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// Pond Precipitation
// Solution by Jacob Plachta
#include <algorithm>
#include <functional>
#include <numeric>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <complex>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <cassert>
#include <string>
#include <vector>
#include <list>
#include <map>
#include <unordered_map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <sstream>
using namespace std;
#define LL long long
#define LD long double
#define PR pair<int,int>
#define Fox(i,n) for (i=0; i<n; i++)
#define Fox1(i,n) for (i=1; i<=n; i++)
#define FoxI(i,a,b) for (i=a; i<=b; i++)
#define FoxR(i,n) for (i=(n)-1; i>=0; i--)
#define FoxR1(i,n) for (i=n; i>0; i--)
#define FoxRI(i,a,b) for (i=b; i>=a; i--)
#define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++)
#define Min(a,b) a=min(a,b)
#define Max(a,b) a=max(a,b)
#define Sz(s) int((s).size())
#define All(s) (s).begin(),(s).end()
#define Fill(s,v) memset(s,v,sizeof(s))
#define pb push_back
#define mp make_pair
#define x first
#define y second
template<typename T> T Abs(T x) { return(x < 0 ? -x : x); }
template<typename T> T Sqr(T x) { return(x * x); }
string plural(string s) { return(Sz(s) && s[Sz(s) - 1] == 'x' ? s + "en" : s + "s"); }
const int INF = (int)1e9;
const LD EPS = 1e-12;
const LD PI = acos(-1.0);
#define GETCHAR getchar_unlocked
bool Read(int& x)
{
char c, r = 0, n = 0;
x = 0;
for (;;)
{
c = GETCHAR();
if ((c < 0) && (!r))
return(0);
if ((c == '-') && (!r))
n = 1;
else
if ((c >= '0') && (c <= '9'))
x = x * 10 + c - '0', r = 1;
else
if (r)
break;
}
if (n)
x = -x;
return(1);
}
#define LIM 32
#define LIM2 500
#define MOD 1000000007
int N;
int D[LIM], C[LIM];
int ch[LIM2][LIM2];
int dyn[LIM][LIM2];
int Exp(int a, int b)
{
LL p = a, v = 1;
while (b)
{
if (b & 1)
v = v * p % MOD;
p = p * p % MOD;
b >>= 1;
}
return(v);
}
void Add(int& a, int b)
{
a = (a + b) % MOD;
}
int Prod(int a, int b)
{
return((LL)a * b % MOD);
}
int Div(int a, int b)
{
return Prod(a, Exp(b, MOD - 2));
}
int ProcessCase()
{
int i, j, k;
// input
Read(N);
Fox(i, N)
Read(D[i]);
// compute max prefix capacities
int p = 0, s = 0;
Fox(i, N + 1)
{
if (i == N || D[i] < D[p])
{
FoxI(j, p, i - 1)
s += D[j] - D[p];
FoxI(j, p + (p ? 1 : 0), i)
C[j] = s;
p = i;
}
}
// DP
Fill(dyn, 0);
dyn[0][0] = 1;
Fox(i, N)
{
Fox(j, (i ? C[i - 1] : 0) + 1)
{
Fox(k, C[i] - j + 1)
Add(dyn[i + 1][j + k], Prod(dyn[i][j], ch[j + k][k]));
}
}
// compute answer
int ans = 0;
Fox(i, s + 1)
Add(ans, Div(dyn[N][i], Exp(N, i)));
return(ans);
}
int main()
{
int T, t;
int i, j;
// precompute choose table
ch[0][0] = 1;
Fox1(i, LIM2 - 1)
{
Fox(j, i + 1)
{
ch[i][j] = ch[i - 1][j];
if (j)
Add(ch[i][j], ch[i - 1][j - 1]);
}
}
// testcase loop
Read(T);
Fox1(t, T)
printf("Case #%d: %d\n", t, ProcessCase());
return(0);
} |