message stringlengths 2 16.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 575 109k | cluster float64 16 16 | __index_level_0__ int64 1.15k 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N pieces of ropes, numbered 1 through N. The length of piece i is a_i.
At first, for each i (1≤i≤N-1), piece i and piece i+1 are tied at the ends, forming one long rope with N-1 knots. Snuke will try to untie all of the knots by performing the following operation repeatedly:
* Choose a (connected) rope with a total length of at least L, then untie one of its knots.
Is it possible to untie all of the N-1 knots by properly applying this operation? If the answer is positive, find one possible order to untie the knots.
Constraints
* 2≤N≤10^5
* 1≤L≤10^9
* 1≤a_i≤10^9
* All input values are integers.
Input
The input is given from Standard Input in the following format:
N L
a_1 a_2 ... a_n
Output
If it is not possible to untie all of the N-1 knots, print `Impossible`.
If it is possible to untie all of the knots, print `Possible`, then print another N-1 lines that describe a possible order to untie the knots. The j-th of those N-1 lines should contain the index of the knot that is untied in the j-th operation. Here, the index of the knot connecting piece i and piece i+1 is i.
If there is more than one solution, output any.
Examples
Input
3 50
30 20 10
Output
Possible
2
1
Input
2 21
10 10
Output
Impossible
Input
5 50
10 20 30 40 50
Output
Possible
1
2
3
4
Submitted Solution:
```
n,l = map(int,input().split())
a = list(map(int,input().split()))
memo = -1
for i in range(n-1):
if a[i]+a[i+1] >= l:
memo = i
if memo == -1:
print("Impossible")
else:
print("Possible")
for i in range(n-1):
if i != memo:
print(i+1)
print(memo+1)
``` | instruction | 0 | 70,650 | 16 | 141,300 |
No | output | 1 | 70,650 | 16 | 141,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N pieces of ropes, numbered 1 through N. The length of piece i is a_i.
At first, for each i (1≤i≤N-1), piece i and piece i+1 are tied at the ends, forming one long rope with N-1 knots. Snuke will try to untie all of the knots by performing the following operation repeatedly:
* Choose a (connected) rope with a total length of at least L, then untie one of its knots.
Is it possible to untie all of the N-1 knots by properly applying this operation? If the answer is positive, find one possible order to untie the knots.
Constraints
* 2≤N≤10^5
* 1≤L≤10^9
* 1≤a_i≤10^9
* All input values are integers.
Input
The input is given from Standard Input in the following format:
N L
a_1 a_2 ... a_n
Output
If it is not possible to untie all of the N-1 knots, print `Impossible`.
If it is possible to untie all of the knots, print `Possible`, then print another N-1 lines that describe a possible order to untie the knots. The j-th of those N-1 lines should contain the index of the knot that is untied in the j-th operation. Here, the index of the knot connecting piece i and piece i+1 is i.
If there is more than one solution, output any.
Examples
Input
3 50
30 20 10
Output
Possible
2
1
Input
2 21
10 10
Output
Impossible
Input
5 50
10 20 30 40 50
Output
Possible
1
2
3
4
Submitted Solution:
```
N,L,*A=map(int,open(0).read().split())
k=-1
for i in range(N-1):
if A[i]+A[i+1]>=L:
k=i
if k>=0:
print('Possible')
for i in range(k+1,N-1):
print(i+1)
for i in range(k+1):
print(i+1)
else:
print('Impossible')
``` | instruction | 0 | 70,651 | 16 | 141,302 |
No | output | 1 | 70,651 | 16 | 141,303 |
Provide a correct Python 3 solution for this coding contest problem.
We have N irregular jigsaw pieces. Each piece is composed of three rectangular parts of width 1 and various heights joined together. More specifically:
* The i-th piece is a part of height H, with another part of height A_i joined to the left, and yet another part of height B_i joined to the right, as shown below. Here, the bottom sides of the left and right parts are respectively at C_i and D_i units length above the bottom side of the center part.
<image>
Snuke is arranging these pieces on a square table of side 10^{100}. Here, the following conditions must be held:
* All pieces must be put on the table.
* The entire bottom side of the center part of each piece must touch the front side of the table.
* The entire bottom side of the non-center parts of each piece must either touch the front side of the table, or touch the top side of a part of some other piece.
* The pieces must not be rotated or flipped.
Determine whether such an arrangement is possible.
Constraints
* 1 \leq N \leq 100000
* 1 \leq H \leq 200
* 1 \leq A_i \leq H
* 1 \leq B_i \leq H
* 0 \leq C_i \leq H - A_i
* 0 \leq D_i \leq H - B_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N H
A_1 B_1 C_1 D_1
A_2 B_2 C_2 D_2
:
A_N B_N C_N D_N
Output
If it is possible to arrange the pieces under the conditions, print `YES`; if it is impossible, print `NO`.
Examples
Input
3 4
1 1 0 0
2 2 0 1
3 3 1 0
Output
YES
Input
4 2
1 1 0 1
1 1 0 1
1 1 0 1
1 1 0 1
Output
NO
Input
10 4
1 1 0 3
2 3 2 0
1 2 3 0
2 1 0 0
3 2 0 2
1 1 3 0
3 2 0 0
1 3 2 0
1 1 1 3
2 3 0 0
Output
YES | instruction | 0 | 71,487 | 16 | 142,974 |
"Correct Solution:
```
import sys
from collections import Counter
# sys.stdin = open('e1.in')
def read_int_list():
return list(map(int, input().split()))
def read_int():
return int(input())
def read_str_list():
return input().split()
def read_str():
return input()
total_in = 0
total_out = 0
def solve():
n, h = read_int_list()
a, b, c, d = zip(*[read_int_list() for _ in range(n)])
n_nodes = 2 * h + 1
n_edges = n
adj = [[] for _ in range(n_nodes)]
r_adj = [[] for _ in range(n_nodes)]
for i in range(n_edges):
if c[i] == 0:
l = a[i]
else:
l = -c[i]
if d[i] == 0:
r = -b[i]
else:
r = d[i]
adj[l + h].append(r + h)
r_adj[r + h].append(l + h)
in_degree = Counter()
out_degree = Counter()
for i in range(n_nodes):
for j in adj[i]:
out_degree[i] += 1
in_degree[j] += 1
for i in range(n_nodes):
if i < h:
if not out_degree[i] <= in_degree[i]:
return 'NO'
if h < i:
if not in_degree[i] <= out_degree[i]:
return 'NO'
state = [0] * n_nodes
def dfs(root):
res = False
state[root] = 1
for i in adj[root]:
if state[i] == 0:
res |= dfs(i)
for i in r_adj[root]:
if state[i] == 0:
res |= dfs(i)
state[root] = 2
if (in_degree[root], out_degree[root]) == (0,0):
return True
res |= in_degree[root] != out_degree[root]
return res
for i in range(n_nodes):
if state[i] == 0:
if not dfs(i):
return 'NO'
return 'YES'
def main():
res = solve()
print(res)
if __name__ == '__main__':
main()
``` | output | 1 | 71,487 | 16 | 142,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N irregular jigsaw pieces. Each piece is composed of three rectangular parts of width 1 and various heights joined together. More specifically:
* The i-th piece is a part of height H, with another part of height A_i joined to the left, and yet another part of height B_i joined to the right, as shown below. Here, the bottom sides of the left and right parts are respectively at C_i and D_i units length above the bottom side of the center part.
<image>
Snuke is arranging these pieces on a square table of side 10^{100}. Here, the following conditions must be held:
* All pieces must be put on the table.
* The entire bottom side of the center part of each piece must touch the front side of the table.
* The entire bottom side of the non-center parts of each piece must either touch the front side of the table, or touch the top side of a part of some other piece.
* The pieces must not be rotated or flipped.
Determine whether such an arrangement is possible.
Constraints
* 1 \leq N \leq 100000
* 1 \leq H \leq 200
* 1 \leq A_i \leq H
* 1 \leq B_i \leq H
* 0 \leq C_i \leq H - A_i
* 0 \leq D_i \leq H - B_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N H
A_1 B_1 C_1 D_1
A_2 B_2 C_2 D_2
:
A_N B_N C_N D_N
Output
If it is possible to arrange the pieces under the conditions, print `YES`; if it is impossible, print `NO`.
Examples
Input
3 4
1 1 0 0
2 2 0 1
3 3 1 0
Output
YES
Input
4 2
1 1 0 1
1 1 0 1
1 1 0 1
1 1 0 1
Output
NO
Input
10 4
1 1 0 3
2 3 2 0
1 2 3 0
2 1 0 0
3 2 0 2
1 1 3 0
3 2 0 0
1 3 2 0
1 1 1 3
2 3 0 0
Output
YES
Submitted Solution:
```
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
a=1
print("YES")
``` | instruction | 0 | 71,488 | 16 | 142,976 |
No | output | 1 | 71,488 | 16 | 142,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N irregular jigsaw pieces. Each piece is composed of three rectangular parts of width 1 and various heights joined together. More specifically:
* The i-th piece is a part of height H, with another part of height A_i joined to the left, and yet another part of height B_i joined to the right, as shown below. Here, the bottom sides of the left and right parts are respectively at C_i and D_i units length above the bottom side of the center part.
<image>
Snuke is arranging these pieces on a square table of side 10^{100}. Here, the following conditions must be held:
* All pieces must be put on the table.
* The entire bottom side of the center part of each piece must touch the front side of the table.
* The entire bottom side of the non-center parts of each piece must either touch the front side of the table, or touch the top side of a part of some other piece.
* The pieces must not be rotated or flipped.
Determine whether such an arrangement is possible.
Constraints
* 1 \leq N \leq 100000
* 1 \leq H \leq 200
* 1 \leq A_i \leq H
* 1 \leq B_i \leq H
* 0 \leq C_i \leq H - A_i
* 0 \leq D_i \leq H - B_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N H
A_1 B_1 C_1 D_1
A_2 B_2 C_2 D_2
:
A_N B_N C_N D_N
Output
If it is possible to arrange the pieces under the conditions, print `YES`; if it is impossible, print `NO`.
Examples
Input
3 4
1 1 0 0
2 2 0 1
3 3 1 0
Output
YES
Input
4 2
1 1 0 1
1 1 0 1
1 1 0 1
1 1 0 1
Output
NO
Input
10 4
1 1 0 3
2 3 2 0
1 2 3 0
2 1 0 0
3 2 0 2
1 1 3 0
3 2 0 0
1 3 2 0
1 1 1 3
2 3 0 0
Output
YES
Submitted Solution:
```
#include <bits/stdc++.h>
using namespace std;
int n, h;
struct node {
int a, b, c, d;
} q[300050];
int ld[233], lu[233], rd[233], ru[233], ds;
int ind[300050], outd[300050], fa[300050], pd[300050];
int getfa(int x) { return fa[x] == x ? x : fa[x] = getfa(fa[x]); }
int main() {
cin >> n >> h;
for (int i = 1; i <= n; ++i)
scanf("%d%d%d%d", &q[i].a, &q[i].b, &q[i].c, &q[i].d);
for (int i = 1; i <= h; ++i) ld[i] = ++ds, fa[ds] = ds;
for (int i = 1; i <= h; ++i) lu[i] = ++ds, fa[ds] = ds;
for (int i = 1; i <= h; ++i) rd[i] = lu[i];
for (int i = 1; i <= h; ++i) ru[i] = ld[i];
for (int i = 1; i <= n; ++i) {
int pol, por;
if (q[i].c)
pol = ld[q[i].c];
else
pol = lu[q[i].a];
if (q[i].d)
por = rd[q[i].d];
else
por = ru[q[i].b];
++outd[pol];
++ind[por];
fa[getfa(pol)] = getfa(por);
}
int al = 0;
for (int i = lu[1]; i <= lu[h]; ++i) {
al += outd[i] - ind[i];
if (outd[i] < ind[i]) return puts("NO"), 0;
}
if (al <= 0) return puts("NO"), 0;
for (int i = ru[1]; i <= ru[h]; ++i) {
if (outd[i] > ind[i]) return puts("NO"), 0;
al += outd[i] - ind[i];
}
if (al) return puts("NO"), 0;
for (int i = 1; i <= ds; ++i)
if (ind[i] != outd[i]) pd[getfa(i)] = 1;
for (int i = 1; i <= ds; ++i)
if ((ind[i] || outd[i]) && !pd[getfa(i)]) return puts("NO"), 0;
return puts("YES"), 0;
}
``` | instruction | 0 | 71,489 | 16 | 142,978 |
No | output | 1 | 71,489 | 16 | 142,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N irregular jigsaw pieces. Each piece is composed of three rectangular parts of width 1 and various heights joined together. More specifically:
* The i-th piece is a part of height H, with another part of height A_i joined to the left, and yet another part of height B_i joined to the right, as shown below. Here, the bottom sides of the left and right parts are respectively at C_i and D_i units length above the bottom side of the center part.
<image>
Snuke is arranging these pieces on a square table of side 10^{100}. Here, the following conditions must be held:
* All pieces must be put on the table.
* The entire bottom side of the center part of each piece must touch the front side of the table.
* The entire bottom side of the non-center parts of each piece must either touch the front side of the table, or touch the top side of a part of some other piece.
* The pieces must not be rotated or flipped.
Determine whether such an arrangement is possible.
Constraints
* 1 \leq N \leq 100000
* 1 \leq H \leq 200
* 1 \leq A_i \leq H
* 1 \leq B_i \leq H
* 0 \leq C_i \leq H - A_i
* 0 \leq D_i \leq H - B_i
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N H
A_1 B_1 C_1 D_1
A_2 B_2 C_2 D_2
:
A_N B_N C_N D_N
Output
If it is possible to arrange the pieces under the conditions, print `YES`; if it is impossible, print `NO`.
Examples
Input
3 4
1 1 0 0
2 2 0 1
3 3 1 0
Output
YES
Input
4 2
1 1 0 1
1 1 0 1
1 1 0 1
1 1 0 1
Output
NO
Input
10 4
1 1 0 3
2 3 2 0
1 2 3 0
2 1 0 0
3 2 0 2
1 1 3 0
3 2 0 0
1 3 2 0
1 1 1 3
2 3 0 0
Output
YES
Submitted Solution:
```
int aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
print("YES")
``` | instruction | 0 | 71,490 | 16 | 142,980 |
No | output | 1 | 71,490 | 16 | 142,981 |
Provide a correct Python 3 solution for this coding contest problem.
We have N camels numbered 1,2,\ldots,N. Snuke has decided to make them line up in a row.
The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise.
Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel.
Solve this problem for each of the T test cases given.
Constraints
* All values in input are integers.
* 1 \leq T \leq 10^5
* 1 \leq N \leq 2 \times 10^{5}
* 1 \leq K_i \leq N
* 1 \leq L_i, R_i \leq 10^9
* The sum of values of N in each input file is at most 2 \times 10^5.
Input
Input is given from Standard Input in the following format:
T
\mathrm{case}_1
\vdots
\mathrm{case}_T
Each case is given in the following format:
N
K_1 L_1 R_1
\vdots
K_N L_N R_N
Output
Print T lines. The i-th line should contain the answer to the i-th test case.
Example
Input
3
2
1 5 10
2 15 5
3
2 93 78
1 71 59
3 57 96
19
19 23 16
5 90 13
12 85 70
19 67 78
12 16 60
18 48 28
5 4 24
12 97 97
4 57 87
19 91 74
18 100 76
7 86 46
9 100 57
3 76 73
6 84 93
1 6 84
11 75 94
19 15 3
12 11 34
Output
25
221
1354 | instruction | 0 | 73,070 | 16 | 146,140 |
"Correct Solution:
```
def main():
from collections import defaultdict
from heapq import heappop, heappush
def greedy(L):
ret = 0
q = []
for k, gain in L:
for g in gain:
ret += g
heappush(q, g)
while len(q) > k:
g = heappop(q)
ret -= g
return ret
for _ in [0]*int(input()):
N = int(input())
ans = 0
pos = defaultdict(list)
neg = defaultdict(list)
for _ in range(N):
K, L, R = map(int, input().split())
if L > R:
ans += R
pos[K].append(L-R)
elif L < R:
ans += L
neg[N-K].append(R-L)
else:
ans += L
pos = sorted(pos.items())
neg = sorted(neg.items())
ans += greedy(pos)
ans += greedy(neg)
print(ans)
if __name__ == '__main__':
main()
``` | output | 1 | 73,070 | 16 | 146,141 |
Provide a correct Python 3 solution for this coding contest problem.
We have N camels numbered 1,2,\ldots,N. Snuke has decided to make them line up in a row.
The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise.
Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel.
Solve this problem for each of the T test cases given.
Constraints
* All values in input are integers.
* 1 \leq T \leq 10^5
* 1 \leq N \leq 2 \times 10^{5}
* 1 \leq K_i \leq N
* 1 \leq L_i, R_i \leq 10^9
* The sum of values of N in each input file is at most 2 \times 10^5.
Input
Input is given from Standard Input in the following format:
T
\mathrm{case}_1
\vdots
\mathrm{case}_T
Each case is given in the following format:
N
K_1 L_1 R_1
\vdots
K_N L_N R_N
Output
Print T lines. The i-th line should contain the answer to the i-th test case.
Example
Input
3
2
1 5 10
2 15 5
3
2 93 78
1 71 59
3 57 96
19
19 23 16
5 90 13
12 85 70
19 67 78
12 16 60
18 48 28
5 4 24
12 97 97
4 57 87
19 91 74
18 100 76
7 86 46
9 100 57
3 76 73
6 84 93
1 6 84
11 75 94
19 15 3
12 11 34
Output
25
221
1354 | instruction | 0 | 73,071 | 16 | 146,142 |
"Correct Solution:
```
from collections import deque
from heapq import heappush, heappushpop
import sys
input = sys.stdin.readline
def calc(camels):
camels = deque(sorted(camels, key=lambda x: x[0]))
N = len(camels)
heap = []
while camels and camels[0][0] == 0:
camels.popleft()
for i in range(1, N+1):
while camels and camels[0][0] == i:
_, x = camels.popleft()
if len(heap) < i:
heappush(heap, x)
elif heap[0] < x:
heappushpop(heap, x)
for _, x in camels:
if len(heap) < N:
heappush(heap, x)
elif heap[0] < x:
heappushpop(heap, x)
return sum(heap)
T = int(input())
for _ in range(T):
N = int(input())
s = 0
ans = 0
first = []
second = []
for i in range(N):
K, L, R = map(int, input().split())
if L >= R:
first.append((K, L - R))
ans += R
else:
second.append((N - K, R - L))
ans += L
ans += calc(first) + calc(second)
print(ans)
``` | output | 1 | 73,071 | 16 | 146,143 |
Provide a correct Python 3 solution for this coding contest problem.
We have N camels numbered 1,2,\ldots,N. Snuke has decided to make them line up in a row.
The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise.
Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel.
Solve this problem for each of the T test cases given.
Constraints
* All values in input are integers.
* 1 \leq T \leq 10^5
* 1 \leq N \leq 2 \times 10^{5}
* 1 \leq K_i \leq N
* 1 \leq L_i, R_i \leq 10^9
* The sum of values of N in each input file is at most 2 \times 10^5.
Input
Input is given from Standard Input in the following format:
T
\mathrm{case}_1
\vdots
\mathrm{case}_T
Each case is given in the following format:
N
K_1 L_1 R_1
\vdots
K_N L_N R_N
Output
Print T lines. The i-th line should contain the answer to the i-th test case.
Example
Input
3
2
1 5 10
2 15 5
3
2 93 78
1 71 59
3 57 96
19
19 23 16
5 90 13
12 85 70
19 67 78
12 16 60
18 48 28
5 4 24
12 97 97
4 57 87
19 91 74
18 100 76
7 86 46
9 100 57
3 76 73
6 84 93
1 6 84
11 75 94
19 15 3
12 11 34
Output
25
221
1354 | instruction | 0 | 73,072 | 16 | 146,144 |
"Correct Solution:
```
import sys
from heapq import heappop, heappush
def main():
def input(): return sys.stdin.readline().rstrip()
n = int(input())
l_camel = [[] for _ in range(n)]
r_camel = [[] for _ in range(n)]
ans = 0
for _ in range(n):
k, l, r = map(int, input().split())
if l >= r:
ans += r
l_camel[k-1].append(l-r)
else:
ans += l
r_camel[~(k-1)].append(r-l)
cur = 0
que = []
for i in range(n):
for el in l_camel[i]:
heappush(que, el)
while len(que) > i+1:
heappop(que)
ans += sum(que)
que = []
for i in range(n):
for el in r_camel[i]:
heappush(que, el)
while len(que) > i:
heappop(que)
ans += sum(que)
print(ans)
if __name__ == '__main__':
t = int(input())
for _ in range(t):
main()
``` | output | 1 | 73,072 | 16 | 146,145 |
Provide a correct Python 3 solution for this coding contest problem.
We have N camels numbered 1,2,\ldots,N. Snuke has decided to make them line up in a row.
The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise.
Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel.
Solve this problem for each of the T test cases given.
Constraints
* All values in input are integers.
* 1 \leq T \leq 10^5
* 1 \leq N \leq 2 \times 10^{5}
* 1 \leq K_i \leq N
* 1 \leq L_i, R_i \leq 10^9
* The sum of values of N in each input file is at most 2 \times 10^5.
Input
Input is given from Standard Input in the following format:
T
\mathrm{case}_1
\vdots
\mathrm{case}_T
Each case is given in the following format:
N
K_1 L_1 R_1
\vdots
K_N L_N R_N
Output
Print T lines. The i-th line should contain the answer to the i-th test case.
Example
Input
3
2
1 5 10
2 15 5
3
2 93 78
1 71 59
3 57 96
19
19 23 16
5 90 13
12 85 70
19 67 78
12 16 60
18 48 28
5 4 24
12 97 97
4 57 87
19 91 74
18 100 76
7 86 46
9 100 57
3 76 73
6 84 93
1 6 84
11 75 94
19 15 3
12 11 34
Output
25
221
1354 | instruction | 0 | 73,073 | 16 | 146,146 |
"Correct Solution:
```
import heapq
def solve():
N = int(input())
klr = [list(map(int,input().split())) for _ in range(N)]
ll = []
lr = []
for i in range(N):
if klr[i][1]>klr[i][2]:
ll.append(klr[i])
else:
lr.append(klr[i])
ansl = 0
ll.sort(key=lambda x:-x[0])
j = 0
cand = []
for i in range(len(ll)-1,-1,-1):
while j<len(ll):
if ll[j][0]-1<i:
break
heapq.heappush(cand,[ll[j][2]-ll[j][1],ll[j][0],ll[j][1],ll[j][2]])
j += 1
if cand:
d,k,l,r = heapq.heappop(cand)
ansl += l
while cand:
d,k,l,r = heapq.heappop(cand)
ansl += r
ansr = 0
lr.sort(key=lambda x:x[0])
j = 0
cand = []
for i in range(len(ll),N+1):
while j<len(lr):
if lr[j][0]>i:
break
heapq.heappush(cand,[-lr[j][2]+lr[j][1],lr[j][0],lr[j][1],lr[j][2]])
j += 1
if cand and i!=N:
d,k,l,r = heapq.heappop(cand)
ansr += r
while cand:
d,k,l,r = heapq.heappop(cand)
ansr += l
ans = ansl+ansr
print(ans)
T = int(input())
for t in range(T):
solve()
``` | output | 1 | 73,073 | 16 | 146,147 |
Provide a correct Python 3 solution for this coding contest problem.
We have N camels numbered 1,2,\ldots,N. Snuke has decided to make them line up in a row.
The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise.
Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel.
Solve this problem for each of the T test cases given.
Constraints
* All values in input are integers.
* 1 \leq T \leq 10^5
* 1 \leq N \leq 2 \times 10^{5}
* 1 \leq K_i \leq N
* 1 \leq L_i, R_i \leq 10^9
* The sum of values of N in each input file is at most 2 \times 10^5.
Input
Input is given from Standard Input in the following format:
T
\mathrm{case}_1
\vdots
\mathrm{case}_T
Each case is given in the following format:
N
K_1 L_1 R_1
\vdots
K_N L_N R_N
Output
Print T lines. The i-th line should contain the answer to the i-th test case.
Example
Input
3
2
1 5 10
2 15 5
3
2 93 78
1 71 59
3 57 96
19
19 23 16
5 90 13
12 85 70
19 67 78
12 16 60
18 48 28
5 4 24
12 97 97
4 57 87
19 91 74
18 100 76
7 86 46
9 100 57
3 76 73
6 84 93
1 6 84
11 75 94
19 15 3
12 11 34
Output
25
221
1354 | instruction | 0 | 73,074 | 16 | 146,148 |
"Correct Solution:
```
import heapq
T = int(input())
for _ in range(T):
N = int(input())
KLR = [list(map(int, input().split())) for i in range(N)]
ans = 0
for i in range(N):
ans += min(KLR[i][1], KLR[i][2])
KX = [(K, L-R) for K, L, R in KLR if L-R >= 0]
KX.sort()
hq = []
id = 0
for i in range(1, N+1):
while(id < len(KX) and KX[id][0] == i):
heapq.heappush(hq, KX[id][1])
id += 1
while(len(hq) > i):
heapq.heappop(hq)
# print(hq)
ans += sum(hq)
L = len(KX)
KX = [(N-K, R-L) for K, L, R in KLR if R-L > 0]
KX.sort()
hq = []
id = 0
for i in range(N):
while(id < len(KX) and KX[id][0] == i):
heapq.heappush(hq, KX[id][1])
id += 1
while(len(hq) > i):
heapq.heappop(hq)
# print(hq)
ans += sum(hq)
print(ans)
``` | output | 1 | 73,074 | 16 | 146,149 |
Provide a correct Python 3 solution for this coding contest problem.
We have N camels numbered 1,2,\ldots,N. Snuke has decided to make them line up in a row.
The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise.
Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel.
Solve this problem for each of the T test cases given.
Constraints
* All values in input are integers.
* 1 \leq T \leq 10^5
* 1 \leq N \leq 2 \times 10^{5}
* 1 \leq K_i \leq N
* 1 \leq L_i, R_i \leq 10^9
* The sum of values of N in each input file is at most 2 \times 10^5.
Input
Input is given from Standard Input in the following format:
T
\mathrm{case}_1
\vdots
\mathrm{case}_T
Each case is given in the following format:
N
K_1 L_1 R_1
\vdots
K_N L_N R_N
Output
Print T lines. The i-th line should contain the answer to the i-th test case.
Example
Input
3
2
1 5 10
2 15 5
3
2 93 78
1 71 59
3 57 96
19
19 23 16
5 90 13
12 85 70
19 67 78
12 16 60
18 48 28
5 4 24
12 97 97
4 57 87
19 91 74
18 100 76
7 86 46
9 100 57
3 76 73
6 84 93
1 6 84
11 75 94
19 15 3
12 11 34
Output
25
221
1354 | instruction | 0 | 73,075 | 16 | 146,150 |
"Correct Solution:
```
#!python3
import sys
iim = lambda: map(int, sys.stdin.readline().rstrip().split())
from heapq import heappush, heappushpop
def resolve():
it = map(int, sys.stdin.read().split())
T = next(it)
ans = []
for i in range(T):
N = next(it)
val = 0
a1 = [[] for i in range(N)];
a2 = [[] for i in range(N)];
for i, v1, v2 in ((next(it)-1, next(it), next(it)) for i in range(N)):
diff = v1 - v2
if diff == 0 or i == N-1:
val += v1
elif diff > 0:
a1[i].append(diff)
val += v2
else:
i = N-1-i-1
a2[i].append(-diff)
val += v1
for ax in (a1, a2):
dq = []
for i, aa in enumerate(ax):
if not aa: continue
for diff in aa:
ld = len(dq)
if ld <= i:
heappush(dq, diff)
else:
diff = heappushpop(dq, diff)
val += sum(dq)
ans.append(val)
print(*ans, sep="\n")
if __name__ == "__main__":
resolve()
``` | output | 1 | 73,075 | 16 | 146,151 |
Provide a correct Python 3 solution for this coding contest problem.
We have N camels numbered 1,2,\ldots,N. Snuke has decided to make them line up in a row.
The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise.
Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel.
Solve this problem for each of the T test cases given.
Constraints
* All values in input are integers.
* 1 \leq T \leq 10^5
* 1 \leq N \leq 2 \times 10^{5}
* 1 \leq K_i \leq N
* 1 \leq L_i, R_i \leq 10^9
* The sum of values of N in each input file is at most 2 \times 10^5.
Input
Input is given from Standard Input in the following format:
T
\mathrm{case}_1
\vdots
\mathrm{case}_T
Each case is given in the following format:
N
K_1 L_1 R_1
\vdots
K_N L_N R_N
Output
Print T lines. The i-th line should contain the answer to the i-th test case.
Example
Input
3
2
1 5 10
2 15 5
3
2 93 78
1 71 59
3 57 96
19
19 23 16
5 90 13
12 85 70
19 67 78
12 16 60
18 48 28
5 4 24
12 97 97
4 57 87
19 91 74
18 100 76
7 86 46
9 100 57
3 76 73
6 84 93
1 6 84
11 75 94
19 15 3
12 11 34
Output
25
221
1354 | instruction | 0 | 73,076 | 16 | 146,152 |
"Correct Solution:
```
from heapq import*
i=input
for s in[0]*int(i()):
n,x,*y=int(i()),[]
for _ in'_'*n:k,l,r=t=[*map(int,i().split())];x+=[t]*(l>r);y+=[[n-k,r,l]]*(l<=r)
for x in x,y:h=[];s+=sum(heappush(h,l-r)or l-(k<len(h)and heappop(h))for k,l,r in sorted(x))
print(s)
``` | output | 1 | 73,076 | 16 | 146,153 |
Provide a correct Python 3 solution for this coding contest problem.
We have N camels numbered 1,2,\ldots,N. Snuke has decided to make them line up in a row.
The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise.
Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel.
Solve this problem for each of the T test cases given.
Constraints
* All values in input are integers.
* 1 \leq T \leq 10^5
* 1 \leq N \leq 2 \times 10^{5}
* 1 \leq K_i \leq N
* 1 \leq L_i, R_i \leq 10^9
* The sum of values of N in each input file is at most 2 \times 10^5.
Input
Input is given from Standard Input in the following format:
T
\mathrm{case}_1
\vdots
\mathrm{case}_T
Each case is given in the following format:
N
K_1 L_1 R_1
\vdots
K_N L_N R_N
Output
Print T lines. The i-th line should contain the answer to the i-th test case.
Example
Input
3
2
1 5 10
2 15 5
3
2 93 78
1 71 59
3 57 96
19
19 23 16
5 90 13
12 85 70
19 67 78
12 16 60
18 48 28
5 4 24
12 97 97
4 57 87
19 91 74
18 100 76
7 86 46
9 100 57
3 76 73
6 84 93
1 6 84
11 75 94
19 15 3
12 11 34
Output
25
221
1354 | instruction | 0 | 73,077 | 16 | 146,154 |
"Correct Solution:
```
from heapq import*
i=input
for s in[0]*int(i()):
n,x,*y=int(i()),[]
for _ in'_'*n:k,l,r=t=[*map(int,i().split())];x+=[t]*(l>r);y+=[[n-k,r,l]]*(l<=r)
for x in x,y:
x.sort();n,*h=len(x),
while n:
while[[n]]<x[-1:]:k,l,r=x.pop();heappush(h,(r-l,l,r))
if h:s+=heappop(h)[1]
n-=1
for*_,r in x+h:s+=r
print(s)
``` | output | 1 | 73,077 | 16 | 146,155 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N camels numbered 1,2,\ldots,N. Snuke has decided to make them line up in a row.
The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise.
Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel.
Solve this problem for each of the T test cases given.
Constraints
* All values in input are integers.
* 1 \leq T \leq 10^5
* 1 \leq N \leq 2 \times 10^{5}
* 1 \leq K_i \leq N
* 1 \leq L_i, R_i \leq 10^9
* The sum of values of N in each input file is at most 2 \times 10^5.
Input
Input is given from Standard Input in the following format:
T
\mathrm{case}_1
\vdots
\mathrm{case}_T
Each case is given in the following format:
N
K_1 L_1 R_1
\vdots
K_N L_N R_N
Output
Print T lines. The i-th line should contain the answer to the i-th test case.
Example
Input
3
2
1 5 10
2 15 5
3
2 93 78
1 71 59
3 57 96
19
19 23 16
5 90 13
12 85 70
19 67 78
12 16 60
18 48 28
5 4 24
12 97 97
4 57 87
19 91 74
18 100 76
7 86 46
9 100 57
3 76 73
6 84 93
1 6 84
11 75 94
19 15 3
12 11 34
Output
25
221
1354
Submitted Solution:
```
#!/usr/bin/env python3
import sys
from heapq import heappush, heappop, heapify
from collections import deque
from collections import namedtuple
sys.setrecursionlimit(10**8)
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
def ZIP(n): return [LIST() for _ in range(n)]
def NSTR(n): return [input() for _ in range(n)]
INF = float("inf")
MOD = 10**9 + 7
def main():
T = INT()
for t in range(T):
solve()
return
def solve():
# 解説PDFの指針
N = INT()
left_camel = []
right_camel = []
camel = namedtuple("camel", "value, kth")
ans = 0
for k, l, r in ZIP(N):
if l >= r:
left_camel.append(camel(l-r, k))
ans += r
else:
right_camel.append(camel(r-l, N-k))
ans += l
left_camel.sort(key=lambda x: x.kth)
right_camel.sort(key=lambda x: x.kth)
left_camel = deque(left_camel)
right_camel = deque(right_camel)
for camels in [left_camel, right_camel]:
S = []
while camels and camels[0].kth == 0:
camels.popleft()
for j in range(1, N+1):
while camels and camels[0].kth == j:
heappush(S, camels.popleft())
while len(S) > j:
heappop(S)
while S:
ans += heappop(S).value
print(ans)
if __name__ == '__main__':
main()
``` | instruction | 0 | 73,078 | 16 | 146,156 |
Yes | output | 1 | 73,078 | 16 | 146,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N camels numbered 1,2,\ldots,N. Snuke has decided to make them line up in a row.
The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise.
Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel.
Solve this problem for each of the T test cases given.
Constraints
* All values in input are integers.
* 1 \leq T \leq 10^5
* 1 \leq N \leq 2 \times 10^{5}
* 1 \leq K_i \leq N
* 1 \leq L_i, R_i \leq 10^9
* The sum of values of N in each input file is at most 2 \times 10^5.
Input
Input is given from Standard Input in the following format:
T
\mathrm{case}_1
\vdots
\mathrm{case}_T
Each case is given in the following format:
N
K_1 L_1 R_1
\vdots
K_N L_N R_N
Output
Print T lines. The i-th line should contain the answer to the i-th test case.
Example
Input
3
2
1 5 10
2 15 5
3
2 93 78
1 71 59
3 57 96
19
19 23 16
5 90 13
12 85 70
19 67 78
12 16 60
18 48 28
5 4 24
12 97 97
4 57 87
19 91 74
18 100 76
7 86 46
9 100 57
3 76 73
6 84 93
1 6 84
11 75 94
19 15 3
12 11 34
Output
25
221
1354
Submitted Solution:
```
#!/usr/bin/env python3
import sys
from heapq import heappush, heappop, heapify
from collections import deque
sys.setrecursionlimit(10**8)
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
def ZIP(n): return [LIST() for _ in range(n)]
def NSTR(n): return [input() for _ in range(n)]
INF = float("inf")
MOD = 10**9 + 7
def main():
T = INT()
for t in range(T):
solve()
return
def solve():
# 解説PDFの指針
N = INT()
left_camel = []
right_camel = []
ans = 0
for k, l, r in ZIP(N):
if l >= r:
left_camel.append((l-r, k))
ans += r
else:
right_camel.append((r-l, N-k))
ans += l
left_camel.sort(key=lambda x: x[1])
right_camel.sort(key=lambda x: x[1])
left_camel = deque(left_camel)
right_camel = deque(right_camel)
# print(left_camel, right_camel)
for camels in [left_camel, right_camel]:
S = []
while len(camels) > 0 and camels[0][1] == 0:
camels.popleft()
for j in range(1, N+1):
while len(camels) > 0 and camels[0][1] == j:
heappush(S, camels.popleft())
while len(S) > j:
heappop(S)
while S:
ans += heappop(S)[0]
print(ans)
if __name__ == '__main__':
main()
``` | instruction | 0 | 73,079 | 16 | 146,158 |
Yes | output | 1 | 73,079 | 16 | 146,159 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N camels numbered 1,2,\ldots,N. Snuke has decided to make them line up in a row.
The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise.
Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel.
Solve this problem for each of the T test cases given.
Constraints
* All values in input are integers.
* 1 \leq T \leq 10^5
* 1 \leq N \leq 2 \times 10^{5}
* 1 \leq K_i \leq N
* 1 \leq L_i, R_i \leq 10^9
* The sum of values of N in each input file is at most 2 \times 10^5.
Input
Input is given from Standard Input in the following format:
T
\mathrm{case}_1
\vdots
\mathrm{case}_T
Each case is given in the following format:
N
K_1 L_1 R_1
\vdots
K_N L_N R_N
Output
Print T lines. The i-th line should contain the answer to the i-th test case.
Example
Input
3
2
1 5 10
2 15 5
3
2 93 78
1 71 59
3 57 96
19
19 23 16
5 90 13
12 85 70
19 67 78
12 16 60
18 48 28
5 4 24
12 97 97
4 57 87
19 91 74
18 100 76
7 86 46
9 100 57
3 76 73
6 84 93
1 6 84
11 75 94
19 15 3
12 11 34
Output
25
221
1354
Submitted Solution:
```
import heapq
t = int(input())
while t:
t-=1
n = int(input())
klr = [list(map(int, input().split())) for i in range(n)]
klr.sort()
q = []
ans = 0
for i in range(n):
K = klr[i][0]
L = klr[i][1]
R = klr[i][2]
if L >= R:
ans += L
heapq.heappush(q,L-R)
while len(q) > K:
dec = heapq.heappop(q)
ans -= dec
q = []
for i in reversed(range(n)):
K = klr[i][0]
L = klr[i][1]
R = klr[i][2]
if L < R:
ans += R
heapq.heappush(q,R-L)
while len(q) > n-K:
dec = heapq.heappop(q)
ans -= dec
print(ans)
``` | instruction | 0 | 73,080 | 16 | 146,160 |
Yes | output | 1 | 73,080 | 16 | 146,161 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N camels numbered 1,2,\ldots,N. Snuke has decided to make them line up in a row.
The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise.
Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel.
Solve this problem for each of the T test cases given.
Constraints
* All values in input are integers.
* 1 \leq T \leq 10^5
* 1 \leq N \leq 2 \times 10^{5}
* 1 \leq K_i \leq N
* 1 \leq L_i, R_i \leq 10^9
* The sum of values of N in each input file is at most 2 \times 10^5.
Input
Input is given from Standard Input in the following format:
T
\mathrm{case}_1
\vdots
\mathrm{case}_T
Each case is given in the following format:
N
K_1 L_1 R_1
\vdots
K_N L_N R_N
Output
Print T lines. The i-th line should contain the answer to the i-th test case.
Example
Input
3
2
1 5 10
2 15 5
3
2 93 78
1 71 59
3 57 96
19
19 23 16
5 90 13
12 85 70
19 67 78
12 16 60
18 48 28
5 4 24
12 97 97
4 57 87
19 91 74
18 100 76
7 86 46
9 100 57
3 76 73
6 84 93
1 6 84
11 75 94
19 15 3
12 11 34
Output
25
221
1354
Submitted Solution:
```
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor, cos, radians, pi, sin
from operator import mul
from functools import reduce
from operator import mul
sys.setrecursionlimit(2147483647)
INF = 10 ** 13
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def I(): return int(sys.stdin.buffer.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 10 ** 9 + 7
t = I()
for _ in range(t):
n = I()
ans = 0
q1 = []
q2 = []
L1 = []
L2 = []
for _ in range(n):
k, l, r = LI()
if l > r:
L1 += [(k, l - r)]
else:
L2 += [(n - k, r - l)]
ans += min(l, r)
L1.sort()
L2.sort()
for ki, d in L1:
heappush(q1, d)
if len(q1) > ki:
heappop(q1)
for ki, d in L2:
heappush(q2, d)
if len(q2) > ki:
heappop(q2)
ans += sum(q1) + sum(q2)
print(ans)
``` | instruction | 0 | 73,081 | 16 | 146,162 |
Yes | output | 1 | 73,081 | 16 | 146,163 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N camels numbered 1,2,\ldots,N. Snuke has decided to make them line up in a row.
The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise.
Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel.
Solve this problem for each of the T test cases given.
Constraints
* All values in input are integers.
* 1 \leq T \leq 10^5
* 1 \leq N \leq 2 \times 10^{5}
* 1 \leq K_i \leq N
* 1 \leq L_i, R_i \leq 10^9
* The sum of values of N in each input file is at most 2 \times 10^5.
Input
Input is given from Standard Input in the following format:
T
\mathrm{case}_1
\vdots
\mathrm{case}_T
Each case is given in the following format:
N
K_1 L_1 R_1
\vdots
K_N L_N R_N
Output
Print T lines. The i-th line should contain the answer to the i-th test case.
Example
Input
3
2
1 5 10
2 15 5
3
2 93 78
1 71 59
3 57 96
19
19 23 16
5 90 13
12 85 70
19 67 78
12 16 60
18 48 28
5 4 24
12 97 97
4 57 87
19 91 74
18 100 76
7 86 46
9 100 57
3 76 73
6 84 93
1 6 84
11 75 94
19 15 3
12 11 34
Output
25
221
1354
Submitted Solution:
```
t=int(input())
import heapq
for _ in range(t):
n=int(input())
L=[];R=[]
ans=0
for _ in range(n):
ki,li,ri = map(int, input().split())
dif=li-ri
if dif>=0:
L.append((ki-1,dif))
ans+=ri
else:
R.append((n-ki-1,-dif))
ans+=li
L.sort();R.sort()
hL=[]
hR=[]
cnt=0
for i in range(n-1,-1,-1):
if L:
while L and L[-1][0]==i:
heapq.heappush(hL,-L[-1][1])
L.pop()
while hL and cnt<=n-i:
ad=heapq.heappop(hL)
ans-=ad
cnt+=1
else:
while hL and cnt<=n-i:
ad=heapq.heappop(hL)
ans-=ad
cnt+=1
cnt=0
for i in range(n-1,-1,-1):
if R:
while R and R[-1][0]==i:
heapq.heappush(hR,-R[-1][1])
R.pop()
while hR and cnt<=n-i:
ad=heapq.heappop(hR)
ans-=ad
cnt+=1
else:
while hR and cnt<=n-i:
ad=heapq.heappop(hR)
ans-=ad
cnt+=1
print(ans)
``` | instruction | 0 | 73,082 | 16 | 146,164 |
No | output | 1 | 73,082 | 16 | 146,165 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N camels numbered 1,2,\ldots,N. Snuke has decided to make them line up in a row.
The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise.
Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel.
Solve this problem for each of the T test cases given.
Constraints
* All values in input are integers.
* 1 \leq T \leq 10^5
* 1 \leq N \leq 2 \times 10^{5}
* 1 \leq K_i \leq N
* 1 \leq L_i, R_i \leq 10^9
* The sum of values of N in each input file is at most 2 \times 10^5.
Input
Input is given from Standard Input in the following format:
T
\mathrm{case}_1
\vdots
\mathrm{case}_T
Each case is given in the following format:
N
K_1 L_1 R_1
\vdots
K_N L_N R_N
Output
Print T lines. The i-th line should contain the answer to the i-th test case.
Example
Input
3
2
1 5 10
2 15 5
3
2 93 78
1 71 59
3 57 96
19
19 23 16
5 90 13
12 85 70
19 67 78
12 16 60
18 48 28
5 4 24
12 97 97
4 57 87
19 91 74
18 100 76
7 86 46
9 100 57
3 76 73
6 84 93
1 6 84
11 75 94
19 15 3
12 11 34
Output
25
221
1354
Submitted Solution:
```
import heapq
T = int(input())
Ns = []
cases = []*T
for i in range(T):
Ns.append(int(input()))
cases.append([list(map(int, input().split())) for _ in range(Ns[-1])])
for i in [1]:
answer = 0
N = Ns[i]
case = cases[i]
sa = [case[j][2]-case[j][1] for j in range(N)]
pluss = []
minuss = []
for j in range(N):
if sa[j] >= 0:
pluss.append([case[j][0],case[j][1],case[j][2],sa[j]])
else:
minuss.append([case[j][0],case[j][1],case[j][2],sa[j]])
pluss = sorted(pluss)
minuss = sorted(minuss,reverse = True)
hq = []
checking = 0
for j in range(N-len(pluss),N):
if checking < len(pluss):
while j > pluss[checking][0]-1:
heapq.heappush(hq, [-pluss[checking][3],pluss[checking][1],pluss[checking][2],pluss[checking][0]])
checking+=1
if checking >= len(pluss):
break
if len(hq)>0:
temp = heapq.heappop(hq)
answer += temp[2]
for j in range(checking,len(pluss)):
heapq.heappush(hq, [-pluss[j][3],pluss[j][1],pluss[j][2],pluss[j][0]])
for j in range(len(hq)):
temp = heapq.heappop(hq)
answer += temp[1]
hq = []
checking = 0
for j in range(len(minuss)-1,-1,-1):
if checking < len(minuss):
while j <= minuss[checking][0]-1:
heapq.heappush(hq, [minuss[checking][3],minuss[checking][1],minuss[checking][2],minuss[checking][0]])
checking+=1
if checking >= len(minuss):
break
if len(hq)>0:
temp = heapq.heappop(hq)
answer += temp[1]
for j in range(checking, len(minuss)):
heapq.heappush(hq, [minuss[j][3],minuss[j][1],minuss[j][2],minuss[j][0]])
for j in range(len(hq)):
temp = heapq.heappop(hq)
answer += temp[2]
print(answer)
``` | instruction | 0 | 73,083 | 16 | 146,166 |
No | output | 1 | 73,083 | 16 | 146,167 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N camels numbered 1,2,\ldots,N. Snuke has decided to make them line up in a row.
The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise.
Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel.
Solve this problem for each of the T test cases given.
Constraints
* All values in input are integers.
* 1 \leq T \leq 10^5
* 1 \leq N \leq 2 \times 10^{5}
* 1 \leq K_i \leq N
* 1 \leq L_i, R_i \leq 10^9
* The sum of values of N in each input file is at most 2 \times 10^5.
Input
Input is given from Standard Input in the following format:
T
\mathrm{case}_1
\vdots
\mathrm{case}_T
Each case is given in the following format:
N
K_1 L_1 R_1
\vdots
K_N L_N R_N
Output
Print T lines. The i-th line should contain the answer to the i-th test case.
Example
Input
3
2
1 5 10
2 15 5
3
2 93 78
1 71 59
3 57 96
19
19 23 16
5 90 13
12 85 70
19 67 78
12 16 60
18 48 28
5 4 24
12 97 97
4 57 87
19 91 74
18 100 76
7 86 46
9 100 57
3 76 73
6 84 93
1 6 84
11 75 94
19 15 3
12 11 34
Output
25
221
1354
Submitted Solution:
```
from heapq import heapify, heappop, heappush
Ncase = int(input())
for case in range(Ncase):
N = int(input())
ans = 0
Left = [];Right = []
for i in range(N):
K,L,R = map(int,input().split())
if L > R:
Left.append((K-1,L-R))
ans += R
elif R > L:
if K != N: #K=Nの時はどこにおいても左と判定される。
Right.append((N-1-K,R-L))
ans += L
else:
ans += L
Left.sort(reverse=True);Right.sort(reverse=True) #Kの小さい順、左にいたときのプラス分が大きい順に並んだ
#print(Left,Right)
now = 0
PQ = []
for loc in reversed(range(N)):
#print(now,Left[now][0],loc)
while now < len(Left) and Left[now][0] == loc:
heappush(PQ,Left[now][1])
now += 1
if now >= len(Left):
break
if PQ:
plus = heappop(PQ)
ans += plus
now = 0
PQ = []
for loc in reversed(range(N)):
while now < len(Right) and Right[now][0] == loc:
heappush(PQ,Right[now][1])
now += 1
if now >= len(Right):
break
if PQ:
plus = heappop(PQ)
ans += plus
print(ans)
``` | instruction | 0 | 73,084 | 16 | 146,168 |
No | output | 1 | 73,084 | 16 | 146,169 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N camels numbered 1,2,\ldots,N. Snuke has decided to make them line up in a row.
The happiness of Camel i will be L_i if it is among the K_i frontmost camels, and R_i otherwise.
Snuke wants to maximize the total happiness of the camels. Find the maximum possible total happiness of the camel.
Solve this problem for each of the T test cases given.
Constraints
* All values in input are integers.
* 1 \leq T \leq 10^5
* 1 \leq N \leq 2 \times 10^{5}
* 1 \leq K_i \leq N
* 1 \leq L_i, R_i \leq 10^9
* The sum of values of N in each input file is at most 2 \times 10^5.
Input
Input is given from Standard Input in the following format:
T
\mathrm{case}_1
\vdots
\mathrm{case}_T
Each case is given in the following format:
N
K_1 L_1 R_1
\vdots
K_N L_N R_N
Output
Print T lines. The i-th line should contain the answer to the i-th test case.
Example
Input
3
2
1 5 10
2 15 5
3
2 93 78
1 71 59
3 57 96
19
19 23 16
5 90 13
12 85 70
19 67 78
12 16 60
18 48 28
5 4 24
12 97 97
4 57 87
19 91 74
18 100 76
7 86 46
9 100 57
3 76 73
6 84 93
1 6 84
11 75 94
19 15 3
12 11 34
Output
25
221
1354
Submitted Solution:
```
t=int(input())
import heapq
for _ in range(t):
n=int(input())
L=[];R=[]
ans=0
for _ in range(n):
ki,li,ri = map(int, input().split())
dif=li-ri
if dif>=0:
L.append((ki-1,-dif))
ans+=ri
else:
R.append((n-ki-1,dif))
ans+=li
L.sort();R.sort()
hL=[]
hR=[]
for i in range(len(L)):
if L[i][0]>len(hL):
heapq.heappush(hL,L[i][1])
elif L[i][0]==len(hL):
heapq.heappushpop(hL,L[i][1])
for i in range(len(R)):
if R[i][0]>len(hR):
heapq.heappush(hR,R[i][1])
elif L[i][0]==len(hR):
heapq.heappushpop(hR,R[i][1])
print(ans-sum(hL)-sum(hR))
``` | instruction | 0 | 73,085 | 16 | 146,170 |
No | output | 1 | 73,085 | 16 | 146,171 |
Provide a correct Python 3 solution for this coding contest problem.
We have N balance beams numbered 1 to N. The length of each beam is 1 meters. Snuke walks on Beam i at a speed of 1/A_i meters per second, and Ringo walks on Beam i at a speed of 1/B_i meters per second.
Snuke and Ringo will play the following game:
* First, Snuke connects the N beams in any order of his choice and makes a long beam of length N meters.
* Then, Snuke starts at the left end of the long beam. At the same time, Ringo starts at a point chosen uniformly at random on the long beam. Both of them walk to the right end of the long beam.
* Snuke wins if and only if he catches up to Ringo before Ringo reaches the right end of the long beam. That is, Snuke wins if there is a moment when Snuke and Ringo stand at the same position, and Ringo wins otherwise.
Find the probability that Snuke wins when Snuke arranges the N beams so that the probability of his winning is maximized.
This probability is a rational number, so we ask you to represent it as an irreducible fraction P/Q (to represent 0, use P=0, Q=1).
Constraints
* 1 \leq N \leq 10^5
* 1 \leq A_i,B_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
A_2 B_2
\vdots
A_N B_N
Output
Print the numerator and denominator of the irreducible fraction that represents the maximum probability of Snuke's winning.
Examples
Input
2
3 2
1 2
Output
1 4
Input
4
1 5
4 7
2 1
8 4
Output
1 2
Input
3
4 1
5 2
6 3
Output
0 1
Input
10
866111664 178537096
705445072 318106937
472381277 579910117
353498483 865935868
383133839 231371336
378371075 681212831
304570952 16537461
955719384 267238505
844917655 218662351
550309930 62731178
Output
697461712 2899550585 | instruction | 0 | 73,102 | 16 | 146,204 |
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
from heapq import heappush, heappushpop, heapify
from fractions import gcd
N = int(readline())
m = map(int,read().split())
AB = sorted(zip(m,m),key=lambda x:(x[1],x[0]))
A,B = zip(*AB)
dp2 = [0]*N # T_B - T_A
for i in range(N-1,0,-1):
a = A[i]; b = B[i]
if a<b:
dp2[i-1] = dp2[i]+(b-a)
else:
dp2[i-1] = dp2[i]
def F(n):
opt_num = -1
opt_den = 1
q = list(-x for x in A[:n])
heapify(q)
S = -sum(q)
for i in range(n,N):
a = A[i]; b = B[i]
x = S+a-dp2[i]
if 0<=x<=b:
den = b
num = (n+1)*den-x
if opt_num*den < opt_den*num:
opt_num = num; opt_den = den
x = a
y = -heappushpop(q,-x)
S += x-y
return opt_num, opt_den
opt_num = 0; opt_den = 1
left = -1 # 値がある
right = N+10 # むり
while left+1 < right:
mid = (left+right)//2
num,den = F(mid)
if num==-1:
right=mid
else:
left=mid
if opt_num*den < opt_den*num:
opt_num = num; opt_den = den
opt_den *= N
g = gcd(opt_num,opt_den)
opt_num//=g; opt_den//=g
print(opt_num,opt_den)
``` | output | 1 | 73,102 | 16 | 146,205 |
Provide a correct Python 3 solution for this coding contest problem.
We have N balance beams numbered 1 to N. The length of each beam is 1 meters. Snuke walks on Beam i at a speed of 1/A_i meters per second, and Ringo walks on Beam i at a speed of 1/B_i meters per second.
Snuke and Ringo will play the following game:
* First, Snuke connects the N beams in any order of his choice and makes a long beam of length N meters.
* Then, Snuke starts at the left end of the long beam. At the same time, Ringo starts at a point chosen uniformly at random on the long beam. Both of them walk to the right end of the long beam.
* Snuke wins if and only if he catches up to Ringo before Ringo reaches the right end of the long beam. That is, Snuke wins if there is a moment when Snuke and Ringo stand at the same position, and Ringo wins otherwise.
Find the probability that Snuke wins when Snuke arranges the N beams so that the probability of his winning is maximized.
This probability is a rational number, so we ask you to represent it as an irreducible fraction P/Q (to represent 0, use P=0, Q=1).
Constraints
* 1 \leq N \leq 10^5
* 1 \leq A_i,B_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
A_2 B_2
\vdots
A_N B_N
Output
Print the numerator and denominator of the irreducible fraction that represents the maximum probability of Snuke's winning.
Examples
Input
2
3 2
1 2
Output
1 4
Input
4
1 5
4 7
2 1
8 4
Output
1 2
Input
3
4 1
5 2
6 3
Output
0 1
Input
10
866111664 178537096
705445072 318106937
472381277 579910117
353498483 865935868
383133839 231371336
378371075 681212831
304570952 16537461
955719384 267238505
844917655 218662351
550309930 62731178
Output
697461712 2899550585 | instruction | 0 | 73,103 | 16 | 146,206 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
def gcd(a, b):
while b: a, b = b, a % b
return a
N = int(input())
S = 0
Y = []
for i in range(N):
a, b = map(int, input().split())
if b > a:
S += b-a
Y.append((b, b, b-a, 0))
else:
Y.append((a, b, 0, a-b))
Y = sorted(Y)
YY = [0] * (N+1)
for i in range(N):
YY[i+1] = YY[i] + Y[i][0]
def f(i, n):
if Y[i][2]:
return S - YY[n] if n <= i else S - (YY[n+1] - Y[i][0])
return S - Y[i][3] - YY[n] if n <= i else S - Y[i][3] - (YY[n+1] - Y[i][0])
maa, mab = 0, 1
for i in range(N):
l = 0
r = N
while r - l > 1:
m = (l+r) // 2
if f(i, m) >= 0:
l = m
else:
r = m
a = l * Y[i][1] + min(f(i, l), Y[i][1])
b = N * Y[i][1]
if a * mab > b * maa:
maa, mab = a, b
g = gcd(maa, mab)
print(maa//g, mab//g)
``` | output | 1 | 73,103 | 16 | 146,207 |
Provide a correct Python 3 solution for this coding contest problem.
We have N balance beams numbered 1 to N. The length of each beam is 1 meters. Snuke walks on Beam i at a speed of 1/A_i meters per second, and Ringo walks on Beam i at a speed of 1/B_i meters per second.
Snuke and Ringo will play the following game:
* First, Snuke connects the N beams in any order of his choice and makes a long beam of length N meters.
* Then, Snuke starts at the left end of the long beam. At the same time, Ringo starts at a point chosen uniformly at random on the long beam. Both of them walk to the right end of the long beam.
* Snuke wins if and only if he catches up to Ringo before Ringo reaches the right end of the long beam. That is, Snuke wins if there is a moment when Snuke and Ringo stand at the same position, and Ringo wins otherwise.
Find the probability that Snuke wins when Snuke arranges the N beams so that the probability of his winning is maximized.
This probability is a rational number, so we ask you to represent it as an irreducible fraction P/Q (to represent 0, use P=0, Q=1).
Constraints
* 1 \leq N \leq 10^5
* 1 \leq A_i,B_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
A_2 B_2
\vdots
A_N B_N
Output
Print the numerator and denominator of the irreducible fraction that represents the maximum probability of Snuke's winning.
Examples
Input
2
3 2
1 2
Output
1 4
Input
4
1 5
4 7
2 1
8 4
Output
1 2
Input
3
4 1
5 2
6 3
Output
0 1
Input
10
866111664 178537096
705445072 318106937
472381277 579910117
353498483 865935868
383133839 231371336
378371075 681212831
304570952 16537461
955719384 267238505
844917655 218662351
550309930 62731178
Output
697461712 2899550585 | instruction | 0 | 73,104 | 16 | 146,208 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
def gcd(a, b):
while b: a, b = b, a % b
return a
N = int(input())
S = 0
Y = []
for i in range(N):
a, b = map(int, input().split())
if b > a:
S += b-a
Y.append((b, b))
else:
Y.append((a, b))
Y = sorted(Y)
YY = [0] * (N+1)
for i in range(N):
YY[i+1] = YY[i] + Y[i][0]
# i番目を除いてn個選ぶときの余裕度
def f(i, n):
return S - Y[i][0] + Y[i][1] - (YY[n] if n <= i else YY[n+1] - Y[i][0])
ma1, ma2 = 0, 1
for i in range(N):
l = 0
r = N
while r - l > 1:
m = (l+r) // 2
if f(i, m) >= 0:
l = m
else:
r = m
a = l * Y[i][1] + min(f(i, l), Y[i][1])
b = N * Y[i][1]
if a * ma2 > b * ma1:
ma1, ma2 = a, b
g = gcd(ma1, ma2)
print(ma1//g, ma2//g)
``` | output | 1 | 73,104 | 16 | 146,209 |
Provide a correct Python 3 solution for this coding contest problem.
We have N balance beams numbered 1 to N. The length of each beam is 1 meters. Snuke walks on Beam i at a speed of 1/A_i meters per second, and Ringo walks on Beam i at a speed of 1/B_i meters per second.
Snuke and Ringo will play the following game:
* First, Snuke connects the N beams in any order of his choice and makes a long beam of length N meters.
* Then, Snuke starts at the left end of the long beam. At the same time, Ringo starts at a point chosen uniformly at random on the long beam. Both of them walk to the right end of the long beam.
* Snuke wins if and only if he catches up to Ringo before Ringo reaches the right end of the long beam. That is, Snuke wins if there is a moment when Snuke and Ringo stand at the same position, and Ringo wins otherwise.
Find the probability that Snuke wins when Snuke arranges the N beams so that the probability of his winning is maximized.
This probability is a rational number, so we ask you to represent it as an irreducible fraction P/Q (to represent 0, use P=0, Q=1).
Constraints
* 1 \leq N \leq 10^5
* 1 \leq A_i,B_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
A_2 B_2
\vdots
A_N B_N
Output
Print the numerator and denominator of the irreducible fraction that represents the maximum probability of Snuke's winning.
Examples
Input
2
3 2
1 2
Output
1 4
Input
4
1 5
4 7
2 1
8 4
Output
1 2
Input
3
4 1
5 2
6 3
Output
0 1
Input
10
866111664 178537096
705445072 318106937
472381277 579910117
353498483 865935868
383133839 231371336
378371075 681212831
304570952 16537461
955719384 267238505
844917655 218662351
550309930 62731178
Output
697461712 2899550585 | instruction | 0 | 73,105 | 16 | 146,210 |
"Correct Solution:
```
import heapq
class Heapq():
def __init__(self,arr):
self.que=[]
for a in arr:
self.que.append(-a)
heapq.heapify(self.que)
def pop(self):
return -heapq.heappop(self.que)
def push(self,a):
heapq.heappush(self.que,-a)
def top(self):
return -self.que[0]
#####segfunc#####
def segfunc(x, y):
return min(x,y)
#################
#####ide_ele#####
ide_ele = 10**15
#################
class SegTree:
def __init__(self, init_val, segfunc, ide_ele):
n = len(init_val)
self.segfunc = segfunc
self.ide_ele = ide_ele
self.num = 1 << (n - 1).bit_length()
self.tree = [ide_ele] * 2 * self.num
for i in range(n):
self.tree[self.num + i] = init_val[i]
for i in range(self.num - 1, 0, -1):
self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])
def update(self, k, x):
k += self.num
self.tree[k] = x
while k > 1:
self.tree[k >> 1] = self.segfunc(self.tree[k], self.tree[k ^ 1])
k >>= 1
def query(self, l, r):
res = self.ide_ele
l += self.num
r += self.num
while l < r:
if l & 1:
res = self.segfunc(res, self.tree[l])
l += 1
if r & 1:
res = self.segfunc(res, self.tree[r - 1])
l >>= 1
r >>= 1
return res
import sys,random
input=sys.stdin.readline
N=int(input())
beam=[tuple(map(int,input().split())) for i in range(N)]
def solve2():
beam.sort(key=lambda x:x[1])
data=[min(0,beam[i][0]-beam[i][1]) for i in range(N)]
for i in range(N-2,-1,-1):
data[i]+=data[i+1]
cummin=[beam[i][1] for i in range(N)]
for i in range(N):
if beam[i][0]-beam[i][1]>0:
cummin[i]=10**15
for i in range(N-2,-1,-1):
cummin[i]=min(cummin[i],cummin[i+1])
def cond(m):
if m==0:
if data[0]<=0:
return True
else:
return False
que=Heapq([])
cnt=0
S=0
for i in range(N-1):
if cnt==m:
test=que.top()
if beam[i][0]<test:
que.pop()
que.push(beam[i][0])
S+=beam[i][0]-test
else:
que.push(beam[i][0])
S+=beam[i][0]
cnt+=1
if cnt==m:
test=S+data[i+1]
if test<=0:
return True
return False
start=0
end=N
while end-start>1:
test=(end+start)//2
if cond(test):
start=test
else:
end=test
if not cond(0):
exit(print(0,1))
m=start
ans=[0,1]
que=Heapq([])
cnt=0
S=0
if m!=0:
for i in range(N-1):
if cnt==m:
test=que.top()
if beam[i][0]<test:
que.pop()
que.push(beam[i][0])
S+=beam[i][0]-test
else:
que.push(beam[i][0])
S+=beam[i][0]
cnt+=1
if cnt==m:
test=S+data[i+1]
B=cummin[i+1]
if test<=0:
t=abs(test)
if B*ans[0]<ans[1]*t:
ans=[t,B]
start=[-1]*N
end=[-1]*N
que=[]
trash=[]
cnt=0
S=0
data1=[ide_ele]*N
data2=[ide_ele]*N
for i in range(N):
if cnt==m:
val,id=que[0]
val=-val
if val>beam[i][0]:
heapq.heappop(que)
heapq.heappush(trash,val)
end[id]=i-1
heapq.heappush(que,(-beam[i][0],i))
start[i]=i
S+=beam[i][0]-val
else:
heapq.heappush(trash,beam[i][0])
else:
heapq.heappush(que,(-beam[i][0],i))
start[i]=i
S+=beam[i][0]
cnt+=1
if cnt==m:
if i!=N-1:
data1[i]=S+data[i+1]
if trash:
data2[i]=S+data[i+1]+trash[0]
else:
data2[i]=S+data[i+1]
else:
data1[i]=S
data2[i]=S+trash[0]
for i in range(N):
if start[i]!=-1 and end[i]==-1:
end[i]=N-1
Seg1=SegTree(data1,segfunc,ide_ele)
Seg2=SegTree(data2,segfunc,ide_ele)
for i in range(m):
if end[i]==m-1:
temp=Seg1.query(m,N)
temp+=beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[temp,B]
else:
L,R=m,end[i]
temp=Seg2.query(L,R+1)-beam[i][0]
temp+=beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[temp,B]
temp=Seg1.query(R+1,N)
temp+=beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[temp,B]
for i in range(m,N):
if beam[i][0]-beam[i][1]<=0:
if start[i]==-1:
temp=Seg1.query(i,N)
temp+=beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[temp,B]
else:
L=start[i]
R=end[i]
temp=Seg2.query(L,R+1)-beam[i][0]
temp+=beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[t,B]
temp=Seg1.query(R+1,N)
temp+=beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[temp,B]
else:
if start[i]==-1:
temp=Seg1.query(m-1,N)
temp+=beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[temp,B]
else:
L=start[i]
R=end[i]
temp=Seg2.query(L,R+1)-beam[i][0]
temp+=beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[temp,B]
temp=min(Seg1.query(R+1,N),Seg1.query(m-1,L))
temp+=beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[temp,B]
else:
for i in range(N):
test=data[i]
B=cummin[i]
if test<=0:
t=abs(test)
if B*ans[0]<ans[1]*t:
ans=[t,B]
M=min(data)
for i in range(N):
if beam[i][0]-beam[i][1]>0:
temp=M+beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[temp,B]
p,q=ans
res=[q*m+p,N*q]
from math import gcd
g=gcd(res[0],res[1])
return (res[0]//g,res[1]//g)
print(*solve2())
``` | output | 1 | 73,105 | 16 | 146,211 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N balance beams numbered 1 to N. The length of each beam is 1 meters. Snuke walks on Beam i at a speed of 1/A_i meters per second, and Ringo walks on Beam i at a speed of 1/B_i meters per second.
Snuke and Ringo will play the following game:
* First, Snuke connects the N beams in any order of his choice and makes a long beam of length N meters.
* Then, Snuke starts at the left end of the long beam. At the same time, Ringo starts at a point chosen uniformly at random on the long beam. Both of them walk to the right end of the long beam.
* Snuke wins if and only if he catches up to Ringo before Ringo reaches the right end of the long beam. That is, Snuke wins if there is a moment when Snuke and Ringo stand at the same position, and Ringo wins otherwise.
Find the probability that Snuke wins when Snuke arranges the N beams so that the probability of his winning is maximized.
This probability is a rational number, so we ask you to represent it as an irreducible fraction P/Q (to represent 0, use P=0, Q=1).
Constraints
* 1 \leq N \leq 10^5
* 1 \leq A_i,B_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
A_2 B_2
\vdots
A_N B_N
Output
Print the numerator and denominator of the irreducible fraction that represents the maximum probability of Snuke's winning.
Examples
Input
2
3 2
1 2
Output
1 4
Input
4
1 5
4 7
2 1
8 4
Output
1 2
Input
3
4 1
5 2
6 3
Output
0 1
Input
10
866111664 178537096
705445072 318106937
472381277 579910117
353498483 865935868
383133839 231371336
378371075 681212831
304570952 16537461
955719384 267238505
844917655 218662351
550309930 62731178
Output
697461712 2899550585
Submitted Solution:
```
import heapq
class Heapq():
def __init__(self,arr):
self.que=[]
for a in arr:
self.que.append(-a)
heapq.heapify(self.que)
def pop(self):
return -heapq.heappop(self.que)
def push(self,a):
heapq.heappush(self.que,-a)
def top(self):
return -self.que[0]
#####segfunc#####
def segfunc(x, y):
return min(x,y)
#################
#####ide_ele#####
ide_ele = 10**15
#################
class SegTree:
def __init__(self, init_val, segfunc, ide_ele):
n = len(init_val)
self.segfunc = segfunc
self.ide_ele = ide_ele
self.num = 1 << (n - 1).bit_length()
self.tree = [ide_ele] * 2 * self.num
for i in range(n):
self.tree[self.num + i] = init_val[i]
for i in range(self.num - 1, 0, -1):
self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])
def update(self, k, x):
k += self.num
self.tree[k] = x
while k > 1:
self.tree[k >> 1] = self.segfunc(self.tree[k], self.tree[k ^ 1])
k >>= 1
def query(self, l, r):
res = self.ide_ele
l += self.num
r += self.num
while l < r:
if l & 1:
res = self.segfunc(res, self.tree[l])
l += 1
if r & 1:
res = self.segfunc(res, self.tree[r - 1])
l >>= 1
r >>= 1
return res
import sys
input=sys.stdin.readline
N=int(input())
beam=[tuple(map(int,input().split())) for i in range(N)]
beam.sort(key=lambda x:x[1])
data=[min(0,beam[i][0]-beam[i][1]) for i in range(N)]
for i in range(N-2,-1,-1):
data[i]+=data[i+1]
cummin=[beam[i][1] for i in range(N)]
for i in range(N):
if beam[i][0]-beam[i][1]>0:
cummin[i]=10**10
for i in range(N-2,-1,-1):
cummin[i]=min(cummin[i],cummin[i+1])
def cond(m):
if m==0:
if data[0]<=0:
return True
else:
return False
que=Heapq([])
cnt=0
S=0
for i in range(N-1):
if cnt==m:
test=que.top()
if beam[i][0]<test:
que.pop()
que.push(beam[i][0])
S+=beam[i][0]-test
else:
que.push(beam[i][0])
S+=beam[i][0]
cnt+=1
if cnt==m:
test=S+data[i+1]
if test<=0:
return True
return False
start=0
end=N
while end-start>1:
test=(end+start)//2
if cond(test):
start=test
else:
end=test
if not cond(0):
exit(print(0,1))
m=start
ans=[0,1]
que=Heapq([])
cnt=0
S=0
if m!=0:
for i in range(N-1):
if cnt==m:
test=que.top()
if beam[i][0]<test:
que.pop()
que.push(beam[i][0])
S+=beam[i][0]-test
else:
que.push(beam[i][0])
S+=beam[i][0]
cnt+=1
if cnt==m:
test=S+data[i+1]
B=cummin[i+1]
if test<=0:
t=abs(test)
if B*ans[0]<ans[1]*t:
ans=[t,B]
start=[-1]*N
end=[-1]*N
que=[]
trash=[]
cnt=0
S=0
data1=[ide_ele]*N
data2=[ide_ele]*N
for i in range(N):
if cnt==m:
val,id=que[0]
val=-val
if val>beam[i][0]:
heapq.heappop(que)
heapq.heappush(trash,val)
end[id]=i-1
heapq.heappush(que,(-beam[i][0],i))
start[i]=i
S+=beam[i][0]-val
else:
heapq.heappush(trash,beam[i][0])
else:
heapq.heappush(que,(-beam[i][0],i))
start[i]=i
S+=beam[i][0]
cnt+=1
if cnt==m:
if i!=N-1:
data1[i]=S+data[i+1]
if trash:
data2[i]=S+data[i+1]+trash[0]
else:
data2[i]=S+data[i+1]
else:
data1[i]=S
data2[i]=S+trash[0]
for i in range(N):
if start[i]!=-1 and end[i]==-1:
end[i]=N-1
Seg1=SegTree(data1,segfunc,ide_ele)
Seg2=SegTree(data2,segfunc,ide_ele)
for i in range(m):
if end[i]==m-1:
temp=Seg1.query(m,N)
temp+=beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[t,B]
else:
L,R=m,end[i]
temp=Seg2.query(L,R+1)-beam[i][0]
temp+=beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[t,B]
temp=Seg1.query(R+1,N)
temp+=beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[t,B]
for i in range(m,N):
if beam[i][0]-beam[i][1]<=0 and 1:
if start[i]==-1:
temp=Seg1.query(i,N)
temp+=beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[t,B]
elif 1:
L=start[i]
R=end[i]
temp=Seg2.query(L,R+1)-beam[i][0]
temp+=beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[t,B]
temp=Seg1.query(R+1,N)
temp+=beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[t,B]
else:
if start[i]==-1 and 1:
temp=Seg1.query(m-1,N)
temp+=beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[t,B]
else:
if 1:
L=start[i]
R=end[i]
temp=Seg2.query(L,R+1)-beam[i][0]
temp+=beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[t,B]
temp=min(Seg1.query(R+1,N),Seg1.query(m-1,L))
temp+=beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[t,B]
else:
for i in range(N):
test=data[i]
B=cummin[i]
if test<=0:
t=abs(test)
if B*ans[0]<ans[1]*t:
ans=[t,B]
M=min(data)
for i in range(N):
if beam[i][0]-beam[i][1]>0:
temp=M+beam[i][0]-beam[i][1]
if temp<=0:
temp=abs(temp)
B=beam[i][1]
if B*ans[0]<ans[1]*temp:
ans=[t,B]
p,q=ans
res=[q*m+p,N*q]
from math import gcd
g=gcd(res[0],res[1])
print(res[0]//g,res[1]//g)
``` | instruction | 0 | 73,106 | 16 | 146,212 |
No | output | 1 | 73,106 | 16 | 146,213 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N balance beams numbered 1 to N. The length of each beam is 1 meters. Snuke walks on Beam i at a speed of 1/A_i meters per second, and Ringo walks on Beam i at a speed of 1/B_i meters per second.
Snuke and Ringo will play the following game:
* First, Snuke connects the N beams in any order of his choice and makes a long beam of length N meters.
* Then, Snuke starts at the left end of the long beam. At the same time, Ringo starts at a point chosen uniformly at random on the long beam. Both of them walk to the right end of the long beam.
* Snuke wins if and only if he catches up to Ringo before Ringo reaches the right end of the long beam. That is, Snuke wins if there is a moment when Snuke and Ringo stand at the same position, and Ringo wins otherwise.
Find the probability that Snuke wins when Snuke arranges the N beams so that the probability of his winning is maximized.
This probability is a rational number, so we ask you to represent it as an irreducible fraction P/Q (to represent 0, use P=0, Q=1).
Constraints
* 1 \leq N \leq 10^5
* 1 \leq A_i,B_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
A_2 B_2
\vdots
A_N B_N
Output
Print the numerator and denominator of the irreducible fraction that represents the maximum probability of Snuke's winning.
Examples
Input
2
3 2
1 2
Output
1 4
Input
4
1 5
4 7
2 1
8 4
Output
1 2
Input
3
4 1
5 2
6 3
Output
0 1
Input
10
866111664 178537096
705445072 318106937
472381277 579910117
353498483 865935868
383133839 231371336
378371075 681212831
304570952 16537461
955719384 267238505
844917655 218662351
550309930 62731178
Output
697461712 2899550585
Submitted Solution:
```
from bisect import bisect_left
def solve_d(n, a_list, b_list):
a_list_sub = [a_list[i] for i in range(n) if a_list[i] < b_list[i]]
b_list_sub = [b_list[i] for i in range(n) if a_list[i] < b_list[i]]
a_list_add = [(a_list[i], i) for i in range(n) if a_list[i] >= b_list[i]]
le = len(a_list_sub)
if le == 0:
return 0, 1
a_sum = sum(a_list_sub)
b_list_sub_s = sorted(b_list_sub, reverse=True)
b_list_sub_cum = [0]
for i, b in enumerate(b_list_sub_s):
b_list_sub_cum.append(b_list_sub_cum[i] + b)
r = bisect_left(b_list_sub_cum, a_sum) - 1
q_max = b_list_sub_s[r]
p_max = q_max * (le - r) - (a_sum - b_list_sub_cum[r])
q_max *= n
# print(le, p_max, q_max, r, a_sum - b_list_sub_cum[r])
a_list_add_s = sorted(a_list_add, key=lambda x:x[0])
for a, i in a_list_add_s:
le += 1
a_sum += a
b_list_sub_cum.append(b_list_sub_cum[-1] + b_list[i])
b_list_sub_s.append(b_list[i])
r = bisect_left(b_list_sub_cum, a_sum) - 1
if r >= len(b_list_sub_s):
q = 1
p = 0
else:
q = b_list_sub_s[r]
p = q * (le - r) - (a_sum - b_list_sub_cum[r])
q *= n
# print(le, p, q)
if p * q_max > p_max * q:
p_max, q_max = p, q
return p_max, q_max
def gcd(x, y):
while y:
x, y = y, x % y
return x
def main():
n = int(input())
a_list = [0] * n
b_list = [0] * n
for i in range(n):
a, b = map(int, input().split())
a_list[i] = a
b_list[i] = b
p, q = solve_d(n, a_list, b_list)
r = gcd(p, q)
print(p // r, q // r)
if __name__ == "__main__":
main()
``` | instruction | 0 | 73,107 | 16 | 146,214 |
No | output | 1 | 73,107 | 16 | 146,215 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N balance beams numbered 1 to N. The length of each beam is 1 meters. Snuke walks on Beam i at a speed of 1/A_i meters per second, and Ringo walks on Beam i at a speed of 1/B_i meters per second.
Snuke and Ringo will play the following game:
* First, Snuke connects the N beams in any order of his choice and makes a long beam of length N meters.
* Then, Snuke starts at the left end of the long beam. At the same time, Ringo starts at a point chosen uniformly at random on the long beam. Both of them walk to the right end of the long beam.
* Snuke wins if and only if he catches up to Ringo before Ringo reaches the right end of the long beam. That is, Snuke wins if there is a moment when Snuke and Ringo stand at the same position, and Ringo wins otherwise.
Find the probability that Snuke wins when Snuke arranges the N beams so that the probability of his winning is maximized.
This probability is a rational number, so we ask you to represent it as an irreducible fraction P/Q (to represent 0, use P=0, Q=1).
Constraints
* 1 \leq N \leq 10^5
* 1 \leq A_i,B_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
A_2 B_2
\vdots
A_N B_N
Output
Print the numerator and denominator of the irreducible fraction that represents the maximum probability of Snuke's winning.
Examples
Input
2
3 2
1 2
Output
1 4
Input
4
1 5
4 7
2 1
8 4
Output
1 2
Input
3
4 1
5 2
6 3
Output
0 1
Input
10
866111664 178537096
705445072 318106937
472381277 579910117
353498483 865935868
383133839 231371336
378371075 681212831
304570952 16537461
955719384 267238505
844917655 218662351
550309930 62731178
Output
697461712 2899550585
Submitted Solution:
```
import sys
input = sys.stdin.readline
def gcd(a, b):
while b: a, b = b, a % b
return abs(a)
N = int(input())
S = 0
Y = []
for i in range(N):
a, b = map(int, input().split())
if b > a:
S += b-a
Y.append((b, b, b-a))
else:
Y.append((a, b, 0))
Y = sorted(Y)
YY = [0] * (N+1)
for i in range(N):
YY[i+1] = YY[i] + Y[i][0]
def f(i, n):
return S - YY[n] if n <= i else S - (YY[n+1] - Y[i][0])
maa, mab = 0, 1
for i in range(N):
if Y[i][2] == 0: continue
l = 0
r = N
while r - l > 1:
m = (l+r) // 2
if f(i, m) >= 0:
l = m
else:
r = m
aa = l * Y[i][1] + f(i, l)
bb = N * Y[i][1]
if aa * mab > bb * maa:
maa, mab = aa, bb
g = gcd(maa, mab)
print(maa//g, mab//g)
``` | instruction | 0 | 73,108 | 16 | 146,216 |
No | output | 1 | 73,108 | 16 | 146,217 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have N balance beams numbered 1 to N. The length of each beam is 1 meters. Snuke walks on Beam i at a speed of 1/A_i meters per second, and Ringo walks on Beam i at a speed of 1/B_i meters per second.
Snuke and Ringo will play the following game:
* First, Snuke connects the N beams in any order of his choice and makes a long beam of length N meters.
* Then, Snuke starts at the left end of the long beam. At the same time, Ringo starts at a point chosen uniformly at random on the long beam. Both of them walk to the right end of the long beam.
* Snuke wins if and only if he catches up to Ringo before Ringo reaches the right end of the long beam. That is, Snuke wins if there is a moment when Snuke and Ringo stand at the same position, and Ringo wins otherwise.
Find the probability that Snuke wins when Snuke arranges the N beams so that the probability of his winning is maximized.
This probability is a rational number, so we ask you to represent it as an irreducible fraction P/Q (to represent 0, use P=0, Q=1).
Constraints
* 1 \leq N \leq 10^5
* 1 \leq A_i,B_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
A_2 B_2
\vdots
A_N B_N
Output
Print the numerator and denominator of the irreducible fraction that represents the maximum probability of Snuke's winning.
Examples
Input
2
3 2
1 2
Output
1 4
Input
4
1 5
4 7
2 1
8 4
Output
1 2
Input
3
4 1
5 2
6 3
Output
0 1
Input
10
866111664 178537096
705445072 318106937
472381277 579910117
353498483 865935868
383133839 231371336
378371075 681212831
304570952 16537461
955719384 267238505
844917655 218662351
550309930 62731178
Output
697461712 2899550585
Submitted Solution:
```
import sys
input = sys.stdin.readline
def gcd(a, b):
while b: a, b = b, a % b
return abs(a)
N = int(input())
S = 0
Y = []
for i in range(N):
a, b = map(int, input().split())
if b > a:
S += b-a
Y.append((b, b, b-a))
else:
Y.append((a, b, 0))
Y = sorted(Y)
YY = [0] * (N+1)
for i in range(N):
YY[i+1] = YY[i] + Y[i][0]
maa, mab = 0, 1
for i in range(N):
if Y[i][2] == 0: continue
l = 0
r = N
while r - l > 1:
m = (l+r) // 2
rem = S - YY[m] if m < i else S - (YY[m+1] - Y[i][0])
if rem >= 0:
l = m
else:
r = m
aa = l * Y[i][1] + S - (YY[l] if l < i else YY[l+1] - Y[i][0])
bb = N * Y[i][1]
if aa * mab > bb * maa:
maa, mab = aa, bb
g = gcd(maa, mab)
print(maa//g, mab//g)
``` | instruction | 0 | 73,109 | 16 | 146,218 |
No | output | 1 | 73,109 | 16 | 146,219 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke loves constructing integer sequences.
There are N piles of stones, numbered 1 through N. The pile numbered i consists of a_i stones.
Snuke will construct an integer sequence s of length Σa_i, as follows:
1. Among the piles with the largest number of stones remaining, let x be the index of the pile with the smallest index. Append x to the end of s.
2. Select a pile with one or more stones remaining, and remove a stone from that pile.
3. If there is a pile with one or more stones remaining, go back to step 1. Otherwise, terminate the process.
We are interested in the lexicographically smallest sequence that can be constructed. For each of the integers 1,2,3,...,N, how many times does it occur in the lexicographically smallest sequence?
Constraints
* 1 ≤ N ≤ 10^{5}
* 1 ≤ a_i ≤ 10^{9}
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{N}
Output
Print N lines. The i-th line should contain the number of the occurrences of the integer i in the lexicographically smallest sequence that can be constructed.
Examples
Input
2
1 2
Output
2
1
Input
10
1 2 1 3 2 4 2 5 8 1
Output
10
7
0
4
0
3
0
2
3
0 | instruction | 0 | 74,072 | 16 | 148,144 |
"Correct Solution:
```
from collections import Counter
N = int(input())
S = list(map(int,input().split()))
T = []
for i in range(N):
T.append((i,S[i]))
U = sorted(T,key=lambda x:x[1], reverse=True)
Ans = [0]*N
c,v = 0,10**10
for i in range(N-1):
c+=1
v = min(v,U[i][0])
if U[i][1]==U[i+1][1]:
continue
else:
Ans[v] += (U[i][1] - U[i+1][1])*c
else:
Ans[0] = 0
Ans[0] = sum(S) - sum(Ans)
for i in range(N):
print(Ans[i])
``` | output | 1 | 74,072 | 16 | 148,145 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke loves constructing integer sequences.
There are N piles of stones, numbered 1 through N. The pile numbered i consists of a_i stones.
Snuke will construct an integer sequence s of length Σa_i, as follows:
1. Among the piles with the largest number of stones remaining, let x be the index of the pile with the smallest index. Append x to the end of s.
2. Select a pile with one or more stones remaining, and remove a stone from that pile.
3. If there is a pile with one or more stones remaining, go back to step 1. Otherwise, terminate the process.
We are interested in the lexicographically smallest sequence that can be constructed. For each of the integers 1,2,3,...,N, how many times does it occur in the lexicographically smallest sequence?
Constraints
* 1 ≤ N ≤ 10^{5}
* 1 ≤ a_i ≤ 10^{9}
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{N}
Output
Print N lines. The i-th line should contain the number of the occurrences of the integer i in the lexicographically smallest sequence that can be constructed.
Examples
Input
2
1 2
Output
2
1
Input
10
1 2 1 3 2 4 2 5 8 1
Output
10
7
0
4
0
3
0
2
3
0 | instruction | 0 | 74,073 | 16 | 148,146 |
"Correct Solution:
```
from operator import itemgetter
import sys
N=int(input())
A=[int(i) for i in input().split()]
B=[(A[i],i) for i in range(N)]
B=sorted(B,key=itemgetter(1))
B=sorted(B,key=itemgetter(0))
T=[0]*N
cos,pt=B.pop()
for i in range(1,N):
cos2,pt2=B.pop()
T[pt]+=(cos-cos2)*i
pt=min(pt,pt2)
cos=cos2
T[pt]+=cos*N
for i in T:
print(i)
``` | output | 1 | 74,073 | 16 | 148,147 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke loves constructing integer sequences.
There are N piles of stones, numbered 1 through N. The pile numbered i consists of a_i stones.
Snuke will construct an integer sequence s of length Σa_i, as follows:
1. Among the piles with the largest number of stones remaining, let x be the index of the pile with the smallest index. Append x to the end of s.
2. Select a pile with one or more stones remaining, and remove a stone from that pile.
3. If there is a pile with one or more stones remaining, go back to step 1. Otherwise, terminate the process.
We are interested in the lexicographically smallest sequence that can be constructed. For each of the integers 1,2,3,...,N, how many times does it occur in the lexicographically smallest sequence?
Constraints
* 1 ≤ N ≤ 10^{5}
* 1 ≤ a_i ≤ 10^{9}
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{N}
Output
Print N lines. The i-th line should contain the number of the occurrences of the integer i in the lexicographically smallest sequence that can be constructed.
Examples
Input
2
1 2
Output
2
1
Input
10
1 2 1 3 2 4 2 5 8 1
Output
10
7
0
4
0
3
0
2
3
0 | instruction | 0 | 74,074 | 16 | 148,148 |
"Correct Solution:
```
import os
import sys
if os.getenv("LOCAL"):
sys.stdin = open("_in.txt", "r")
sys.setrecursionlimit(2147483647)
INF = float("inf")
IINF = 10 ** 18
MOD = 10 ** 9 + 7
N = int(sys.stdin.readline())
A = list(map(int, sys.stdin.readline().split()))
XY = [(a, i) for i, a in enumerate(A)]
XY.sort(reverse=True)
# 解説
ans = [0] * N
min_y = IINF
for i, (x, y) in enumerate(XY):
min_y = min(min_y, y)
x2 = XY[i + 1][0] if i + 1 < N else 0
ans[min_y] += (x - x2) * (i + 1)
print(*ans, sep='\n')
``` | output | 1 | 74,074 | 16 | 148,149 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke loves constructing integer sequences.
There are N piles of stones, numbered 1 through N. The pile numbered i consists of a_i stones.
Snuke will construct an integer sequence s of length Σa_i, as follows:
1. Among the piles with the largest number of stones remaining, let x be the index of the pile with the smallest index. Append x to the end of s.
2. Select a pile with one or more stones remaining, and remove a stone from that pile.
3. If there is a pile with one or more stones remaining, go back to step 1. Otherwise, terminate the process.
We are interested in the lexicographically smallest sequence that can be constructed. For each of the integers 1,2,3,...,N, how many times does it occur in the lexicographically smallest sequence?
Constraints
* 1 ≤ N ≤ 10^{5}
* 1 ≤ a_i ≤ 10^{9}
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{N}
Output
Print N lines. The i-th line should contain the number of the occurrences of the integer i in the lexicographically smallest sequence that can be constructed.
Examples
Input
2
1 2
Output
2
1
Input
10
1 2 1 3 2 4 2 5 8 1
Output
10
7
0
4
0
3
0
2
3
0 | instruction | 0 | 74,075 | 16 | 148,150 |
"Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
b = list(range(n))
s = list(map(list,zip(a,b)))
s.sort(reverse=True)
s.append([0,0])
ans = [0]*n
for i in range(n):
dif = s[i][0]-s[i+1][0]
ans[s[i][1]] += dif*(i+1)
if s[i+1][1]>s[i][1]:
s[i+1][1] = s[i][1]
print(*ans,sep="\n")
``` | output | 1 | 74,075 | 16 | 148,151 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke loves constructing integer sequences.
There are N piles of stones, numbered 1 through N. The pile numbered i consists of a_i stones.
Snuke will construct an integer sequence s of length Σa_i, as follows:
1. Among the piles with the largest number of stones remaining, let x be the index of the pile with the smallest index. Append x to the end of s.
2. Select a pile with one or more stones remaining, and remove a stone from that pile.
3. If there is a pile with one or more stones remaining, go back to step 1. Otherwise, terminate the process.
We are interested in the lexicographically smallest sequence that can be constructed. For each of the integers 1,2,3,...,N, how many times does it occur in the lexicographically smallest sequence?
Constraints
* 1 ≤ N ≤ 10^{5}
* 1 ≤ a_i ≤ 10^{9}
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{N}
Output
Print N lines. The i-th line should contain the number of the occurrences of the integer i in the lexicographically smallest sequence that can be constructed.
Examples
Input
2
1 2
Output
2
1
Input
10
1 2 1 3 2 4 2 5 8 1
Output
10
7
0
4
0
3
0
2
3
0 | instruction | 0 | 74,076 | 16 | 148,152 |
"Correct Solution:
```
from collections import defaultdict
import sys
class BIT():
def __init__(self, number):
self.n = number
self.list = [0] * (number + 1)
def add(self, i, x): # ith added x 1indexed
while i <= self.n:
self.list[i] += x
i += i & -i
def search(self, i): # 1-i sum
s = 0
while i > 0:
s += self.list[i]
i -= i & -i
return s
def suma(self, i, j): # i,i+1,..j sum
return self.search(j) - self.search(i - 1)
N=int(input())
A=[int(i) for i in input().split()]
C=sorted(set(A))
ndd=defaultdict(int)
for i in range(len(C)):
ndd[i+1]=C[i]
dd=defaultdict(int)
for i in range(len(C)):
dd[C[i]]=i+1
#print(ndd,dd)
visit=[0]*N
visit[0]=1
s=A[0]
H=[]
H.append(s)
for i in range(1,N):
if s<A[i]:
s=A[i]
H.append(s)
visit[i]=1
BITI=BIT(N+1)
BITI2=BIT(N+1)
j=len(H)-1
l=dd[H[j]]
num=sum(A)
T=[0]*N
ans=0
for i in range(N-1,-1,-1):
if l==dd[A[0]]:
break
BITI.add(dd[A[i]],A[i])
BITI2.add(dd[A[i]],1)
if dd[A[i]]==l and visit[i]==1:
T[i]=BITI.suma(dd[H[j-1]],N+1)-BITI2.suma(dd[H[j-1]],N+1)*H[j-1]-ans
ans+=T[i]
#print(i,j,l,T[i],ans)
j-=1
l=dd[H[j]]
x=num-sum(T)
T[0]=x
#print(visit,T)
for t in T:
print(t)
``` | output | 1 | 74,076 | 16 | 148,153 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke loves constructing integer sequences.
There are N piles of stones, numbered 1 through N. The pile numbered i consists of a_i stones.
Snuke will construct an integer sequence s of length Σa_i, as follows:
1. Among the piles with the largest number of stones remaining, let x be the index of the pile with the smallest index. Append x to the end of s.
2. Select a pile with one or more stones remaining, and remove a stone from that pile.
3. If there is a pile with one or more stones remaining, go back to step 1. Otherwise, terminate the process.
We are interested in the lexicographically smallest sequence that can be constructed. For each of the integers 1,2,3,...,N, how many times does it occur in the lexicographically smallest sequence?
Constraints
* 1 ≤ N ≤ 10^{5}
* 1 ≤ a_i ≤ 10^{9}
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{N}
Output
Print N lines. The i-th line should contain the number of the occurrences of the integer i in the lexicographically smallest sequence that can be constructed.
Examples
Input
2
1 2
Output
2
1
Input
10
1 2 1 3 2 4 2 5 8 1
Output
10
7
0
4
0
3
0
2
3
0 | instruction | 0 | 74,077 | 16 | 148,154 |
"Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
tmp = [[e, n - i] for i, e in enumerate(a)]
tmp.sort(reverse=True)
aa = [[e, n - i] for e, i in tmp] + [[0, -1]]
v_prev, i_prev = aa[0]
i = 0
ans = [0] * n
sm = 0
while i < n:
while aa[i][1] >= i_prev:
sm += aa[i][0]
i += 1
ans[i_prev] += sm - aa[i][0] * i
sm = aa[i][0] * i
v_prev, i_prev = aa[i]
print(*ans, sep="\n")
``` | output | 1 | 74,077 | 16 | 148,155 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke loves constructing integer sequences.
There are N piles of stones, numbered 1 through N. The pile numbered i consists of a_i stones.
Snuke will construct an integer sequence s of length Σa_i, as follows:
1. Among the piles with the largest number of stones remaining, let x be the index of the pile with the smallest index. Append x to the end of s.
2. Select a pile with one or more stones remaining, and remove a stone from that pile.
3. If there is a pile with one or more stones remaining, go back to step 1. Otherwise, terminate the process.
We are interested in the lexicographically smallest sequence that can be constructed. For each of the integers 1,2,3,...,N, how many times does it occur in the lexicographically smallest sequence?
Constraints
* 1 ≤ N ≤ 10^{5}
* 1 ≤ a_i ≤ 10^{9}
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{N}
Output
Print N lines. The i-th line should contain the number of the occurrences of the integer i in the lexicographically smallest sequence that can be constructed.
Examples
Input
2
1 2
Output
2
1
Input
10
1 2 1 3 2 4 2 5 8 1
Output
10
7
0
4
0
3
0
2
3
0 | instruction | 0 | 74,078 | 16 | 148,156 |
"Correct Solution:
```
def main():
n = int(input())
a = list(enumerate(map(int, input().split())))
a.sort(key=lambda x: x[1], reverse=True)
a.append((-1, 0))
# min_ind = min(a[:i+1])[0]
min_ind = n
ans = [0] * n
for i in range(n):
ind, value = a[i]
if ind < min_ind:
min_ind = ind
ans[min_ind] += (value - a[i+1][1]) * (i+1)
for i in ans:
print(i)
main()
``` | output | 1 | 74,078 | 16 | 148,157 |
Provide a correct Python 3 solution for this coding contest problem.
Snuke loves constructing integer sequences.
There are N piles of stones, numbered 1 through N. The pile numbered i consists of a_i stones.
Snuke will construct an integer sequence s of length Σa_i, as follows:
1. Among the piles with the largest number of stones remaining, let x be the index of the pile with the smallest index. Append x to the end of s.
2. Select a pile with one or more stones remaining, and remove a stone from that pile.
3. If there is a pile with one or more stones remaining, go back to step 1. Otherwise, terminate the process.
We are interested in the lexicographically smallest sequence that can be constructed. For each of the integers 1,2,3,...,N, how many times does it occur in the lexicographically smallest sequence?
Constraints
* 1 ≤ N ≤ 10^{5}
* 1 ≤ a_i ≤ 10^{9}
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{N}
Output
Print N lines. The i-th line should contain the number of the occurrences of the integer i in the lexicographically smallest sequence that can be constructed.
Examples
Input
2
1 2
Output
2
1
Input
10
1 2 1 3 2 4 2 5 8 1
Output
10
7
0
4
0
3
0
2
3
0 | instruction | 0 | 74,079 | 16 | 148,158 |
"Correct Solution:
```
#1は一意
#2は小さい番号の石の数が大きくなるように、大きい番号の石を除きたい
#個数が同順のものを、番号小以外が個数が時点になるまで除いてから、番号少を除く
n = int(input())
a = list(map(int, input().split( )))
cnt = {}#keyは個数、要素は最小番号
for i in range(n):
if a[i] in cnt:
cnt[a[i]][0] = min(cnt[a[i]][0],i)
cnt[a[i]][1]+=1
else:
cnt[a[i]] = [i,1]
#個数、最小番号、番号の数
mins = [[k,cnt[k][0],cnt[k][1]] for k in cnt]
mins.append([0,0,0])
#末端処理になっていない
mins.sort(reverse = True)
#print(mins)
ans = [0]*n
for i in range(len(mins)-1):
k,cntk,l = mins[i][0],mins[i][1],mins[i][2]
ans[cntk] += (k-mins[i+1][0])*l
mins[i+1][2]+=l###=じゃおかしい
if mins[i+1][1]>mins[i][1]:##追加
mins[i+1][1] = mins[i][1]
#ans[mins[-1][1]]=n*mins[-1][0]
for ai in ans:
print(ai)
``` | output | 1 | 74,079 | 16 | 148,159 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke loves constructing integer sequences.
There are N piles of stones, numbered 1 through N. The pile numbered i consists of a_i stones.
Snuke will construct an integer sequence s of length Σa_i, as follows:
1. Among the piles with the largest number of stones remaining, let x be the index of the pile with the smallest index. Append x to the end of s.
2. Select a pile with one or more stones remaining, and remove a stone from that pile.
3. If there is a pile with one or more stones remaining, go back to step 1. Otherwise, terminate the process.
We are interested in the lexicographically smallest sequence that can be constructed. For each of the integers 1,2,3,...,N, how many times does it occur in the lexicographically smallest sequence?
Constraints
* 1 ≤ N ≤ 10^{5}
* 1 ≤ a_i ≤ 10^{9}
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{N}
Output
Print N lines. The i-th line should contain the number of the occurrences of the integer i in the lexicographically smallest sequence that can be constructed.
Examples
Input
2
1 2
Output
2
1
Input
10
1 2 1 3 2 4 2 5 8 1
Output
10
7
0
4
0
3
0
2
3
0
Submitted Solution:
```
N = int(input())
A = list(map(int, input().split()))
# 現在末尾に追加される山より若い山で，最も個数の大きいものが次の山になる
# 他の山を次の山の個数以下まで減らす
ans = [0] * N
numList = [(a, i) for i, a in enumerate(A)]
numList.sort(reverse=True)
numList.append((0, 0))
y = float('inf')
for c, (a, i) in enumerate(numList[: N], start=1):
y = min(y, i)
ans[y] += c * (a - numList[c][0])
print(*ans, sep='\n')
``` | instruction | 0 | 74,080 | 16 | 148,160 |
Yes | output | 1 | 74,080 | 16 | 148,161 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke loves constructing integer sequences.
There are N piles of stones, numbered 1 through N. The pile numbered i consists of a_i stones.
Snuke will construct an integer sequence s of length Σa_i, as follows:
1. Among the piles with the largest number of stones remaining, let x be the index of the pile with the smallest index. Append x to the end of s.
2. Select a pile with one or more stones remaining, and remove a stone from that pile.
3. If there is a pile with one or more stones remaining, go back to step 1. Otherwise, terminate the process.
We are interested in the lexicographically smallest sequence that can be constructed. For each of the integers 1,2,3,...,N, how many times does it occur in the lexicographically smallest sequence?
Constraints
* 1 ≤ N ≤ 10^{5}
* 1 ≤ a_i ≤ 10^{9}
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{N}
Output
Print N lines. The i-th line should contain the number of the occurrences of the integer i in the lexicographically smallest sequence that can be constructed.
Examples
Input
2
1 2
Output
2
1
Input
10
1 2 1 3 2 4 2 5 8 1
Output
10
7
0
4
0
3
0
2
3
0
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
def readln():
_res = list(map(int,str(input()).split(' ')))
return _res
n = int(input())
a = readln()
c = sorted([[-a[i],i]for i in range(0,n)])
ans = [0 for i in range(0,n)]
front = c[0][1]
for i in range(1,n):
ans[front] = ans[front] + i * (c[i][0] - c[i-1][0])
if front > c[i][1]:
front = c[i][1]
ans[front] = ans[front] + n * (-c[n-1][0])
for a in ans:
print(a)
``` | instruction | 0 | 74,081 | 16 | 148,162 |
Yes | output | 1 | 74,081 | 16 | 148,163 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke loves constructing integer sequences.
There are N piles of stones, numbered 1 through N. The pile numbered i consists of a_i stones.
Snuke will construct an integer sequence s of length Σa_i, as follows:
1. Among the piles with the largest number of stones remaining, let x be the index of the pile with the smallest index. Append x to the end of s.
2. Select a pile with one or more stones remaining, and remove a stone from that pile.
3. If there is a pile with one or more stones remaining, go back to step 1. Otherwise, terminate the process.
We are interested in the lexicographically smallest sequence that can be constructed. For each of the integers 1,2,3,...,N, how many times does it occur in the lexicographically smallest sequence?
Constraints
* 1 ≤ N ≤ 10^{5}
* 1 ≤ a_i ≤ 10^{9}
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{N}
Output
Print N lines. The i-th line should contain the number of the occurrences of the integer i in the lexicographically smallest sequence that can be constructed.
Examples
Input
2
1 2
Output
2
1
Input
10
1 2 1 3 2 4 2 5 8 1
Output
10
7
0
4
0
3
0
2
3
0
Submitted Solution:
```
n = int(input())
A = list(map(int, input().split()))
from collections import defaultdict
d = defaultdict(lambda: (n, 0))
for i, a in enumerate(A):
min_id, cnt = d[a]
d[a] = (min(min_id, i), cnt+1)
B = list(d.items())
B.sort()
#print(B)
ans = [0]*n
while True:
a1, (m1, c1) = B.pop()
if not B:
ans[m1] += c1*a1
break
a2, (m2, c2) = B[-1]
ans[m1] += c1*(a1-a2)
B[-1] = (a2, (min(m1, m2), c1+c2))
print(*ans, sep='\n')
``` | instruction | 0 | 74,082 | 16 | 148,164 |
Yes | output | 1 | 74,082 | 16 | 148,165 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke loves constructing integer sequences.
There are N piles of stones, numbered 1 through N. The pile numbered i consists of a_i stones.
Snuke will construct an integer sequence s of length Σa_i, as follows:
1. Among the piles with the largest number of stones remaining, let x be the index of the pile with the smallest index. Append x to the end of s.
2. Select a pile with one or more stones remaining, and remove a stone from that pile.
3. If there is a pile with one or more stones remaining, go back to step 1. Otherwise, terminate the process.
We are interested in the lexicographically smallest sequence that can be constructed. For each of the integers 1,2,3,...,N, how many times does it occur in the lexicographically smallest sequence?
Constraints
* 1 ≤ N ≤ 10^{5}
* 1 ≤ a_i ≤ 10^{9}
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{N}
Output
Print N lines. The i-th line should contain the number of the occurrences of the integer i in the lexicographically smallest sequence that can be constructed.
Examples
Input
2
1 2
Output
2
1
Input
10
1 2 1 3 2 4 2 5 8 1
Output
10
7
0
4
0
3
0
2
3
0
Submitted Solution:
```
from collections import deque
def inpl(): return list(map(int, input().split()))
N = int(input())
A = inpl()
X = sorted([(a, i) for i, a in enumerate(A, start=1)], key=lambda x: [x[0], -x[1]])
answer = [0]*(N+1)
a, i = X.pop()
for c in range(1, N):
b, j = X.pop()
answer[i] += (a-b)*c
i = min(i, j)
a = b
answer[i] += a * N
print(*answer[1:], sep="\n")
``` | instruction | 0 | 74,083 | 16 | 148,166 |
Yes | output | 1 | 74,083 | 16 | 148,167 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke loves constructing integer sequences.
There are N piles of stones, numbered 1 through N. The pile numbered i consists of a_i stones.
Snuke will construct an integer sequence s of length Σa_i, as follows:
1. Among the piles with the largest number of stones remaining, let x be the index of the pile with the smallest index. Append x to the end of s.
2. Select a pile with one or more stones remaining, and remove a stone from that pile.
3. If there is a pile with one or more stones remaining, go back to step 1. Otherwise, terminate the process.
We are interested in the lexicographically smallest sequence that can be constructed. For each of the integers 1,2,3,...,N, how many times does it occur in the lexicographically smallest sequence?
Constraints
* 1 ≤ N ≤ 10^{5}
* 1 ≤ a_i ≤ 10^{9}
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{N}
Output
Print N lines. The i-th line should contain the number of the occurrences of the integer i in the lexicographically smallest sequence that can be constructed.
Examples
Input
2
1 2
Output
2
1
Input
10
1 2 1 3 2 4 2 5 8 1
Output
10
7
0
4
0
3
0
2
3
0
Submitted Solution:
```
import sys
import math
from collections import defaultdict
from bisect import bisect_left, bisect_right
sys.setrecursionlimit(10**7)
def input():
return sys.stdin.readline()[:-1]
mod = 10**9 + 7
def I(): return int(input())
def LI(): return list(map(int, input().split()))
def LIR(row,col):
if row <= 0:
return [[] for _ in range(col)]
elif col == 1:
return [I() for _ in range(row)]
else:
read_all = [LI() for _ in range(row)]
return map(list, zip(*read_all))
#################
N = I()
a = LI()
asort = sorted(a)
ans = [0]*N
max_ = 0
for i in range(N):
if a[i] > max_:
diff = a[i]-max_
num = N-bisect_left(asort,a[i])
ans[i] = num*diff
max_ = a[i]
ans[0] = sum(a)-sum(ans[1:])
for i in range(N):
print(ans[i])
``` | instruction | 0 | 74,084 | 16 | 148,168 |
No | output | 1 | 74,084 | 16 | 148,169 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke loves constructing integer sequences.
There are N piles of stones, numbered 1 through N. The pile numbered i consists of a_i stones.
Snuke will construct an integer sequence s of length Σa_i, as follows:
1. Among the piles with the largest number of stones remaining, let x be the index of the pile with the smallest index. Append x to the end of s.
2. Select a pile with one or more stones remaining, and remove a stone from that pile.
3. If there is a pile with one or more stones remaining, go back to step 1. Otherwise, terminate the process.
We are interested in the lexicographically smallest sequence that can be constructed. For each of the integers 1,2,3,...,N, how many times does it occur in the lexicographically smallest sequence?
Constraints
* 1 ≤ N ≤ 10^{5}
* 1 ≤ a_i ≤ 10^{9}
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{N}
Output
Print N lines. The i-th line should contain the number of the occurrences of the integer i in the lexicographically smallest sequence that can be constructed.
Examples
Input
2
1 2
Output
2
1
Input
10
1 2 1 3 2 4 2 5 8 1
Output
10
7
0
4
0
3
0
2
3
0
Submitted Solution:
```
from functools import cmp_to_key
n = int(input())
a = list(map(int, input().split()))
def cmp(x, y):
return x[1] - y[1]
def group(a):
d = {}
for i, x in enumerate(a):
d.setdefault(x, []).append(i)
return list(sorted(d.items(), key=cmp_to_key(lambda x, y: x[0] - y[0]), reverse=True))
ans = [0] * len(a)
g = group(a)
g.append((0, []))
for c, n in zip(g[:-1], g[1:]):
ans[c[1][0]] = (c[0] - n[0]) * len(c[1])
if n[1] and n[1][0] < c[1][0]:
tmp = n[1].copy()
n[1].clear()
n[1].extend(tmp + c[1])
else:
tmp = n[1].copy()
n[1].clear()
n[1].extend(c[1] + tmp)
[print(a) for a in ans]
``` | instruction | 0 | 74,085 | 16 | 148,170 |
No | output | 1 | 74,085 | 16 | 148,171 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke loves constructing integer sequences.
There are N piles of stones, numbered 1 through N. The pile numbered i consists of a_i stones.
Snuke will construct an integer sequence s of length Σa_i, as follows:
1. Among the piles with the largest number of stones remaining, let x be the index of the pile with the smallest index. Append x to the end of s.
2. Select a pile with one or more stones remaining, and remove a stone from that pile.
3. If there is a pile with one or more stones remaining, go back to step 1. Otherwise, terminate the process.
We are interested in the lexicographically smallest sequence that can be constructed. For each of the integers 1,2,3,...,N, how many times does it occur in the lexicographically smallest sequence?
Constraints
* 1 ≤ N ≤ 10^{5}
* 1 ≤ a_i ≤ 10^{9}
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{N}
Output
Print N lines. The i-th line should contain the number of the occurrences of the integer i in the lexicographically smallest sequence that can be constructed.
Examples
Input
2
1 2
Output
2
1
Input
10
1 2 1 3 2 4 2 5 8 1
Output
10
7
0
4
0
3
0
2
3
0
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
df = []
ans = []
m = 0
for i in range(n):
if m < a[i]:
ans.append([i,0])
df.append([i,a[i]-m])
m = a[i]
tmp = a[i]
j = 0
while tmp > 0:
ans[j][1] += min(tmp,df[j][1])
tmp -= df[j][1]
j += 1
if j == len(df):
break
else:
j = 0
tmp = a[i]
while tmp > 0:
ans[j][1] += min(tmp,df[j][1])
tmp -= df[j][1]
j += 1
if j == len(df):
break
j = 0
for i in range(n):
if i == ans[j][0]:
print(ans[j][1])
if j < len(ans)-1:
j += 1
else:
print(0)
``` | instruction | 0 | 74,086 | 16 | 148,172 |
No | output | 1 | 74,086 | 16 | 148,173 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke loves constructing integer sequences.
There are N piles of stones, numbered 1 through N. The pile numbered i consists of a_i stones.
Snuke will construct an integer sequence s of length Σa_i, as follows:
1. Among the piles with the largest number of stones remaining, let x be the index of the pile with the smallest index. Append x to the end of s.
2. Select a pile with one or more stones remaining, and remove a stone from that pile.
3. If there is a pile with one or more stones remaining, go back to step 1. Otherwise, terminate the process.
We are interested in the lexicographically smallest sequence that can be constructed. For each of the integers 1,2,3,...,N, how many times does it occur in the lexicographically smallest sequence?
Constraints
* 1 ≤ N ≤ 10^{5}
* 1 ≤ a_i ≤ 10^{9}
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{N}
Output
Print N lines. The i-th line should contain the number of the occurrences of the integer i in the lexicographically smallest sequence that can be constructed.
Examples
Input
2
1 2
Output
2
1
Input
10
1 2 1 3 2 4 2 5 8 1
Output
10
7
0
4
0
3
0
2
3
0
Submitted Solution:
```
N = int(input())
A = map(int, input().split())
S = sorted([[a, i] for i, a in enumerate(A)], key=lambda x: x[0], reverse=True)
B = [0 for i in range(N)]
B[0] = N
num = S[0][1]
for i in range(1, N):
if S[i - 1][0] > S[i][0]:
B[num] += (S[i - 1][0] - S[i][0]) * i
num = min(num, S[i][1])
for b in B:
print(b)
``` | instruction | 0 | 74,087 | 16 | 148,174 |
No | output | 1 | 74,087 | 16 | 148,175 |
Provide a correct Python 3 solution for this coding contest problem.
N cells are arranged in a row. Some of them may contain tokens. You are given a string s that consists of `0`s and `1`s. If the i-th character of s is `1`, the i-th cell (from left) contains a token. Otherwise, it doesn't contain a token.
Snuke wants to perform the following operation as many times as possible. In each operation, he chooses three consecutive cells. Let's call the cells X, Y, Z from left to right. In order for the operation to be valid, both X and Z must contain tokens and Y must not contain a token. Then, he removes these two tokens and puts a new token on Y.
How many operations can he perform if he performs operations in the optimal way?
Constraints
* 1 \leq N \leq 500,000
* |s| = N
* Each character in s is either `0` or `1`.
Input
Input is given from Standard Input in the following format:
N
s
Output
Print the answer.
Examples
Input
7
1010101
Output
2
Input
50
10101000010011011110001001111110000101010111100110
Output
10 | instruction | 0 | 77,437 | 16 | 154,874 |
"Correct Solution:
```
N = int(input())
S = [x == '1' for x in input()]
S.append(0)
ans = 0
now = 0
def dp(arr):
L = len(arr)
if L == 1:
return 0
dp = [[0] * 4 for _ in range(L)]
dp[0] = [-1, arr[0], -1, 0]
ret = max(dp[0])
for i in range(L - 1):
dp[i + 1][0] = dp[i][3] + arr[i + 1]
if dp[i][0] != -1 and arr[i] > 1:
dp[i + 1][0] = max(dp[i + 1][0], dp[i][0] + arr[i + 1] - 1)
if dp[i][2] != -1 and arr[i] > 1:
dp[i + 1][0] = max(dp[i + 1][0], dp[i][2] + arr[i + 1])
dp[i + 1][1] = dp[i][3] + arr[i + 1]
if dp[i][0] != -1 and i != L - 2:
dp[i + 1][1] = max(dp[i + 1][1], dp[i][0] + arr[i + 1])
if dp[i][1] != -1 and arr[i + 1] > 1 and i != L - 2:
dp[i + 1][1] = max(dp[i + 1][1], dp[i][1] + arr[i + 1] - 1)
if dp[i][2] != -1 and i != L - 2:
dp[i + 1][1] = max(dp[i + 1][1], dp[i][2] + arr[i + 1])
dp[i + 1][2] = dp[i][1]
dp[i + 1][3] = max([dp[i][0], dp[i][3], dp[i][2]])
ret = max(ret, max(dp[i + 1]))
#print(*dp, sep='\n')
return ret
while now < N:
if S[now] == 1:
left = now
cnt_1 = 1
prev = 1
now += 1
arr = []
while now < N and (prev != 0 or S[now] != 0):
if S[now] == 0:
arr.append(cnt_1)
cnt_1 = 0
else:
cnt_1 += 1
prev = S[now]
now += 1
if cnt_1:
arr.append(cnt_1)
if now == N:
if prev == 0:
right = N - 2
else:
right = N - 1
else:
right = now - 2
#print('#', arr)
ans += dp(arr)
while now < N and S[now] == 0:
now += 1
else:
now += 1
print(ans)
``` | output | 1 | 77,437 | 16 | 154,875 |
Provide a correct Python 3 solution for this coding contest problem.
N cells are arranged in a row. Some of them may contain tokens. You are given a string s that consists of `0`s and `1`s. If the i-th character of s is `1`, the i-th cell (from left) contains a token. Otherwise, it doesn't contain a token.
Snuke wants to perform the following operation as many times as possible. In each operation, he chooses three consecutive cells. Let's call the cells X, Y, Z from left to right. In order for the operation to be valid, both X and Z must contain tokens and Y must not contain a token. Then, he removes these two tokens and puts a new token on Y.
How many operations can he perform if he performs operations in the optimal way?
Constraints
* 1 \leq N \leq 500,000
* |s| = N
* Each character in s is either `0` or `1`.
Input
Input is given from Standard Input in the following format:
N
s
Output
Print the answer.
Examples
Input
7
1010101
Output
2
Input
50
10101000010011011110001001111110000101010111100110
Output
10 | instruction | 0 | 77,438 | 16 | 154,876 |
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
N = int(readline())
S = readline().rstrip().decode('utf-8')
def solve_partial(S):
INF = 10**18
"""
・Sは1から始まり、1で終わる
・Sは00を含まない
・したがって、Sは1,01に分解可能
・残る最小個数を調べるdp。これは、1, 101111,111101 の3種を数えることと同じ
・a, b0cccc, dddd0e として、「現在の1がどれであるか -> 最小個数」でdp
・個数はa,b,eのときに数える
"""
S = S.replace('01','2')
a,b,c,d,e = 1,1,INF,0,INF
for x in S[1:]:
if x == '1':
a2 = min(a,c,e)+1
b2 = min(a,c,e)+1
c2 = c
d2 = min(a,c,d,e)
e2 = INF
else:
a2 = min(a,c,e)+1
b2 = min(a,c,e)+1
c2 = b
d2 = min(a,c,e)
e2 = d+1
a,b,c,d,e = a2,b2,c2,d2,e2
return len(S)-min(a,c,e)
answer = 0
for x in S.split('00'):
x = x.strip('0')
if x:
answer += solve_partial(x)
print(answer)
``` | output | 1 | 77,438 | 16 | 154,877 |
Provide a correct Python 3 solution for this coding contest problem.
N cells are arranged in a row. Some of them may contain tokens. You are given a string s that consists of `0`s and `1`s. If the i-th character of s is `1`, the i-th cell (from left) contains a token. Otherwise, it doesn't contain a token.
Snuke wants to perform the following operation as many times as possible. In each operation, he chooses three consecutive cells. Let's call the cells X, Y, Z from left to right. In order for the operation to be valid, both X and Z must contain tokens and Y must not contain a token. Then, he removes these two tokens and puts a new token on Y.
How many operations can he perform if he performs operations in the optimal way?
Constraints
* 1 \leq N \leq 500,000
* |s| = N
* Each character in s is either `0` or `1`.
Input
Input is given from Standard Input in the following format:
N
s
Output
Print the answer.
Examples
Input
7
1010101
Output
2
Input
50
10101000010011011110001001111110000101010111100110
Output
10 | instruction | 0 | 77,439 | 16 | 154,878 |
"Correct Solution:
```
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor
from operator import mul
from functools import reduce
sys.setrecursionlimit(2147483647)
INF = 10 ** 13
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.readline().rstrip().split()
def S(): return sys.stdin.readline().rstrip()
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 10 ** 9 + 7
n = I()
s = list(S())
ans = 0
one_cnt = 0
z_cnt = 0
dp = []
flg = 0
for i in range(n):
if s[i] == "1":
one_cnt += 1
z_cnt = 0
if flg:
if dp:
ans += max(dp)
dp = []
flg = 0
if i == n - 1 or s[i + 1] == "0":
# 手付かず、左端だけ消した、右端だけ残す、全て消した。
if dp:
if one_cnt >= 2:
ndp = [max(dp), max(dp[0] + pre, dp[1] + pre - 1, dp[2] + 1),
max(dp[0], dp[1], dp[2]) + one_cnt - 1,
max(dp[0], dp[1], dp[2]) + one_cnt]
else:
ndp = [-INF, -INF, max(dp),
max(dp[0] + pre, dp[1] + pre - 1, dp[2] + 1)]
dp = ndp
else:
dp = [0, -INF, -INF, -INF]
pre = one_cnt
else:
z_cnt += 1
one_cnt = 0
if z_cnt >= 2:
flg = 1
if dp:
ans += max(dp)
print(ans)
``` | output | 1 | 77,439 | 16 | 154,879 |
Provide a correct Python 3 solution for this coding contest problem.
N cells are arranged in a row. Some of them may contain tokens. You are given a string s that consists of `0`s and `1`s. If the i-th character of s is `1`, the i-th cell (from left) contains a token. Otherwise, it doesn't contain a token.
Snuke wants to perform the following operation as many times as possible. In each operation, he chooses three consecutive cells. Let's call the cells X, Y, Z from left to right. In order for the operation to be valid, both X and Z must contain tokens and Y must not contain a token. Then, he removes these two tokens and puts a new token on Y.
How many operations can he perform if he performs operations in the optimal way?
Constraints
* 1 \leq N \leq 500,000
* |s| = N
* Each character in s is either `0` or `1`.
Input
Input is given from Standard Input in the following format:
N
s
Output
Print the answer.
Examples
Input
7
1010101
Output
2
Input
50
10101000010011011110001001111110000101010111100110
Output
10 | instruction | 0 | 77,440 | 16 | 154,880 |
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
N = int(readline())
S = readline().rstrip().decode('utf-8')
def solve_partial(S):
INF = 10**18
"""
・Sは1から始まり、1で終わる
・Sは00を含まない
・したがって、Sは1,01に分解可能
・残る最小個数を調べるdp。これは、1, 101111,111101 の3種を数えることと同じ
・a, b0cccc, dddd0e として、「現在の1がどれであるか -> 最小個数」でdp
・個数はa,b,eのときに数える
"""
S = S.replace('01','2')
a,b,c,d,e = 1,1,INF,0,INF
for x in S[1:]:
if x == '1':
a2 = min(a,c,e)+1
c2 = c
d2 = min(a,c,d,e)
e2 = INF
else:
a2 = min(a,c,e)+1
c2 = a
d2 = min(a,c,e)
e2 = d+1
a,c,d,e = a2,c2,d2,e2
return len(S)-min(a,c,e)
answer = 0
for x in S.split('00'):
x = x.strip('0')
if x:
answer += solve_partial(x)
print(answer)
``` | output | 1 | 77,440 | 16 | 154,881 |
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