |
+
<p>
|
27 |
+
While you wait for the progress pie to fill in, you find yourself thinking about whether certain points would be white or black at different amounts of progress.
|
28 |
+
</p>
|
29 |
+
|
30 |
+
|
31 |
+
<h3>Input</h3>
|
32 |
+
|
33 |
+
<p>
|
34 |
+
Input begins with an integer <strong>T</strong>, the number of points you're curious about.
|
35 |
+
For each point, there is a line containing three space-separated integers, <strong>P</strong>, the amount of progress as a percentage,
|
36 |
+
and <strong>X</strong> and <strong>Y</strong>, the coordinates of the point.
|
37 |
+
</p>
|
38 |
+
|
39 |
+
<h3>Output</h3>
|
40 |
+
|
41 |
+
<p>
|
42 |
+
For the <strong>i</strong>th point, print a line containing "Case #<strong>i</strong>: " followed by the color of the point, either
|
43 |
+
"black" or "white".
|
44 |
+
</p>
|
45 |
+
|
46 |
+
<h3>Constraints</h3>
|
47 |
+
|
48 |
+
<p>
|
49 |
+
1 ≤ <strong>T</strong> ≤ 1,000 <br />
|
50 |
+
0 ≤ <strong>P</strong>, <strong>X</strong>, <strong>Y</strong> ≤ 100 <br />
|
51 |
+
</p>
|
52 |
+
|
53 |
+
<p>
|
54 |
+
Whenever a point (<strong>X</strong>, <strong>Y</strong>) is queried, it's guaranteed that all points within a distance of 10<sup>-6</sup> of
|
55 |
+
(<strong>X</strong>, <strong>Y</strong>) are the same color as (<strong>X</strong>, <strong>Y</strong>).
|
56 |
+
</p>
|
57 |
+
|
58 |
+
|
59 |
+
<h3>Explanation of Sample</h3>
|
60 |
+
|
61 |
+
<p>
|
62 |
+
In the first case all of the points are white, so the point at (55, 55) is of course white.
|
63 |
+
</p>
|
64 |
+
|
65 |
+
<p>
|
66 |
+
In the second case, (55, 55) is close to the filled-in sector of the circle, but it's still white.
|
67 |
+
</p>
|
68 |
+
|
69 |
+
<p>
|
70 |
+
In the third case, the filled-in sector of the circle now covers (55, 55), coloring it black.
|
71 |
+
</p>
|
72 |
+
|
2017/quals/progresspie.in
ADDED
@@ -0,0 +1,2006 @@
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+
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|
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+
14 46 40
|
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|
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+
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|
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|
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|
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|
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|
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|
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|
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54 18 72
|
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+
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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52 58 41
|
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|
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|
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|
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|
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79 26 26
|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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94 24 53
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19 57 99
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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55 58 79
|
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32 47 87
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|
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|
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|
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19 87 86
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97 58 73
|
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|
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|
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|
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|
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67 58 20
|
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33 58 9
|
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|
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|
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|
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5 4 81
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|
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|
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9 67 98
|
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44 26 6
|
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|
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|
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|
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|
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100 89 76
|
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|
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|
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|
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|
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|
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1 81 66
|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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6 26 84
|
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4 47 1
|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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|
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16 58 42
|
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|
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|
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|
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|
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|
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|
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4 64 27
|
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56 30 6
|
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83 83 38
|
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14 60 30
|
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+
3 98 22
|
1905 |
+
53 100 79
|
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14 63 94
|
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+
53 48 96
|
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+
59 72 79
|
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|
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|
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|
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51 9 46
|
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95 85 81
|
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32 77 66
|
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28 37 95
|
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|
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|
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28 12 87
|
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19 54 59
|
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64 82 82
|
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4 99 54
|
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36 32 85
|
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38 68 94
|
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+
2 77 57
|
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+
61 88 67
|
1926 |
+
88 17 100
|
1927 |
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26 26 94
|
1928 |
+
77 31 9
|
1929 |
+
55 2 71
|
1930 |
+
80 53 38
|
1931 |
+
75 2 67
|
1932 |
+
5 36 8
|
1933 |
+
44 7 12
|
1934 |
+
92 55 47
|
1935 |
+
52 10 20
|
1936 |
+
51 46 69
|
1937 |
+
77 28 9
|
1938 |
+
46 26 68
|
1939 |
+
19 71 93
|
1940 |
+
58 27 76
|
1941 |
+
68 85 59
|
1942 |
+
81 53 85
|
1943 |
+
18 89 10
|
1944 |
+
93 40 71
|
1945 |
+
85 16 71
|
1946 |
+
39 62 100
|
1947 |
+
6 97 77
|
1948 |
+
17 26 74
|
1949 |
+
86 14 22
|
1950 |
+
11 5 82
|
1951 |
+
21 94 36
|
1952 |
+
56 43 8
|
1953 |
+
20 60 5
|
1954 |
+
29 11 42
|
1955 |
+
37 65 24
|
1956 |
+
3 83 27
|
1957 |
+
26 91 40
|
1958 |
+
23 66 83
|
1959 |
+
4 87 25
|
1960 |
+
24 29 11
|
1961 |
+
91 86 25
|
1962 |
+
59 96 91
|
1963 |
+
0 89 69
|
1964 |
+
95 95 17
|
1965 |
+
3 22 39
|
1966 |
+
44 88 41
|
1967 |
+
29 36 86
|
1968 |
+
24 98 37
|
1969 |
+
2 42 20
|
1970 |
+
43 99 74
|
1971 |
+
91 10 57
|
1972 |
+
47 68 48
|
1973 |
+
18 18 95
|
1974 |
+
76 87 22
|
1975 |
+
15 81 62
|
1976 |
+
41 85 26
|
1977 |
+
38 4 1
|
1978 |
+
60 100 78
|
1979 |
+
54 98 4
|
1980 |
+
60 57 42
|
1981 |
+
98 13 29
|
1982 |
+
28 58 46
|
1983 |
+
90 16 86
|
1984 |
+
0 20 21
|
1985 |
+
37 43 86
|
1986 |
+
20 62 88
|
1987 |
+
33 55 85
|
1988 |
+
82 42 37
|
1989 |
+
54 53 91
|
1990 |
+
82 97 70
|
1991 |
+
99 70 86
|
1992 |
+
77 75 53
|
1993 |
+
29 71 99
|
1994 |
+
63 74 95
|
1995 |
+
58 20 22
|
1996 |
+
89 48 39
|
1997 |
+
25 2 9
|
1998 |
+
54 83 44
|
1999 |
+
8 64 34
|
2000 |
+
33 70 76
|
2001 |
+
31 11 98
|
2002 |
+
87 93 35
|
2003 |
+
58 85 42
|
2004 |
+
96 74 56
|
2005 |
+
96 28 17
|
2006 |
+
2 100 54
|
2017/quals/progresspie.java
ADDED
@@ -0,0 +1,46 @@
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
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|
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|
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|
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|
|
|
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|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
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|
|
|
|
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|
|
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|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
1 |
+
// Hacker Cup 2017
|
2 |
+
// Qualification Round
|
3 |
+
// Progress Pie
|
4 |
+
// Wesley May
|
5 |
+
|
6 |
+
import java.util.*;
|
7 |
+
|
8 |
+
public class ProgressPie
|
9 |
+
{
|
10 |
+
public static void main(String args[])
|
11 |
+
{
|
12 |
+
double EPS = 0.000001;
|
13 |
+
Scanner scan = new Scanner(System.in);
|
14 |
+
int T = scan.nextInt();
|
15 |
+
|
16 |
+
for(int ca=1;ca <= T;ca++)
|
17 |
+
{
|
18 |
+
double p = scan.nextDouble() / 100;
|
19 |
+
double x = scan.nextDouble();
|
20 |
+
double y = scan.nextDouble();
|
21 |
+
boolean ans = solve(p, x, y);
|
22 |
+
|
23 |
+
System.out.println("Case #" + ca + ": " + (ans ? "black" : "white"));
|
24 |
+
}
|
25 |
+
}
|
26 |
+
|
27 |
+
public static boolean solve(double p, double x, double y)
|
28 |
+
{
|
29 |
+
if(p == 0) return false;
|
30 |
+
|
31 |
+
double dx = x - 50;
|
32 |
+
double dy = y - 50;
|
33 |
+
double d = Math.sqrt(dx*dx + dy*dy);
|
34 |
+
if (d > 50) return false;
|
35 |
+
|
36 |
+
if(dx == 0 && dy == 0) return true;
|
37 |
+
|
38 |
+
double theta = Math.atan2(dy, dx);
|
39 |
+
if (theta < 0) theta += 2 * Math.PI;
|
40 |
+
double myp = (2 * Math.PI - theta) / (2 * Math.PI);
|
41 |
+
myp += 0.25;
|
42 |
+
if (myp >= 1) myp -= 1;
|
43 |
+
|
44 |
+
return myp <= p;
|
45 |
+
}
|
46 |
+
}
|
2017/quals/progresspie.md
ADDED
@@ -0,0 +1,56 @@
|
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|
1 |
+
Some progress bars fill you with anticipation. Some are finished before you
|
2 |
+
know it and make you wonder why there was a progress bar at all.
|
3 |
+
|
4 |
+
Some progress bars progress at a pleasant, steady rate. Some are chaotic,
|
5 |
+
lurching forward and then pausing for long periods. Some seem to slow down as
|
6 |
+
they go, never quite reaching 100%.
|
7 |
+
|
8 |
+
Some progress bars are in fact not bars at all, but circles.
|
9 |
+
|
10 |
+
On your screen is a progress pie, a sort of progress bar that shows its
|
11 |
+
progress as a sector of a circle. Envision your screen as a square on the
|
12 |
+
plane with its bottom-left corner at (0, 0), and its upper-right corner at
|
13 |
+
(100, 100). Every point on the screen is either white or black. Initially, the
|
14 |
+
progress is 0%, and all points on the screen are white. When the progress
|
15 |
+
percentage, **P**, is greater than 0%, a sector of angle (**P**% * 360)
|
16 |
+
degrees is colored black, anchored by the line segment from the center of the
|
17 |
+
square to the center of the top side, and proceeding clockwise.
|
18 |
+
|
19 |
+
![]({{PHOTO_ID:592385891651966}})
|
20 |
+
|
21 |
+
While you wait for the progress pie to fill in, you find yourself thinking
|
22 |
+
about whether certain points would be white or black at different amounts of
|
23 |
+
progress.
|
24 |
+
|
25 |
+
### Input
|
26 |
+
|
27 |
+
Input begins with an integer **T**, the number of points you're curious about.
|
28 |
+
For each point, there is a line containing three space-separated integers,
|
29 |
+
**P**, the amount of progress as a percentage, and **X** and **Y**, the
|
30 |
+
coordinates of the point.
|
31 |
+
|
32 |
+
### Output
|
33 |
+
|
34 |
+
For the **i**th point, print a line containing "Case #**i**: " followed by the
|
35 |
+
color of the point, either "black" or "white".
|
36 |
+
|
37 |
+
### Constraints
|
38 |
+
|
39 |
+
1 ≤ **T** ≤ 1,000
|
40 |
+
0 ≤ **P**, **X**, **Y** ≤ 100
|
41 |
+
|
42 |
+
Whenever a point (**X**, **Y**) is queried, it's guaranteed that all points
|
43 |
+
within a distance of 10-6 of (**X**, **Y**) are the same color as (**X**,
|
44 |
+
**Y**).
|
45 |
+
|
46 |
+
### Explanation of Sample
|
47 |
+
|
48 |
+
In the first case all of the points are white, so the point at (55, 55) is of
|
49 |
+
course white.
|
50 |
+
|
51 |
+
In the second case, (55, 55) is close to the filled-in sector of the circle,
|
52 |
+
but it's still white.
|
53 |
+
|
54 |
+
In the third case, the filled-in sector of the circle now covers (55, 55),
|
55 |
+
coloring it black.
|
56 |
+
|
2017/quals/progresspie.out
ADDED
@@ -0,0 +1,2005 @@
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1 |
+
Case #1: white
|
2 |
+
Case #2: white
|
3 |
+
Case #3: black
|
4 |
+
Case #4: white
|
5 |
+
Case #5: black
|
6 |
+
Case #6: white
|
7 |
+
Case #7: white
|
8 |
+
Case #8: white
|
9 |
+
Case #9: black
|
10 |
+
Case #10: white
|
11 |
+
Case #11: black
|
12 |
+
Case #12: white
|
13 |
+
Case #13: black
|
14 |
+
Case #14: white
|
15 |
+
Case #15: white
|
16 |
+
Case #16: black
|
17 |
+
Case #17: white
|
18 |
+
Case #18: black
|
19 |
+
Case #19: black
|
20 |
+
Case #20: white
|
21 |
+
Case #21: white
|
22 |
+
Case #22: white
|
23 |
+
Case #23: black
|
24 |
+
Case #24: black
|
25 |
+
Case #25: white
|
26 |
+
Case #26: white
|
27 |
+
Case #27: black
|
28 |
+
Case #28: black
|
29 |
+
Case #29: white
|
30 |
+
Case #30: black
|
31 |
+
Case #31: black
|
32 |
+
Case #32: white
|
33 |
+
Case #33: white
|
34 |
+
Case #34: white
|
35 |
+
Case #35: white
|
36 |
+
Case #36: white
|
37 |
+
Case #37: black
|
38 |
+
Case #38: black
|
39 |
+
Case #39: white
|
40 |
+
Case #40: white
|
41 |
+
Case #41: white
|
42 |
+
Case #42: white
|
43 |
+
Case #43: white
|
44 |
+
Case #44: black
|
45 |
+
Case #45: white
|
46 |
+
Case #46: white
|
47 |
+
Case #47: black
|
48 |
+
Case #48: black
|
49 |
+
Case #49: white
|
50 |
+
Case #50: white
|
51 |
+
Case #51: white
|
52 |
+
Case #52: white
|
53 |
+
Case #53: white
|
54 |
+
Case #54: white
|
55 |
+
Case #55: white
|
56 |
+
Case #56: white
|
57 |
+
Case #57: black
|
58 |
+
Case #58: white
|
59 |
+
Case #59: white
|
60 |
+
Case #60: black
|
61 |
+
Case #61: black
|
62 |
+
Case #62: white
|
63 |
+
Case #63: black
|
64 |
+
Case #64: white
|
65 |
+
Case #65: black
|
66 |
+
Case #66: white
|
67 |
+
Case #67: black
|
68 |
+
Case #68: white
|
69 |
+
Case #69: white
|
70 |
+
Case #70: white
|
71 |
+
Case #71: white
|
72 |
+
Case #72: black
|
73 |
+
Case #73: white
|
74 |
+
Case #74: black
|
75 |
+
Case #75: black
|
76 |
+
Case #76: black
|
77 |
+
Case #77: black
|
78 |
+
Case #78: white
|
79 |
+
Case #79: black
|
80 |
+
Case #80: black
|
81 |
+
Case #81: white
|
82 |
+
Case #82: white
|
83 |
+
Case #83: black
|
84 |
+
Case #84: black
|
85 |
+
Case #85: white
|
86 |
+
Case #86: black
|
87 |
+
Case #87: white
|
88 |
+
Case #88: white
|
89 |
+
Case #89: white
|
90 |
+
Case #90: white
|
91 |
+
Case #91: white
|
92 |
+
Case #92: white
|
93 |
+
Case #93: white
|
94 |
+
Case #94: white
|
95 |
+
Case #95: white
|
96 |
+
Case #96: black
|
97 |
+
Case #97: white
|
98 |
+
Case #98: black
|
99 |
+
Case #99: black
|
100 |
+
Case #100: black
|
101 |
+
Case #101: white
|
102 |
+
Case #102: black
|
103 |
+
Case #103: black
|
104 |
+
Case #104: white
|
105 |
+
Case #105: black
|
106 |
+
Case #106: white
|
107 |
+
Case #107: white
|
108 |
+
Case #108: white
|
109 |
+
Case #109: white
|
110 |
+
Case #110: white
|
111 |
+
Case #111: black
|
112 |
+
Case #112: white
|
113 |
+
Case #113: black
|
114 |
+
Case #114: black
|
115 |
+
Case #115: white
|
116 |
+
Case #116: black
|
117 |
+
Case #117: white
|
118 |
+
Case #118: black
|
119 |
+
Case #119: black
|
120 |
+
Case #120: black
|
121 |
+
Case #121: white
|
122 |
+
Case #122: black
|
123 |
+
Case #123: white
|
124 |
+
Case #124: white
|
125 |
+
Case #125: black
|
126 |
+
Case #126: white
|
127 |
+
Case #127: white
|
128 |
+
Case #128: white
|
129 |
+
Case #129: white
|
130 |
+
Case #130: black
|
131 |
+
Case #131: white
|
132 |
+
Case #132: white
|
133 |
+
Case #133: black
|
134 |
+
Case #134: white
|
135 |
+
Case #135: black
|
136 |
+
Case #136: black
|
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1941 |
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Case #1941: black
|
1942 |
+
Case #1942: white
|
1943 |
+
Case #1943: black
|
1944 |
+
Case #1944: black
|
1945 |
+
Case #1945: white
|
1946 |
+
Case #1946: white
|
1947 |
+
Case #1947: white
|
1948 |
+
Case #1948: black
|
1949 |
+
Case #1949: white
|
1950 |
+
Case #1950: white
|
1951 |
+
Case #1951: black
|
1952 |
+
Case #1952: white
|
1953 |
+
Case #1953: white
|
1954 |
+
Case #1954: white
|
1955 |
+
Case #1955: white
|
1956 |
+
Case #1956: white
|
1957 |
+
Case #1957: black
|
1958 |
+
Case #1958: white
|
1959 |
+
Case #1959: white
|
1960 |
+
Case #1960: black
|
1961 |
+
Case #1961: white
|
1962 |
+
Case #1962: white
|
1963 |
+
Case #1963: white
|
1964 |
+
Case #1964: white
|
1965 |
+
Case #1965: black
|
1966 |
+
Case #1966: white
|
1967 |
+
Case #1967: white
|
1968 |
+
Case #1968: white
|
1969 |
+
Case #1969: white
|
1970 |
+
Case #1970: black
|
1971 |
+
Case #1971: black
|
1972 |
+
Case #1972: white
|
1973 |
+
Case #1973: black
|
1974 |
+
Case #1974: white
|
1975 |
+
Case #1975: black
|
1976 |
+
Case #1976: white
|
1977 |
+
Case #1977: white
|
1978 |
+
Case #1978: white
|
1979 |
+
Case #1979: black
|
1980 |
+
Case #1980: black
|
1981 |
+
Case #1981: white
|
1982 |
+
Case #1982: black
|
1983 |
+
Case #1983: white
|
1984 |
+
Case #1984: white
|
1985 |
+
Case #1985: black
|
1986 |
+
Case #1986: black
|
1987 |
+
Case #1987: black
|
1988 |
+
Case #1988: black
|
1989 |
+
Case #1989: white
|
1990 |
+
Case #1990: black
|
1991 |
+
Case #1991: black
|
1992 |
+
Case #1992: white
|
1993 |
+
Case #1993: white
|
1994 |
+
Case #1994: white
|
1995 |
+
Case #1995: black
|
1996 |
+
Case #1996: white
|
1997 |
+
Case #1997: black
|
1998 |
+
Case #1998: white
|
1999 |
+
Case #1999: black
|
2000 |
+
Case #2000: white
|
2001 |
+
Case #2001: black
|
2002 |
+
Case #2002: black
|
2003 |
+
Case #2003: black
|
2004 |
+
Case #2004: black
|
2005 |
+
Case #2005: white
|
2017/round1/268425761045248.jpg
ADDED
Git LFS Details
|
2017/round1/336634350719298.jpg
ADDED
Git LFS Details
|