message stringlengths 2 45.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 254 108k | cluster float64 3 3 | __index_level_0__ int64 508 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are developing frog-shaped robots, and decided to race them against each other.
First, you placed N robots onto a number line. These robots are numbered 1 through N. The current coordinate of robot i is x_i. Here, all x_i are integers, and 0 < x_1 < x_2 < ... < x_N.
You will repeatedly perform the following operation:
* Select a robot on the number line. Let the coordinate of the robot be x. Select the destination coordinate, either x-1 or x-2, that is not occupied by another robot. The robot now jumps to the selected coordinate.
When the coordinate of a robot becomes 0 or less, the robot is considered finished and will be removed from the number line immediately. You will repeat the operation until all the robots finish the race.
Depending on your choice in the operation, the N robots can finish the race in different orders. In how many different orders can the N robots finish the race? Find the answer modulo 10^9+7.
Constraints
* 2 ≤ N ≤ 10^5
* x_i is an integer.
* 0 < x_1 < x_2 < ... < x_N ≤ 10^9
Input
The input is given from Standard Input in the following format:
N
x_1 x_2 ... x_N
Output
Print the number of the different orders in which the N robots can finish the race, modulo 10^9+7.
Examples
Input
3
1 2 3
Output
4
Input
3
2 3 4
Output
6
Input
8
1 2 3 5 7 11 13 17
Output
10080
Input
13
4 6 8 9 10 12 14 15 16 18 20 21 22
Output
311014372
Submitted Solution:
```
N = int(input())
X = list(map(int,input().split()))
MOD = 10**9+7
ans = 1
robots = 0
for x in X:
if x > 2*robots:
robots += 1
else:
ans *= (robots+1)
ans %= MOD
for n in range(1,robots+1):
ans *= n
ans %= MOD
print(ans)
``` | instruction | 0 | 64,698 | 3 | 129,396 |
Yes | output | 1 | 64,698 | 3 | 129,397 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are developing frog-shaped robots, and decided to race them against each other.
First, you placed N robots onto a number line. These robots are numbered 1 through N. The current coordinate of robot i is x_i. Here, all x_i are integers, and 0 < x_1 < x_2 < ... < x_N.
You will repeatedly perform the following operation:
* Select a robot on the number line. Let the coordinate of the robot be x. Select the destination coordinate, either x-1 or x-2, that is not occupied by another robot. The robot now jumps to the selected coordinate.
When the coordinate of a robot becomes 0 or less, the robot is considered finished and will be removed from the number line immediately. You will repeat the operation until all the robots finish the race.
Depending on your choice in the operation, the N robots can finish the race in different orders. In how many different orders can the N robots finish the race? Find the answer modulo 10^9+7.
Constraints
* 2 ≤ N ≤ 10^5
* x_i is an integer.
* 0 < x_1 < x_2 < ... < x_N ≤ 10^9
Input
The input is given from Standard Input in the following format:
N
x_1 x_2 ... x_N
Output
Print the number of the different orders in which the N robots can finish the race, modulo 10^9+7.
Examples
Input
3
1 2 3
Output
4
Input
3
2 3 4
Output
6
Input
8
1 2 3 5 7 11 13 17
Output
10080
Input
13
4 6 8 9 10 12 14 15 16 18 20 21 22
Output
311014372
Submitted Solution:
```
n = int(input())
count = [0]
cur = 0
data = list(map(int, input().split()))
for i,v in enumerate(data):
count[-1] += 1
if v < 2*(i-cur)+1:
cur += 1
count.append(0)
res = 1
cur = 1
for i in range(n):
cur -= 1
if count: cur += count.pop(0)
res *= cur
res %= 1000000007
print(res)
``` | instruction | 0 | 64,699 | 3 | 129,398 |
Yes | output | 1 | 64,699 | 3 | 129,399 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are developing frog-shaped robots, and decided to race them against each other.
First, you placed N robots onto a number line. These robots are numbered 1 through N. The current coordinate of robot i is x_i. Here, all x_i are integers, and 0 < x_1 < x_2 < ... < x_N.
You will repeatedly perform the following operation:
* Select a robot on the number line. Let the coordinate of the robot be x. Select the destination coordinate, either x-1 or x-2, that is not occupied by another robot. The robot now jumps to the selected coordinate.
When the coordinate of a robot becomes 0 or less, the robot is considered finished and will be removed from the number line immediately. You will repeat the operation until all the robots finish the race.
Depending on your choice in the operation, the N robots can finish the race in different orders. In how many different orders can the N robots finish the race? Find the answer modulo 10^9+7.
Constraints
* 2 ≤ N ≤ 10^5
* x_i is an integer.
* 0 < x_1 < x_2 < ... < x_N ≤ 10^9
Input
The input is given from Standard Input in the following format:
N
x_1 x_2 ... x_N
Output
Print the number of the different orders in which the N robots can finish the race, modulo 10^9+7.
Examples
Input
3
1 2 3
Output
4
Input
3
2 3 4
Output
6
Input
8
1 2 3 5 7 11 13 17
Output
10080
Input
13
4 6 8 9 10 12 14 15 16 18 20 21 22
Output
311014372
Submitted Solution:
```
MOD = 10**9+7
n = int(input())
a = list(map(int, input().split()))
if n == 2:
print(2)
else:
ans = 1
s = []
for x in a:
if x < len(s)*2 + 1:
ans *= len(s)+1
ans %= MOD
else:
s.append(x)
for i in range(len(s)):
ans *= i+1
ans %= MOD
print(ans)
``` | instruction | 0 | 64,700 | 3 | 129,400 |
Yes | output | 1 | 64,700 | 3 | 129,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are developing frog-shaped robots, and decided to race them against each other.
First, you placed N robots onto a number line. These robots are numbered 1 through N. The current coordinate of robot i is x_i. Here, all x_i are integers, and 0 < x_1 < x_2 < ... < x_N.
You will repeatedly perform the following operation:
* Select a robot on the number line. Let the coordinate of the robot be x. Select the destination coordinate, either x-1 or x-2, that is not occupied by another robot. The robot now jumps to the selected coordinate.
When the coordinate of a robot becomes 0 or less, the robot is considered finished and will be removed from the number line immediately. You will repeat the operation until all the robots finish the race.
Depending on your choice in the operation, the N robots can finish the race in different orders. In how many different orders can the N robots finish the race? Find the answer modulo 10^9+7.
Constraints
* 2 ≤ N ≤ 10^5
* x_i is an integer.
* 0 < x_1 < x_2 < ... < x_N ≤ 10^9
Input
The input is given from Standard Input in the following format:
N
x_1 x_2 ... x_N
Output
Print the number of the different orders in which the N robots can finish the race, modulo 10^9+7.
Examples
Input
3
1 2 3
Output
4
Input
3
2 3 4
Output
6
Input
8
1 2 3 5 7 11 13 17
Output
10080
Input
13
4 6 8 9 10 12 14 15 16 18 20 21 22
Output
311014372
Submitted Solution:
```
import math
MOD = 10**9+7
N = int(input())
X = list(map(int,input().split()))
Y = [0]*N
Y[0] += 1
X[0] = 1
hoge = 0
for i in range(1,N):
X[i] = min(X[i-1]+2,X[i])
Y[hoge] += 1
if X[i]-X[i-1] == 1:
hoge += 1
X[i] = X[i-1]
ans = 1
mem = 0
for i in range(N):
mem += Y[i]
ans *= mem
ans %= MOD
mem -= 1
print(ans)
``` | instruction | 0 | 64,701 | 3 | 129,402 |
Yes | output | 1 | 64,701 | 3 | 129,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are developing frog-shaped robots, and decided to race them against each other.
First, you placed N robots onto a number line. These robots are numbered 1 through N. The current coordinate of robot i is x_i. Here, all x_i are integers, and 0 < x_1 < x_2 < ... < x_N.
You will repeatedly perform the following operation:
* Select a robot on the number line. Let the coordinate of the robot be x. Select the destination coordinate, either x-1 or x-2, that is not occupied by another robot. The robot now jumps to the selected coordinate.
When the coordinate of a robot becomes 0 or less, the robot is considered finished and will be removed from the number line immediately. You will repeat the operation until all the robots finish the race.
Depending on your choice in the operation, the N robots can finish the race in different orders. In how many different orders can the N robots finish the race? Find the answer modulo 10^9+7.
Constraints
* 2 ≤ N ≤ 10^5
* x_i is an integer.
* 0 < x_1 < x_2 < ... < x_N ≤ 10^9
Input
The input is given from Standard Input in the following format:
N
x_1 x_2 ... x_N
Output
Print the number of the different orders in which the N robots can finish the race, modulo 10^9+7.
Examples
Input
3
1 2 3
Output
4
Input
3
2 3 4
Output
6
Input
8
1 2 3 5 7 11 13 17
Output
10080
Input
13
4 6 8 9 10 12 14 15 16 18 20 21 22
Output
311014372
Submitted Solution:
```
mod = 10**9+7
N = int(input())
x = list(map(int, input().split()))
ans = 1
cnt = 1
now = 0
while x[now]>=cnt*2-1:
cnt+=1
now+=1
ans*=cnt
ans%=mod
for i in range(1, N):
cnt-=1
while now<N-1 and x[now]>=cnt*2-1:
cnt+=1
now+=1
ans*=cnt
ans%=mod
print(ans)
``` | instruction | 0 | 64,702 | 3 | 129,404 |
No | output | 1 | 64,702 | 3 | 129,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are developing frog-shaped robots, and decided to race them against each other.
First, you placed N robots onto a number line. These robots are numbered 1 through N. The current coordinate of robot i is x_i. Here, all x_i are integers, and 0 < x_1 < x_2 < ... < x_N.
You will repeatedly perform the following operation:
* Select a robot on the number line. Let the coordinate of the robot be x. Select the destination coordinate, either x-1 or x-2, that is not occupied by another robot. The robot now jumps to the selected coordinate.
When the coordinate of a robot becomes 0 or less, the robot is considered finished and will be removed from the number line immediately. You will repeat the operation until all the robots finish the race.
Depending on your choice in the operation, the N robots can finish the race in different orders. In how many different orders can the N robots finish the race? Find the answer modulo 10^9+7.
Constraints
* 2 ≤ N ≤ 10^5
* x_i is an integer.
* 0 < x_1 < x_2 < ... < x_N ≤ 10^9
Input
The input is given from Standard Input in the following format:
N
x_1 x_2 ... x_N
Output
Print the number of the different orders in which the N robots can finish the race, modulo 10^9+7.
Examples
Input
3
1 2 3
Output
4
Input
3
2 3 4
Output
6
Input
8
1 2 3 5 7 11 13 17
Output
10080
Input
13
4 6 8 9 10 12 14 15 16 18 20 21 22
Output
311014372
Submitted Solution:
```
N = int(input())
p_list = [int(x) for x in input().strip().split()]
pat = 1
sum = 0
for i, r_pos in enumerate(p_list):
pat *= sum + 1
if r_pos >= 2 * sum + 1:
sum += 1
print(pat % 1000000007)
``` | instruction | 0 | 64,703 | 3 | 129,406 |
No | output | 1 | 64,703 | 3 | 129,407 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are developing frog-shaped robots, and decided to race them against each other.
First, you placed N robots onto a number line. These robots are numbered 1 through N. The current coordinate of robot i is x_i. Here, all x_i are integers, and 0 < x_1 < x_2 < ... < x_N.
You will repeatedly perform the following operation:
* Select a robot on the number line. Let the coordinate of the robot be x. Select the destination coordinate, either x-1 or x-2, that is not occupied by another robot. The robot now jumps to the selected coordinate.
When the coordinate of a robot becomes 0 or less, the robot is considered finished and will be removed from the number line immediately. You will repeat the operation until all the robots finish the race.
Depending on your choice in the operation, the N robots can finish the race in different orders. In how many different orders can the N robots finish the race? Find the answer modulo 10^9+7.
Constraints
* 2 ≤ N ≤ 10^5
* x_i is an integer.
* 0 < x_1 < x_2 < ... < x_N ≤ 10^9
Input
The input is given from Standard Input in the following format:
N
x_1 x_2 ... x_N
Output
Print the number of the different orders in which the N robots can finish the race, modulo 10^9+7.
Examples
Input
3
1 2 3
Output
4
Input
3
2 3 4
Output
6
Input
8
1 2 3 5 7 11 13 17
Output
10080
Input
13
4 6 8 9 10 12 14 15 16 18 20 21 22
Output
311014372
Submitted Solution:
```
def rec(x, xofs, goal):
num_of_rest_x = len(x) - (xofs + 1)
if num_of_rest_x == 0:
return 1
elif num_of_rest_x == 1:
return 2
elif x[xofs] == goal + 1 and x[xofs + 1] == goal + 2:
return num_of_rest_x * rec(x, xofs + 2, goal + 2) * 2
elif x[xofs] == goal + 1 or x[xofs] == goal + 2:
return (num_of_rest_x + 1) * rec(x, xofs + 1, goal + 2)
else:
return rec(x, xofs, goal + 1)
n = int(input())
x = list(map(int, input().split()))
if n <= 8:
print(rec(x, 0, 0) % (1000000000 + 7))
else:
print(1)
``` | instruction | 0 | 64,704 | 3 | 129,408 |
No | output | 1 | 64,704 | 3 | 129,409 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are developing frog-shaped robots, and decided to race them against each other.
First, you placed N robots onto a number line. These robots are numbered 1 through N. The current coordinate of robot i is x_i. Here, all x_i are integers, and 0 < x_1 < x_2 < ... < x_N.
You will repeatedly perform the following operation:
* Select a robot on the number line. Let the coordinate of the robot be x. Select the destination coordinate, either x-1 or x-2, that is not occupied by another robot. The robot now jumps to the selected coordinate.
When the coordinate of a robot becomes 0 or less, the robot is considered finished and will be removed from the number line immediately. You will repeat the operation until all the robots finish the race.
Depending on your choice in the operation, the N robots can finish the race in different orders. In how many different orders can the N robots finish the race? Find the answer modulo 10^9+7.
Constraints
* 2 ≤ N ≤ 10^5
* x_i is an integer.
* 0 < x_1 < x_2 < ... < x_N ≤ 10^9
Input
The input is given from Standard Input in the following format:
N
x_1 x_2 ... x_N
Output
Print the number of the different orders in which the N robots can finish the race, modulo 10^9+7.
Examples
Input
3
1 2 3
Output
4
Input
3
2 3 4
Output
6
Input
8
1 2 3 5 7 11 13 17
Output
10080
Input
13
4 6 8 9 10 12 14 15 16 18 20 21 22
Output
311014372
Submitted Solution:
```
def rec(x, xofs, goal):
num_of_x = len(x) - xofs
if num_of_x == 1:
return 1
elif num_of_x == 2:
return 2
elif x[xofs] == goal + 1 and x[xofs + 1] == goal + 2:
return (num_of_x - 2 + 1) * rec(x, xofs + 2, goal + 2) * 2
elif x[xofs] == goal + 1 or x[xofs] == goal + 2:
return (num_of_x - 1 + 1) * rec(x, xofs + 1, goal + 2)
else:
return rec(x, xofs, x[xofs] - 2)
n = int(input())
x = list(map(int, input().split()))
if n <= 8:
print(rec(x, 0, 0) % (1000000000 + 7))
else:
print(1)
``` | instruction | 0 | 64,705 | 3 | 129,410 |
No | output | 1 | 64,705 | 3 | 129,411 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0.
Submitted Solution:
```
from collections import defaultdict
for i in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
a.sort()
ans=0
presum=0
for j in range(1,n):
presum+=a[j-1]
x=a[j]-a[j-1]
y=(a[j]*j)-presum
ans+=x-y
print(ans)
``` | instruction | 0 | 65,116 | 3 | 130,232 |
Yes | output | 1 | 65,116 | 3 | 130,233 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0.
Submitted Solution:
```
if __name__ == '__main__':
t = int(input())
for i in range(t):
n = int(input())
lis = list(map(int,input().split()))
lis.sort()
s = 0
for i in range(n-1):
x = lis[i+1] -lis[i]
s -= (i+1)*(n-i -1)*x
s += x
print(s)
``` | instruction | 0 | 65,117 | 3 | 130,234 |
Yes | output | 1 | 65,117 | 3 | 130,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0.
Submitted Solution:
```
# region fastio # from https://codeforces.com/contest/1333/submission/75948789
import sys, io, os
BUFSIZE = 8192
class FastIO(io.IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = io.BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(io.IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#endregion
T = int(input())
for _ in range(T):
N = int(input())
D = sorted(map(int, input().split()))
c = 0
n = 0
ans = 0
for i in range(2, N):
d = D[i]
c += D[i-2]
n += 1
ans += n * d - c
print(-ans)
``` | instruction | 0 | 65,118 | 3 | 130,236 |
Yes | output | 1 | 65,118 | 3 | 130,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0.
Submitted Solution:
```
import traceback
import math
from collections import defaultdict, Counter
from functools import lru_cache
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def geti():
return int(input())
def gets():
return input()
def getil():
return list(map(int, input().split()))
def getsl():
return input().split()
def get2d(nrows, ncols, n=0):
return [[n] * ncols for r in range(nrows)]
from itertools import accumulate
def get_acc(a):
return list(accumulate(a))
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
inf = float('inf')
mod = 10 ** 9 + 7
def main():
N = geti()
a = getil()
a.sort()
acc = get_acc(a)
# tot = acc[-1]
ans = a[-1] - a[0]
for i in range(N):
ans += acc[i] - (i+1) * a[i]
# print(ans)
return ans
try:
Test = geti()
answers = []
for _ in range(Test):
ans = main()
# ans = 'Yes' if ans else 'No'
ans = str(ans)
answers.append(ans)
print('\n'.join(answers))
except Exception as e:
print(e)
traceback.print_exc()
``` | instruction | 0 | 65,119 | 3 | 130,238 |
Yes | output | 1 | 65,119 | 3 | 130,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0.
Submitted Solution:
```
# _
#####################################################################################################################
def main():
return f'-3\n0\n0'
if __name__ == '__main__':
print(main())
# main()
``` | instruction | 0 | 65,120 | 3 | 130,240 |
No | output | 1 | 65,120 | 3 | 130,241 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0.
Submitted Solution:
```
# region fastio # from https://codeforces.com/contest/1333/submission/75948789
import sys, io, os
BUFSIZE = 8192
class FastIO(io.IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = io.BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(io.IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#endregion
T = int(input())
for _ in range(T):
N = int(input())
D = list(map(int, input().split()))
c = 0
n = 0
ans = 0
for i in range(2, N):
d = D[i]
c += D[i-2]
n += 1
ans += n * d - c
print(-ans)
``` | instruction | 0 | 65,121 | 3 | 130,242 |
No | output | 1 | 65,121 | 3 | 130,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0.
Submitted Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
import io
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
from collections import Counter
import threading
#sys.setrecursionlimit(300000)
#threading.stack_size(10**8)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
#mod = 9223372036854775807
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: max(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
class SegmentTree1:
def __init__(self, data, default=10**6, func=lambda a, b: min(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
MOD=10**9+7
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
mod=10**9+7
omod=998244353
#-------------------------------------------------------------------------
prime = [True for i in range(10001)]
prime[0]=prime[1]=False
#pp=[0]*10000
def SieveOfEratosthenes(n=10000):
p = 2
c=0
while (p <= n):
if (prime[p] == True):
c+=1
for i in range(p, n+1, p):
#pp[i]=1
prime[i] = False
p += 1
#-----------------------------------DSU--------------------------------------------------
class DSU:
def __init__(self, R, C):
#R * C is the source, and isn't a grid square
self.par = range(R*C + 1)
self.rnk = [0] * (R*C + 1)
self.sz = [1] * (R*C + 1)
def find(self, x):
if self.par[x] != x:
self.par[x] = self.find(self.par[x])
return self.par[x]
def union(self, x, y):
xr, yr = self.find(x), self.find(y)
if xr == yr: return
if self.rnk[xr] < self.rnk[yr]:
xr, yr = yr, xr
if self.rnk[xr] == self.rnk[yr]:
self.rnk[xr] += 1
self.par[yr] = xr
self.sz[xr] += self.sz[yr]
def size(self, x):
return self.sz[self.find(x)]
def top(self):
# Size of component at ephemeral "source" node at index R*C,
# minus 1 to not count the source itself in the size
return self.size(len(self.sz) - 1) - 1
#---------------------------------Lazy Segment Tree--------------------------------------
# https://github.com/atcoder/ac-library/blob/master/atcoder/lazysegtree.hpp
class LazySegTree:
def __init__(self, _op, _e, _mapping, _composition, _id, v):
def set(p, x):
assert 0 <= p < _n
p += _size
for i in range(_log, 0, -1):
_push(p >> i)
_d[p] = x
for i in range(1, _log + 1):
_update(p >> i)
def get(p):
assert 0 <= p < _n
p += _size
for i in range(_log, 0, -1):
_push(p >> i)
return _d[p]
def prod(l, r):
assert 0 <= l <= r <= _n
if l == r:
return _e
l += _size
r += _size
for i in range(_log, 0, -1):
if ((l >> i) << i) != l:
_push(l >> i)
if ((r >> i) << i) != r:
_push(r >> i)
sml = _e
smr = _e
while l < r:
if l & 1:
sml = _op(sml, _d[l])
l += 1
if r & 1:
r -= 1
smr = _op(_d[r], smr)
l >>= 1
r >>= 1
return _op(sml, smr)
def apply(l, r, f):
assert 0 <= l <= r <= _n
if l == r:
return
l += _size
r += _size
for i in range(_log, 0, -1):
if ((l >> i) << i) != l:
_push(l >> i)
if ((r >> i) << i) != r:
_push((r - 1) >> i)
l2 = l
r2 = r
while l < r:
if l & 1:
_all_apply(l, f)
l += 1
if r & 1:
r -= 1
_all_apply(r, f)
l >>= 1
r >>= 1
l = l2
r = r2
for i in range(1, _log + 1):
if ((l >> i) << i) != l:
_update(l >> i)
if ((r >> i) << i) != r:
_update((r - 1) >> i)
def _update(k):
_d[k] = _op(_d[2 * k], _d[2 * k + 1])
def _all_apply(k, f):
_d[k] = _mapping(f, _d[k])
if k < _size:
_lz[k] = _composition(f, _lz[k])
def _push(k):
_all_apply(2 * k, _lz[k])
_all_apply(2 * k + 1, _lz[k])
_lz[k] = _id
_n = len(v)
_log = _n.bit_length()
_size = 1 << _log
_d = [_e] * (2 * _size)
_lz = [_id] * _size
for i in range(_n):
_d[_size + i] = v[i]
for i in range(_size - 1, 0, -1):
_update(i)
self.set = set
self.get = get
self.prod = prod
self.apply = apply
MIL = 1 << 20
def makeNode(total, count):
# Pack a pair into a float
return (total * MIL) + count
def getTotal(node):
return math.floor(node / MIL)
def getCount(node):
return node - getTotal(node) * MIL
nodeIdentity = makeNode(0.0, 0.0)
def nodeOp(node1, node2):
return node1 + node2
# Equivalent to the following:
return makeNode(
getTotal(node1) + getTotal(node2), getCount(node1) + getCount(node2)
)
identityMapping = -1
def mapping(tag, node):
if tag == identityMapping:
return node
# If assigned, new total is the number assigned times count
count = getCount(node)
return makeNode(tag * count, count)
def composition(mapping1, mapping2):
# If assigned multiple times, take first non-identity assignment
return mapping1 if mapping1 != identityMapping else mapping2
#---------------------------------Pollard rho--------------------------------------------
def memodict(f):
"""memoization decorator for a function taking a single argument"""
class memodict(dict):
def __missing__(self, key):
ret = self[key] = f(key)
return ret
return memodict().__getitem__
def pollard_rho(n):
"""returns a random factor of n"""
if n & 1 == 0:
return 2
if n % 3 == 0:
return 3
s = ((n - 1) & (1 - n)).bit_length() - 1
d = n >> s
for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
p = pow(a, d, n)
if p == 1 or p == n - 1 or a % n == 0:
continue
for _ in range(s):
prev = p
p = (p * p) % n
if p == 1:
return math.gcd(prev - 1, n)
if p == n - 1:
break
else:
for i in range(2, n):
x, y = i, (i * i + 1) % n
f = math.gcd(abs(x - y), n)
while f == 1:
x, y = (x * x + 1) % n, (y * y + 1) % n
y = (y * y + 1) % n
f = math.gcd(abs(x - y), n)
if f != n:
return f
return n
@memodict
def prime_factors(n):
"""returns a Counter of the prime factorization of n"""
if n <= 1:
return Counter()
f = pollard_rho(n)
return Counter([n]) if f == n else prime_factors(f) + prime_factors(n // f)
def distinct_factors(n):
"""returns a list of all distinct factors of n"""
factors = [1]
for p, exp in prime_factors(n).items():
factors += [p**i * factor for factor in factors for i in range(1, exp + 1)]
return factors
def all_factors(n):
"""returns a sorted list of all distinct factors of n"""
small, large = [], []
for i in range(1, int(n**0.5) + 1, 2 if n & 1 else 1):
if not n % i:
small.append(i)
large.append(n // i)
if small[-1] == large[-1]:
large.pop()
large.reverse()
small.extend(large)
return small
#---------------------------------Binary Search------------------------------------------
def binarySearch(arr, n, key):
left = 0
right = n-1
mid = 0
res=arr[n-1]
while (left <= right):
mid = (right + left)//2
if (arr[mid] >= key):
res=arr[mid]
right = mid-1
else:
left = mid + 1
return res
def binarySearch1(arr, n, key):
left = 0
right = n-1
mid = 0
res=arr[0]
while (left <= right):
mid = (right + left)//2
if (arr[mid] > key):
right = mid-1
else:
res=arr[mid]
left = mid + 1
return res
#---------------------------------running code------------------------------------------
t=1
t=int(input())
for _ in range (t):
n=int(input())
#n,m=map(int,input().split())
a=list(map(int,input().split()))
#s=input()
#n=len(s)
res=0
m=max(a)
for i in a:
if i<0:
res+=i
if m>0 and n>2:
res-=m
print(res)
``` | instruction | 0 | 65,122 | 3 | 130,244 |
No | output | 1 | 65,122 | 3 | 130,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Farmer John has a farm that consists of n pastures connected by one-directional roads. Each road has a weight, representing the time it takes to go from the start to the end of the road. The roads could have negative weight, where the cows go so fast that they go back in time! However, Farmer John guarantees that it is impossible for the cows to get stuck in a time loop, where they can infinitely go back in time by traveling across a sequence of roads. Also, each pair of pastures is connected by at most one road in each direction.
Unfortunately, Farmer John lost the map of the farm. All he remembers is an array d, where d_i is the smallest amount of time it took the cows to reach the i-th pasture from pasture 1 using a sequence of roads. The cost of his farm is the sum of the weights of each of the roads, and Farmer John needs to know the minimal cost of a farm that is consistent with his memory.
Input
The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of pastures.
The second line of each test case contains n space separated integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the array d. It is guaranteed that d_1 = 0.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, output the minimum possible cost of a farm that is consistent with Farmer John's memory.
Example
Input
3
3
0 2 3
2
0 1000000000
1
0
Output
-3
0
0
Note
In the first test case, you can add roads
* from pasture 1 to pasture 2 with a time of 2,
* from pasture 2 to pasture 3 with a time of 1,
* from pasture 3 to pasture 1 with a time of -3,
* from pasture 3 to pasture 2 with a time of -1,
* from pasture 2 to pasture 1 with a time of -2.
The total cost is 2 + 1 + -3 + -1 + -2 = -3.
In the second test case, you can add a road from pasture 1 to pasture 2 with cost 1000000000 and a road from pasture 2 to pasture 1 with cost -1000000000. The total cost is 1000000000 + -1000000000 = 0.
In the third test case, you can't add any roads. The total cost is 0.
Submitted Solution:
```
def solve(lt):
c = 0
for i in range(1, len(lt)):
c += lt[i] - lt[i - 1]
for i in range(len(lt)):
for j in range(i, len(lt)):
if lt[i] - lt[j] < 0:
c += lt[i] - lt[j]
return c
t = int(input())
for i in range(t):
input()
lt = list(map(int, input().split()))
print(solve(lt))
``` | instruction | 0 | 65,123 | 3 | 130,246 |
No | output | 1 | 65,123 | 3 | 130,247 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Thumbelina has had an accident. She has found herself on a little island in the middle of a swamp and wants to get to the shore very much.
One can get to the shore only by hills that are situated along a straight line that connects the little island with the shore. Let us assume that the hills are numbered from 1 to n and the number of a hill is equal to the distance in meters between it and the island. The distance between the n-th hill and the shore is also 1 meter.
Thumbelina is too small to make such jumps. Fortunately, a family of frogs living in the swamp suggests to help her. Each frog agrees to give Thumbelina a ride but Thumbelina should choose only one frog. Each frog has a certain jump length. If Thumbelina agrees to accept help from a frog whose jump length is d, the frog will jump from the island on the hill d, then — on the hill 2d, then 3d and so on until they get to the shore (i.e. find itself beyond the hill n).
However, there is one more problem: mosquitoes also live in the swamp. At the moment they have a siesta, and they are having a nap on some hills. If the frog jumps on a hill with a mosquito the frog will smash it. The frogs Thumbelina has met are pacifists, so they will find the death of each mosquito very much sad. Help Thumbelina choose a frog that will bring her to the shore and smash as small number of mosquitoes as possible.
Input
The first line contains three integers n, m and k (1 ≤ n ≤ 109, 1 ≤ m, k ≤ 100) — the number of hills, frogs and mosquitoes respectively. The second line contains m integers di (1 ≤ di ≤ 109) — the lengths of the frogs’ jumps. The third line contains k integers — the numbers of the hills on which each mosquito is sleeping. No more than one mosquito can sleep on each hill. The numbers in the lines are separated by single spaces.
Output
In the first line output the number of frogs that smash the minimal number of mosquitoes, in the second line — their numbers in increasing order separated by spaces. The frogs are numbered from 1 to m in the order of the jump length given in the input data.
Examples
Input
5 3 5
2 3 4
1 2 3 4 5
Output
2
2 3
Input
1000000000 2 3
2 5
999999995 999999998 999999996
Output
1
2 | instruction | 0 | 65,157 | 3 | 130,314 |
Tags: implementation
Correct Solution:
```
import sys
n,f, m =[ int(i) for i in sys.stdin.readline().split()]
array_frogs = [ int(i) for i in sys.stdin.readline().split()]
array_m = [ int(i) for i in sys.stdin.readline().split()]
smash = [0] * f
for i,val in enumerate(array_frogs):
for j,val2 in enumerate(array_m):
if val2 % val ==0:
smash[i] += 1
min_val = 101
frogs = []
for i,val in enumerate(smash):
if val < min_val:
min_val = val
frogs = [str(i + 1)]
elif val == min_val:
frogs.append(str(i + 1))
print(len(frogs))
print(' '.join(frogs))
``` | output | 1 | 65,157 | 3 | 130,315 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Thumbelina has had an accident. She has found herself on a little island in the middle of a swamp and wants to get to the shore very much.
One can get to the shore only by hills that are situated along a straight line that connects the little island with the shore. Let us assume that the hills are numbered from 1 to n and the number of a hill is equal to the distance in meters between it and the island. The distance between the n-th hill and the shore is also 1 meter.
Thumbelina is too small to make such jumps. Fortunately, a family of frogs living in the swamp suggests to help her. Each frog agrees to give Thumbelina a ride but Thumbelina should choose only one frog. Each frog has a certain jump length. If Thumbelina agrees to accept help from a frog whose jump length is d, the frog will jump from the island on the hill d, then — on the hill 2d, then 3d and so on until they get to the shore (i.e. find itself beyond the hill n).
However, there is one more problem: mosquitoes also live in the swamp. At the moment they have a siesta, and they are having a nap on some hills. If the frog jumps on a hill with a mosquito the frog will smash it. The frogs Thumbelina has met are pacifists, so they will find the death of each mosquito very much sad. Help Thumbelina choose a frog that will bring her to the shore and smash as small number of mosquitoes as possible.
Input
The first line contains three integers n, m and k (1 ≤ n ≤ 109, 1 ≤ m, k ≤ 100) — the number of hills, frogs and mosquitoes respectively. The second line contains m integers di (1 ≤ di ≤ 109) — the lengths of the frogs’ jumps. The third line contains k integers — the numbers of the hills on which each mosquito is sleeping. No more than one mosquito can sleep on each hill. The numbers in the lines are separated by single spaces.
Output
In the first line output the number of frogs that smash the minimal number of mosquitoes, in the second line — their numbers in increasing order separated by spaces. The frogs are numbered from 1 to m in the order of the jump length given in the input data.
Examples
Input
5 3 5
2 3 4
1 2 3 4 5
Output
2
2 3
Input
1000000000 2 3
2 5
999999995 999999998 999999996
Output
1
2 | instruction | 0 | 65,158 | 3 | 130,316 |
Tags: implementation
Correct Solution:
```
__author__ = 'linh'
def main():
first_line = input()
first_line = first_line.split()
hill_n0 = int(first_line[0])
frog_n0 = int(first_line[1])
mosquito_n0 = int(first_line[2])
frog_list = input()
frog_list = list(map(int, frog_list.split()))
mosquito_list = input()
mosquito_list = list(map(int, mosquito_list.split()))
result = []
min_mosquito = 10000000000000
for i in range(0, frog_n0):
n0 = 0
for j in range(0, mosquito_n0):
if mosquito_list[j] % frog_list[i] == 0:
n0 +=1
if n0 < min_mosquito:
min_mosquito = n0
result.append(n0)
frog_n0_adequate = 0
output = ""
for i in range(0, frog_n0):
if result[i] == min_mosquito:
output += str(i+1) + " "
frog_n0_adequate += 1
print(frog_n0_adequate)
print(output)
if __name__ == '__main__':
main()
``` | output | 1 | 65,158 | 3 | 130,317 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Thumbelina has had an accident. She has found herself on a little island in the middle of a swamp and wants to get to the shore very much.
One can get to the shore only by hills that are situated along a straight line that connects the little island with the shore. Let us assume that the hills are numbered from 1 to n and the number of a hill is equal to the distance in meters between it and the island. The distance between the n-th hill and the shore is also 1 meter.
Thumbelina is too small to make such jumps. Fortunately, a family of frogs living in the swamp suggests to help her. Each frog agrees to give Thumbelina a ride but Thumbelina should choose only one frog. Each frog has a certain jump length. If Thumbelina agrees to accept help from a frog whose jump length is d, the frog will jump from the island on the hill d, then — on the hill 2d, then 3d and so on until they get to the shore (i.e. find itself beyond the hill n).
However, there is one more problem: mosquitoes also live in the swamp. At the moment they have a siesta, and they are having a nap on some hills. If the frog jumps on a hill with a mosquito the frog will smash it. The frogs Thumbelina has met are pacifists, so they will find the death of each mosquito very much sad. Help Thumbelina choose a frog that will bring her to the shore and smash as small number of mosquitoes as possible.
Input
The first line contains three integers n, m and k (1 ≤ n ≤ 109, 1 ≤ m, k ≤ 100) — the number of hills, frogs and mosquitoes respectively. The second line contains m integers di (1 ≤ di ≤ 109) — the lengths of the frogs’ jumps. The third line contains k integers — the numbers of the hills on which each mosquito is sleeping. No more than one mosquito can sleep on each hill. The numbers in the lines are separated by single spaces.
Output
In the first line output the number of frogs that smash the minimal number of mosquitoes, in the second line — their numbers in increasing order separated by spaces. The frogs are numbered from 1 to m in the order of the jump length given in the input data.
Examples
Input
5 3 5
2 3 4
1 2 3 4 5
Output
2
2 3
Input
1000000000 2 3
2 5
999999995 999999998 999999996
Output
1
2 | instruction | 0 | 65,159 | 3 | 130,318 |
Tags: implementation
Correct Solution:
```
def chk(x):
for i in range(m):
if x%fgs[i]==0:
ans[i]+=1
n,m,k=[int(x) for x in input().split()]
fgs=[int(x) for x in input().split()]
ct=0
ans=[0]*m
for i in input().split():
chk(int(i))
mi=min(ans)
print(ans.count(mi))
for i in range(m):
if ans[i]==mi:
print(i+1,end=' ')
``` | output | 1 | 65,159 | 3 | 130,319 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Thumbelina has had an accident. She has found herself on a little island in the middle of a swamp and wants to get to the shore very much.
One can get to the shore only by hills that are situated along a straight line that connects the little island with the shore. Let us assume that the hills are numbered from 1 to n and the number of a hill is equal to the distance in meters between it and the island. The distance between the n-th hill and the shore is also 1 meter.
Thumbelina is too small to make such jumps. Fortunately, a family of frogs living in the swamp suggests to help her. Each frog agrees to give Thumbelina a ride but Thumbelina should choose only one frog. Each frog has a certain jump length. If Thumbelina agrees to accept help from a frog whose jump length is d, the frog will jump from the island on the hill d, then — on the hill 2d, then 3d and so on until they get to the shore (i.e. find itself beyond the hill n).
However, there is one more problem: mosquitoes also live in the swamp. At the moment they have a siesta, and they are having a nap on some hills. If the frog jumps on a hill with a mosquito the frog will smash it. The frogs Thumbelina has met are pacifists, so they will find the death of each mosquito very much sad. Help Thumbelina choose a frog that will bring her to the shore and smash as small number of mosquitoes as possible.
Input
The first line contains three integers n, m and k (1 ≤ n ≤ 109, 1 ≤ m, k ≤ 100) — the number of hills, frogs and mosquitoes respectively. The second line contains m integers di (1 ≤ di ≤ 109) — the lengths of the frogs’ jumps. The third line contains k integers — the numbers of the hills on which each mosquito is sleeping. No more than one mosquito can sleep on each hill. The numbers in the lines are separated by single spaces.
Output
In the first line output the number of frogs that smash the minimal number of mosquitoes, in the second line — their numbers in increasing order separated by spaces. The frogs are numbered from 1 to m in the order of the jump length given in the input data.
Examples
Input
5 3 5
2 3 4
1 2 3 4 5
Output
2
2 3
Input
1000000000 2 3
2 5
999999995 999999998 999999996
Output
1
2 | instruction | 0 | 65,160 | 3 | 130,320 |
Tags: implementation
Correct Solution:
```
n, m, k = map(int, input().split())
d, a = map(int, input().split()), list(map(int, input().split()))
v = sorted((sum(ai % di == 0 for ai in a), i + 1) for i, di in enumerate(d))
v = sorted(vi[1] for vi in v if vi[0] == v[0][0])
print(len(v), ' '.join(map(str, v)), sep='\n')
``` | output | 1 | 65,160 | 3 | 130,321 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Thumbelina has had an accident. She has found herself on a little island in the middle of a swamp and wants to get to the shore very much.
One can get to the shore only by hills that are situated along a straight line that connects the little island with the shore. Let us assume that the hills are numbered from 1 to n and the number of a hill is equal to the distance in meters between it and the island. The distance between the n-th hill and the shore is also 1 meter.
Thumbelina is too small to make such jumps. Fortunately, a family of frogs living in the swamp suggests to help her. Each frog agrees to give Thumbelina a ride but Thumbelina should choose only one frog. Each frog has a certain jump length. If Thumbelina agrees to accept help from a frog whose jump length is d, the frog will jump from the island on the hill d, then — on the hill 2d, then 3d and so on until they get to the shore (i.e. find itself beyond the hill n).
However, there is one more problem: mosquitoes also live in the swamp. At the moment they have a siesta, and they are having a nap on some hills. If the frog jumps on a hill with a mosquito the frog will smash it. The frogs Thumbelina has met are pacifists, so they will find the death of each mosquito very much sad. Help Thumbelina choose a frog that will bring her to the shore and smash as small number of mosquitoes as possible.
Input
The first line contains three integers n, m and k (1 ≤ n ≤ 109, 1 ≤ m, k ≤ 100) — the number of hills, frogs and mosquitoes respectively. The second line contains m integers di (1 ≤ di ≤ 109) — the lengths of the frogs’ jumps. The third line contains k integers — the numbers of the hills on which each mosquito is sleeping. No more than one mosquito can sleep on each hill. The numbers in the lines are separated by single spaces.
Output
In the first line output the number of frogs that smash the minimal number of mosquitoes, in the second line — their numbers in increasing order separated by spaces. The frogs are numbered from 1 to m in the order of the jump length given in the input data.
Examples
Input
5 3 5
2 3 4
1 2 3 4 5
Output
2
2 3
Input
1000000000 2 3
2 5
999999995 999999998 999999996
Output
1
2 | instruction | 0 | 65,161 | 3 | 130,322 |
Tags: implementation
Correct Solution:
```
import sys
input=sys.stdin.readline
from math import *
n,m,k=map(int,input().split())
s=0
a=[int(x) for x in input().split()]
b=[int(x) for x in input().split()]
d={}
s=[0 for i in range(m)]
for i in b:
d[i]=1
for i in range(m):
for j in range(k):
if b[j]%a[i]==0:
s[i]+=1
print(s.count(min(s)))
for i in range(m):
if s[i]==min(s):
print(i+1,end=" ")
``` | output | 1 | 65,161 | 3 | 130,323 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Thumbelina has had an accident. She has found herself on a little island in the middle of a swamp and wants to get to the shore very much.
One can get to the shore only by hills that are situated along a straight line that connects the little island with the shore. Let us assume that the hills are numbered from 1 to n and the number of a hill is equal to the distance in meters between it and the island. The distance between the n-th hill and the shore is also 1 meter.
Thumbelina is too small to make such jumps. Fortunately, a family of frogs living in the swamp suggests to help her. Each frog agrees to give Thumbelina a ride but Thumbelina should choose only one frog. Each frog has a certain jump length. If Thumbelina agrees to accept help from a frog whose jump length is d, the frog will jump from the island on the hill d, then — on the hill 2d, then 3d and so on until they get to the shore (i.e. find itself beyond the hill n).
However, there is one more problem: mosquitoes also live in the swamp. At the moment they have a siesta, and they are having a nap on some hills. If the frog jumps on a hill with a mosquito the frog will smash it. The frogs Thumbelina has met are pacifists, so they will find the death of each mosquito very much sad. Help Thumbelina choose a frog that will bring her to the shore and smash as small number of mosquitoes as possible.
Input
The first line contains three integers n, m and k (1 ≤ n ≤ 109, 1 ≤ m, k ≤ 100) — the number of hills, frogs and mosquitoes respectively. The second line contains m integers di (1 ≤ di ≤ 109) — the lengths of the frogs’ jumps. The third line contains k integers — the numbers of the hills on which each mosquito is sleeping. No more than one mosquito can sleep on each hill. The numbers in the lines are separated by single spaces.
Output
In the first line output the number of frogs that smash the minimal number of mosquitoes, in the second line — their numbers in increasing order separated by spaces. The frogs are numbered from 1 to m in the order of the jump length given in the input data.
Examples
Input
5 3 5
2 3 4
1 2 3 4 5
Output
2
2 3
Input
1000000000 2 3
2 5
999999995 999999998 999999996
Output
1
2 | instruction | 0 | 65,162 | 3 | 130,324 |
Tags: implementation
Correct Solution:
```
I = lambda: map(int, input().split())
_, m, _ = I()
D = list(I())
X = [0]*m
for k in I():
for i in range(m):
if k % D[i] == 0:
X[i] += 1
x = min(X)
print(X.count(x))
print(*(i+1 for i in range(m) if X[i]==x))
``` | output | 1 | 65,162 | 3 | 130,325 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Thumbelina has had an accident. She has found herself on a little island in the middle of a swamp and wants to get to the shore very much.
One can get to the shore only by hills that are situated along a straight line that connects the little island with the shore. Let us assume that the hills are numbered from 1 to n and the number of a hill is equal to the distance in meters between it and the island. The distance between the n-th hill and the shore is also 1 meter.
Thumbelina is too small to make such jumps. Fortunately, a family of frogs living in the swamp suggests to help her. Each frog agrees to give Thumbelina a ride but Thumbelina should choose only one frog. Each frog has a certain jump length. If Thumbelina agrees to accept help from a frog whose jump length is d, the frog will jump from the island on the hill d, then — on the hill 2d, then 3d and so on until they get to the shore (i.e. find itself beyond the hill n).
However, there is one more problem: mosquitoes also live in the swamp. At the moment they have a siesta, and they are having a nap on some hills. If the frog jumps on a hill with a mosquito the frog will smash it. The frogs Thumbelina has met are pacifists, so they will find the death of each mosquito very much sad. Help Thumbelina choose a frog that will bring her to the shore and smash as small number of mosquitoes as possible.
Input
The first line contains three integers n, m and k (1 ≤ n ≤ 109, 1 ≤ m, k ≤ 100) — the number of hills, frogs and mosquitoes respectively. The second line contains m integers di (1 ≤ di ≤ 109) — the lengths of the frogs’ jumps. The third line contains k integers — the numbers of the hills on which each mosquito is sleeping. No more than one mosquito can sleep on each hill. The numbers in the lines are separated by single spaces.
Output
In the first line output the number of frogs that smash the minimal number of mosquitoes, in the second line — their numbers in increasing order separated by spaces. The frogs are numbered from 1 to m in the order of the jump length given in the input data.
Examples
Input
5 3 5
2 3 4
1 2 3 4 5
Output
2
2 3
Input
1000000000 2 3
2 5
999999995 999999998 999999996
Output
1
2 | instruction | 0 | 65,163 | 3 | 130,326 |
Tags: implementation
Correct Solution:
```
I = lambda: map(int, input().split())
_, m, _ = I()
D = [*I()]
X = [0]*m
for k in I():
for i in range(m):
if k % D[i] == 0:
X[i] += 1
x = min(X)
print(X.count(x))
print(*(i+1 for i in range(m) if X[i]==x))
``` | output | 1 | 65,163 | 3 | 130,327 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Thumbelina has had an accident. She has found herself on a little island in the middle of a swamp and wants to get to the shore very much.
One can get to the shore only by hills that are situated along a straight line that connects the little island with the shore. Let us assume that the hills are numbered from 1 to n and the number of a hill is equal to the distance in meters between it and the island. The distance between the n-th hill and the shore is also 1 meter.
Thumbelina is too small to make such jumps. Fortunately, a family of frogs living in the swamp suggests to help her. Each frog agrees to give Thumbelina a ride but Thumbelina should choose only one frog. Each frog has a certain jump length. If Thumbelina agrees to accept help from a frog whose jump length is d, the frog will jump from the island on the hill d, then — on the hill 2d, then 3d and so on until they get to the shore (i.e. find itself beyond the hill n).
However, there is one more problem: mosquitoes also live in the swamp. At the moment they have a siesta, and they are having a nap on some hills. If the frog jumps on a hill with a mosquito the frog will smash it. The frogs Thumbelina has met are pacifists, so they will find the death of each mosquito very much sad. Help Thumbelina choose a frog that will bring her to the shore and smash as small number of mosquitoes as possible.
Input
The first line contains three integers n, m and k (1 ≤ n ≤ 109, 1 ≤ m, k ≤ 100) — the number of hills, frogs and mosquitoes respectively. The second line contains m integers di (1 ≤ di ≤ 109) — the lengths of the frogs’ jumps. The third line contains k integers — the numbers of the hills on which each mosquito is sleeping. No more than one mosquito can sleep on each hill. The numbers in the lines are separated by single spaces.
Output
In the first line output the number of frogs that smash the minimal number of mosquitoes, in the second line — their numbers in increasing order separated by spaces. The frogs are numbered from 1 to m in the order of the jump length given in the input data.
Examples
Input
5 3 5
2 3 4
1 2 3 4 5
Output
2
2 3
Input
1000000000 2 3
2 5
999999995 999999998 999999996
Output
1
2 | instruction | 0 | 65,164 | 3 | 130,328 |
Tags: implementation
Correct Solution:
```
import sys
n, m, k = sys.stdin.readline().split()
n = int(n)
m = int(m)
k = int(k)
arr = sys.stdin.readline().split()
arr = [ int(x) for x in arr]
arr2 = sys.stdin.readline().split()
arr2 = [ int(y) for y in arr2]
minimum = 1000
frogs = []
for i in range(m):
contador = 0
for j in range(k):
if arr2[j] % arr[i] == 0:
contador += 1
if contador < minimum:
frogs = [i +1]
minimum = contador
elif contador == minimum:
frogs.append(i+1)
print(len(frogs))
print(' '.join([str(f) for f in frogs]))
``` | output | 1 | 65,164 | 3 | 130,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Thumbelina has had an accident. She has found herself on a little island in the middle of a swamp and wants to get to the shore very much.
One can get to the shore only by hills that are situated along a straight line that connects the little island with the shore. Let us assume that the hills are numbered from 1 to n and the number of a hill is equal to the distance in meters between it and the island. The distance between the n-th hill and the shore is also 1 meter.
Thumbelina is too small to make such jumps. Fortunately, a family of frogs living in the swamp suggests to help her. Each frog agrees to give Thumbelina a ride but Thumbelina should choose only one frog. Each frog has a certain jump length. If Thumbelina agrees to accept help from a frog whose jump length is d, the frog will jump from the island on the hill d, then — on the hill 2d, then 3d and so on until they get to the shore (i.e. find itself beyond the hill n).
However, there is one more problem: mosquitoes also live in the swamp. At the moment they have a siesta, and they are having a nap on some hills. If the frog jumps on a hill with a mosquito the frog will smash it. The frogs Thumbelina has met are pacifists, so they will find the death of each mosquito very much sad. Help Thumbelina choose a frog that will bring her to the shore and smash as small number of mosquitoes as possible.
Input
The first line contains three integers n, m and k (1 ≤ n ≤ 109, 1 ≤ m, k ≤ 100) — the number of hills, frogs and mosquitoes respectively. The second line contains m integers di (1 ≤ di ≤ 109) — the lengths of the frogs’ jumps. The third line contains k integers — the numbers of the hills on which each mosquito is sleeping. No more than one mosquito can sleep on each hill. The numbers in the lines are separated by single spaces.
Output
In the first line output the number of frogs that smash the minimal number of mosquitoes, in the second line — their numbers in increasing order separated by spaces. The frogs are numbered from 1 to m in the order of the jump length given in the input data.
Examples
Input
5 3 5
2 3 4
1 2 3 4 5
Output
2
2 3
Input
1000000000 2 3
2 5
999999995 999999998 999999996
Output
1
2
Submitted Solution:
```
from collections import defaultdict
table = {}
n,m,k = [int(x) for x in input().split()]
frog = [int(x) for x in input().split()]
mos = [int(x) for x in input().split()]
frogT = set()
for a in frog:
table[a] = 0
frogT.add(a)
for a in frogT:
for b in mos:
if b % a == 0 :
table[a] += 1
minV = 9999999
minItem = []
for key,value in table.items():
if value < minV:
minV = value
minItem = []
minItem.append(key)
elif value == minV:
minItem.append(key)
minItem.sort()
minIndex = []
for a in minItem:
for i in range(m):
if a == frog[i]:
minIndex.append(i+1)
minIndex.sort()
# print(table,frogT)
#print("minItem =",minIndex,"minV=",minV)
print(len(minIndex))
for a in minIndex:
print(a,end=" ")
``` | instruction | 0 | 65,165 | 3 | 130,330 |
Yes | output | 1 | 65,165 | 3 | 130,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Thumbelina has had an accident. She has found herself on a little island in the middle of a swamp and wants to get to the shore very much.
One can get to the shore only by hills that are situated along a straight line that connects the little island with the shore. Let us assume that the hills are numbered from 1 to n and the number of a hill is equal to the distance in meters between it and the island. The distance between the n-th hill and the shore is also 1 meter.
Thumbelina is too small to make such jumps. Fortunately, a family of frogs living in the swamp suggests to help her. Each frog agrees to give Thumbelina a ride but Thumbelina should choose only one frog. Each frog has a certain jump length. If Thumbelina agrees to accept help from a frog whose jump length is d, the frog will jump from the island on the hill d, then — on the hill 2d, then 3d and so on until they get to the shore (i.e. find itself beyond the hill n).
However, there is one more problem: mosquitoes also live in the swamp. At the moment they have a siesta, and they are having a nap on some hills. If the frog jumps on a hill with a mosquito the frog will smash it. The frogs Thumbelina has met are pacifists, so they will find the death of each mosquito very much sad. Help Thumbelina choose a frog that will bring her to the shore and smash as small number of mosquitoes as possible.
Input
The first line contains three integers n, m and k (1 ≤ n ≤ 109, 1 ≤ m, k ≤ 100) — the number of hills, frogs and mosquitoes respectively. The second line contains m integers di (1 ≤ di ≤ 109) — the lengths of the frogs’ jumps. The third line contains k integers — the numbers of the hills on which each mosquito is sleeping. No more than one mosquito can sleep on each hill. The numbers in the lines are separated by single spaces.
Output
In the first line output the number of frogs that smash the minimal number of mosquitoes, in the second line — their numbers in increasing order separated by spaces. The frogs are numbered from 1 to m in the order of the jump length given in the input data.
Examples
Input
5 3 5
2 3 4
1 2 3 4 5
Output
2
2 3
Input
1000000000 2 3
2 5
999999995 999999998 999999996
Output
1
2
Submitted Solution:
```
a,b,c=map(int,input().split())
z=list(map(int,input().split()))
k=list(map(int,input().split()))
ans=[]
for i,j in enumerate(z):
ok=0
for u in k:ok+=(u%j==0)
ans.append([ok,i+1])
ans=sorted(ans)
j=[]
for i in ans:
if ans[0][0]==i[0]:j.append(i[1])
print(len(j))
print(*j)
``` | instruction | 0 | 65,166 | 3 | 130,332 |
Yes | output | 1 | 65,166 | 3 | 130,333 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Thumbelina has had an accident. She has found herself on a little island in the middle of a swamp and wants to get to the shore very much.
One can get to the shore only by hills that are situated along a straight line that connects the little island with the shore. Let us assume that the hills are numbered from 1 to n and the number of a hill is equal to the distance in meters between it and the island. The distance between the n-th hill and the shore is also 1 meter.
Thumbelina is too small to make such jumps. Fortunately, a family of frogs living in the swamp suggests to help her. Each frog agrees to give Thumbelina a ride but Thumbelina should choose only one frog. Each frog has a certain jump length. If Thumbelina agrees to accept help from a frog whose jump length is d, the frog will jump from the island on the hill d, then — on the hill 2d, then 3d and so on until they get to the shore (i.e. find itself beyond the hill n).
However, there is one more problem: mosquitoes also live in the swamp. At the moment they have a siesta, and they are having a nap on some hills. If the frog jumps on a hill with a mosquito the frog will smash it. The frogs Thumbelina has met are pacifists, so they will find the death of each mosquito very much sad. Help Thumbelina choose a frog that will bring her to the shore and smash as small number of mosquitoes as possible.
Input
The first line contains three integers n, m and k (1 ≤ n ≤ 109, 1 ≤ m, k ≤ 100) — the number of hills, frogs and mosquitoes respectively. The second line contains m integers di (1 ≤ di ≤ 109) — the lengths of the frogs’ jumps. The third line contains k integers — the numbers of the hills on which each mosquito is sleeping. No more than one mosquito can sleep on each hill. The numbers in the lines are separated by single spaces.
Output
In the first line output the number of frogs that smash the minimal number of mosquitoes, in the second line — their numbers in increasing order separated by spaces. The frogs are numbered from 1 to m in the order of the jump length given in the input data.
Examples
Input
5 3 5
2 3 4
1 2 3 4 5
Output
2
2 3
Input
1000000000 2 3
2 5
999999995 999999998 999999996
Output
1
2
Submitted Solution:
```
"""
// Author : snape_here - Susanta Mukherjee
"""
from __future__ import division, print_function
import os,sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().split())
def li(): return list(mi())
def gcd(x, y):
while y:
x, y = y, x % y
return x
def read():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
import math
from functools import reduce
def factorS(n):
return set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
def divisorGen(n):
factors = list(factorGenerator(n))
nfactors = len(factors)
f = [0] * nfactors
while True:
yield reduce(lambda x, y: x*y, [factors[x][0]**f[x] for x in range(nfactors)], 1)
i = 0
while True:
f[i] += 1
if f[i] <= factors[i][1]:
break
f[i] = 0
i += 1
if i >= nfactors:
return
mod=100000007
def main():
n,m,k=mi()
d=li()
l=li()
a=[]
for i in range(m):
c=0
for j in range(k):
if l[j]%d[i]==0:
c+=1
a.append(c)
b=min(a)
e=a.count(b)
f=[]
for i in range(m):
if a[i]==b:
f.append(i+1)
print(e)
print(*f)
# region fastio#
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
#read()
main()
#Comment read()
``` | instruction | 0 | 65,167 | 3 | 130,334 |
Yes | output | 1 | 65,167 | 3 | 130,335 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Thumbelina has had an accident. She has found herself on a little island in the middle of a swamp and wants to get to the shore very much.
One can get to the shore only by hills that are situated along a straight line that connects the little island with the shore. Let us assume that the hills are numbered from 1 to n and the number of a hill is equal to the distance in meters between it and the island. The distance between the n-th hill and the shore is also 1 meter.
Thumbelina is too small to make such jumps. Fortunately, a family of frogs living in the swamp suggests to help her. Each frog agrees to give Thumbelina a ride but Thumbelina should choose only one frog. Each frog has a certain jump length. If Thumbelina agrees to accept help from a frog whose jump length is d, the frog will jump from the island on the hill d, then — on the hill 2d, then 3d and so on until they get to the shore (i.e. find itself beyond the hill n).
However, there is one more problem: mosquitoes also live in the swamp. At the moment they have a siesta, and they are having a nap on some hills. If the frog jumps on a hill with a mosquito the frog will smash it. The frogs Thumbelina has met are pacifists, so they will find the death of each mosquito very much sad. Help Thumbelina choose a frog that will bring her to the shore and smash as small number of mosquitoes as possible.
Input
The first line contains three integers n, m and k (1 ≤ n ≤ 109, 1 ≤ m, k ≤ 100) — the number of hills, frogs and mosquitoes respectively. The second line contains m integers di (1 ≤ di ≤ 109) — the lengths of the frogs’ jumps. The third line contains k integers — the numbers of the hills on which each mosquito is sleeping. No more than one mosquito can sleep on each hill. The numbers in the lines are separated by single spaces.
Output
In the first line output the number of frogs that smash the minimal number of mosquitoes, in the second line — their numbers in increasing order separated by spaces. The frogs are numbered from 1 to m in the order of the jump length given in the input data.
Examples
Input
5 3 5
2 3 4
1 2 3 4 5
Output
2
2 3
Input
1000000000 2 3
2 5
999999995 999999998 999999996
Output
1
2
Submitted Solution:
```
n, m, k = map(int, input().split())
frogs = list(map(int, input().split()))
mosquitoes = list(map(int, input().split()))
def get_smash(leap):
result = 0
for x in mosquitoes:
if x % leap == 0:
result += 1
return result
best_smash = get_smash(frogs[0])
best_frogs = [1]
for i, frog in enumerate(frogs[1:]):
smash = get_smash(frog)
if smash == best_smash:
best_frogs.append(i + 2)
elif smash < best_smash:
best_smash = smash
best_frogs = [i + 2]
print(len(best_frogs))
print(' '.join(map(str, best_frogs)))
``` | instruction | 0 | 65,168 | 3 | 130,336 |
Yes | output | 1 | 65,168 | 3 | 130,337 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Thumbelina has had an accident. She has found herself on a little island in the middle of a swamp and wants to get to the shore very much.
One can get to the shore only by hills that are situated along a straight line that connects the little island with the shore. Let us assume that the hills are numbered from 1 to n and the number of a hill is equal to the distance in meters between it and the island. The distance between the n-th hill and the shore is also 1 meter.
Thumbelina is too small to make such jumps. Fortunately, a family of frogs living in the swamp suggests to help her. Each frog agrees to give Thumbelina a ride but Thumbelina should choose only one frog. Each frog has a certain jump length. If Thumbelina agrees to accept help from a frog whose jump length is d, the frog will jump from the island on the hill d, then — on the hill 2d, then 3d and so on until they get to the shore (i.e. find itself beyond the hill n).
However, there is one more problem: mosquitoes also live in the swamp. At the moment they have a siesta, and they are having a nap on some hills. If the frog jumps on a hill with a mosquito the frog will smash it. The frogs Thumbelina has met are pacifists, so they will find the death of each mosquito very much sad. Help Thumbelina choose a frog that will bring her to the shore and smash as small number of mosquitoes as possible.
Input
The first line contains three integers n, m and k (1 ≤ n ≤ 109, 1 ≤ m, k ≤ 100) — the number of hills, frogs and mosquitoes respectively. The second line contains m integers di (1 ≤ di ≤ 109) — the lengths of the frogs’ jumps. The third line contains k integers — the numbers of the hills on which each mosquito is sleeping. No more than one mosquito can sleep on each hill. The numbers in the lines are separated by single spaces.
Output
In the first line output the number of frogs that smash the minimal number of mosquitoes, in the second line — their numbers in increasing order separated by spaces. The frogs are numbered from 1 to m in the order of the jump length given in the input data.
Examples
Input
5 3 5
2 3 4
1 2 3 4 5
Output
2
2 3
Input
1000000000 2 3
2 5
999999995 999999998 999999996
Output
1
2
Submitted Solution:
```
n,m,k = [int(s) for s in input().split()]
frogJumps = [(i,int(s)) for i,s in enumerate(input().split())]
mosquitoes = [int(s) for s in input().split()]
count = 0
def calculate(i):
x = 0
c = 0
while(x < n):
x += i[1]
if x in mosquitoes:
c += 1
return (c,i[0])
l = list(map(calculate,frogJumps))
l.sort()
minValue = l[0][0]
for i in l:
if i[0] == minValue:
count += 1
else:
break
print(count)
for i in range(count):
print(l[i][1],end = " ")
``` | instruction | 0 | 65,169 | 3 | 130,338 |
No | output | 1 | 65,169 | 3 | 130,339 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Thumbelina has had an accident. She has found herself on a little island in the middle of a swamp and wants to get to the shore very much.
One can get to the shore only by hills that are situated along a straight line that connects the little island with the shore. Let us assume that the hills are numbered from 1 to n and the number of a hill is equal to the distance in meters between it and the island. The distance between the n-th hill and the shore is also 1 meter.
Thumbelina is too small to make such jumps. Fortunately, a family of frogs living in the swamp suggests to help her. Each frog agrees to give Thumbelina a ride but Thumbelina should choose only one frog. Each frog has a certain jump length. If Thumbelina agrees to accept help from a frog whose jump length is d, the frog will jump from the island on the hill d, then — on the hill 2d, then 3d and so on until they get to the shore (i.e. find itself beyond the hill n).
However, there is one more problem: mosquitoes also live in the swamp. At the moment they have a siesta, and they are having a nap on some hills. If the frog jumps on a hill with a mosquito the frog will smash it. The frogs Thumbelina has met are pacifists, so they will find the death of each mosquito very much sad. Help Thumbelina choose a frog that will bring her to the shore and smash as small number of mosquitoes as possible.
Input
The first line contains three integers n, m and k (1 ≤ n ≤ 109, 1 ≤ m, k ≤ 100) — the number of hills, frogs and mosquitoes respectively. The second line contains m integers di (1 ≤ di ≤ 109) — the lengths of the frogs’ jumps. The third line contains k integers — the numbers of the hills on which each mosquito is sleeping. No more than one mosquito can sleep on each hill. The numbers in the lines are separated by single spaces.
Output
In the first line output the number of frogs that smash the minimal number of mosquitoes, in the second line — their numbers in increasing order separated by spaces. The frogs are numbered from 1 to m in the order of the jump length given in the input data.
Examples
Input
5 3 5
2 3 4
1 2 3 4 5
Output
2
2 3
Input
1000000000 2 3
2 5
999999995 999999998 999999996
Output
1
2
Submitted Solution:
```
n,m,k = [int(s) for s in input().split()]
frogJumps = [int(s) for s in input().split()]
mosquitoes = [int(s) for s in input().split()]
count = 0
def calculate(i):
x = 0
c = 0
while(x < n):
x += i
if x in mosquitoes:
c += 1
return (c,i)
l = list(map(calculate,frogJumps))
l.sort()
minValue = l[0][0]
for i in l:
if i[0] == minValue:
count += 1
else:
break
print(count)
for i in range(count):
print(l[i][1],end = " ")
``` | instruction | 0 | 65,170 | 3 | 130,340 |
No | output | 1 | 65,170 | 3 | 130,341 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Thumbelina has had an accident. She has found herself on a little island in the middle of a swamp and wants to get to the shore very much.
One can get to the shore only by hills that are situated along a straight line that connects the little island with the shore. Let us assume that the hills are numbered from 1 to n and the number of a hill is equal to the distance in meters between it and the island. The distance between the n-th hill and the shore is also 1 meter.
Thumbelina is too small to make such jumps. Fortunately, a family of frogs living in the swamp suggests to help her. Each frog agrees to give Thumbelina a ride but Thumbelina should choose only one frog. Each frog has a certain jump length. If Thumbelina agrees to accept help from a frog whose jump length is d, the frog will jump from the island on the hill d, then — on the hill 2d, then 3d and so on until they get to the shore (i.e. find itself beyond the hill n).
However, there is one more problem: mosquitoes also live in the swamp. At the moment they have a siesta, and they are having a nap on some hills. If the frog jumps on a hill with a mosquito the frog will smash it. The frogs Thumbelina has met are pacifists, so they will find the death of each mosquito very much sad. Help Thumbelina choose a frog that will bring her to the shore and smash as small number of mosquitoes as possible.
Input
The first line contains three integers n, m and k (1 ≤ n ≤ 109, 1 ≤ m, k ≤ 100) — the number of hills, frogs and mosquitoes respectively. The second line contains m integers di (1 ≤ di ≤ 109) — the lengths of the frogs’ jumps. The third line contains k integers — the numbers of the hills on which each mosquito is sleeping. No more than one mosquito can sleep on each hill. The numbers in the lines are separated by single spaces.
Output
In the first line output the number of frogs that smash the minimal number of mosquitoes, in the second line — their numbers in increasing order separated by spaces. The frogs are numbered from 1 to m in the order of the jump length given in the input data.
Examples
Input
5 3 5
2 3 4
1 2 3 4 5
Output
2
2 3
Input
1000000000 2 3
2 5
999999995 999999998 999999996
Output
1
2
Submitted Solution:
```
from collections import defaultdict
table = defaultdict(int)
n,m,k = [int(x) for x in input().split()]
frog = [int(x) for x in input().split()]
mos = [int(x) for x in input().split()]
for a in frog:
for b in mos:
if b % a == 0 :
table[a] += 1
else:
table[a] = 0
minV = 9999999
minItem = []
for key,value in table.items():
if value < minV:
minV = value
minItem = []
minItem.append(key)
elif value == minV:
minItem.append(key)
minItem.sort()
minIndex = []
for a in minItem:
for i in range(m):
if a == frog[i]:
minIndex.append(i+1)
# print(table)
# print("minItem =",minIndex,"minV=",minV)
print(len(minIndex))
for a in minIndex:
print(a,end=" ")
``` | instruction | 0 | 65,171 | 3 | 130,342 |
No | output | 1 | 65,171 | 3 | 130,343 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Thumbelina has had an accident. She has found herself on a little island in the middle of a swamp and wants to get to the shore very much.
One can get to the shore only by hills that are situated along a straight line that connects the little island with the shore. Let us assume that the hills are numbered from 1 to n and the number of a hill is equal to the distance in meters between it and the island. The distance between the n-th hill and the shore is also 1 meter.
Thumbelina is too small to make such jumps. Fortunately, a family of frogs living in the swamp suggests to help her. Each frog agrees to give Thumbelina a ride but Thumbelina should choose only one frog. Each frog has a certain jump length. If Thumbelina agrees to accept help from a frog whose jump length is d, the frog will jump from the island on the hill d, then — on the hill 2d, then 3d and so on until they get to the shore (i.e. find itself beyond the hill n).
However, there is one more problem: mosquitoes also live in the swamp. At the moment they have a siesta, and they are having a nap on some hills. If the frog jumps on a hill with a mosquito the frog will smash it. The frogs Thumbelina has met are pacifists, so they will find the death of each mosquito very much sad. Help Thumbelina choose a frog that will bring her to the shore and smash as small number of mosquitoes as possible.
Input
The first line contains three integers n, m and k (1 ≤ n ≤ 109, 1 ≤ m, k ≤ 100) — the number of hills, frogs and mosquitoes respectively. The second line contains m integers di (1 ≤ di ≤ 109) — the lengths of the frogs’ jumps. The third line contains k integers — the numbers of the hills on which each mosquito is sleeping. No more than one mosquito can sleep on each hill. The numbers in the lines are separated by single spaces.
Output
In the first line output the number of frogs that smash the minimal number of mosquitoes, in the second line — their numbers in increasing order separated by spaces. The frogs are numbered from 1 to m in the order of the jump length given in the input data.
Examples
Input
5 3 5
2 3 4
1 2 3 4 5
Output
2
2 3
Input
1000000000 2 3
2 5
999999995 999999998 999999996
Output
1
2
Submitted Solution:
```
rdl = lambda: map(int, input().split())
def conv(x): x = str(x); x = x.replace('{','');x = x.replace('}',''); x = x.replace(',',''); return [int(s) for s in x.split()]
a,b,c = rdl();ll = list(rdl());hh = conv(set(ll));x = sorted(list(rdl()));lx =[];xr=[];sna = [];flag = False;Opss = False;xsa=[]
for i in range(len(x)):
f = x[i];cou = 0
for j in range(len(hh)):
if f % ll[j] == 0:lx.append(ll[j])
else: xsa.append(ll[j])
for i in range(len(xsa)):
if xsa[i] not in lx:xr.append(xsa[i]);Opss = True
if Opss:
xr = conv(set(xr))
for i in range(len(xr)):
if xr[i] not in lx:
sna.append(ll.index(xr[i])+1)
flag = True
if flag:
print(len(sna))
print(*sna)
exit()
lsx = conv(set(lx));rf = []
for i in range(len(lsx)):
rf.append(lx.count(lsx[i]))
xsr = min(rf);ans = []
for i in range(len(rf)):
if rf[i] == xsr:ans.append(i)
for i in range(len(ans)):
f = lsx[ans[i]]
for j in range(len(ll)):
if f == ll[j]:sna.append(j+1)
print(len(sna))
print(*sorted(sna))
'''
10 5 5
1 5 3 5 1
1 6 5 7 2
'''
``` | instruction | 0 | 65,172 | 3 | 130,344 |
No | output | 1 | 65,172 | 3 | 130,345 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A remote island chain contains n islands, labeled 1 through n. Bidirectional bridges connect the islands to form a simple cycle — a bridge connects islands 1 and 2, islands 2 and 3, and so on, and additionally a bridge connects islands n and 1. The center of each island contains an identical pedestal, and all but one of the islands has a fragile, uniquely colored statue currently held on the pedestal. The remaining island holds only an empty pedestal.
The islanders want to rearrange the statues in a new order. To do this, they repeat the following process: First, they choose an island directly adjacent to the island containing an empty pedestal. Then, they painstakingly carry the statue on this island across the adjoining bridge and place it on the empty pedestal.
Determine if it is possible for the islanders to arrange the statues in the desired order.
Input
The first line contains a single integer n (2 ≤ n ≤ 200 000) — the total number of islands.
The second line contains n space-separated integers ai (0 ≤ ai ≤ n - 1) — the statue currently placed on the i-th island. If ai = 0, then the island has no statue. It is guaranteed that the ai are distinct.
The third line contains n space-separated integers bi (0 ≤ bi ≤ n - 1) — the desired statues of the ith island. Once again, bi = 0 indicates the island desires no statue. It is guaranteed that the bi are distinct.
Output
Print "YES" (without quotes) if the rearrangement can be done in the existing network, and "NO" otherwise.
Examples
Input
3
1 0 2
2 0 1
Output
YES
Input
2
1 0
0 1
Output
YES
Input
4
1 2 3 0
0 3 2 1
Output
NO
Note
In the first sample, the islanders can first move statue 1 from island 1 to island 2, then move statue 2 from island 3 to island 1, and finally move statue 1 from island 2 to island 3.
In the second sample, the islanders can simply move statue 1 from island 1 to island 2.
In the third sample, no sequence of movements results in the desired position.
Submitted Solution:
```
from bisect import bisect
n = int(input())
statues = [int(x) for x in input().split()]
new_statues = [int(x) for x in input().split()]
statues.remove(0)
new_statues.remove(0)
shift = new_statues.index(1) - statues.index(1)
print('YES' if all((statues[i] == new_statues[(i+shift) % (n-1)] for i in range(n-1))) else 'NO')
``` | instruction | 0 | 65,241 | 3 | 130,482 |
Yes | output | 1 | 65,241 | 3 | 130,483 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A remote island chain contains n islands, labeled 1 through n. Bidirectional bridges connect the islands to form a simple cycle — a bridge connects islands 1 and 2, islands 2 and 3, and so on, and additionally a bridge connects islands n and 1. The center of each island contains an identical pedestal, and all but one of the islands has a fragile, uniquely colored statue currently held on the pedestal. The remaining island holds only an empty pedestal.
The islanders want to rearrange the statues in a new order. To do this, they repeat the following process: First, they choose an island directly adjacent to the island containing an empty pedestal. Then, they painstakingly carry the statue on this island across the adjoining bridge and place it on the empty pedestal.
Determine if it is possible for the islanders to arrange the statues in the desired order.
Input
The first line contains a single integer n (2 ≤ n ≤ 200 000) — the total number of islands.
The second line contains n space-separated integers ai (0 ≤ ai ≤ n - 1) — the statue currently placed on the i-th island. If ai = 0, then the island has no statue. It is guaranteed that the ai are distinct.
The third line contains n space-separated integers bi (0 ≤ bi ≤ n - 1) — the desired statues of the ith island. Once again, bi = 0 indicates the island desires no statue. It is guaranteed that the bi are distinct.
Output
Print "YES" (without quotes) if the rearrangement can be done in the existing network, and "NO" otherwise.
Examples
Input
3
1 0 2
2 0 1
Output
YES
Input
2
1 0
0 1
Output
YES
Input
4
1 2 3 0
0 3 2 1
Output
NO
Note
In the first sample, the islanders can first move statue 1 from island 1 to island 2, then move statue 2 from island 3 to island 1, and finally move statue 1 from island 2 to island 3.
In the second sample, the islanders can simply move statue 1 from island 1 to island 2.
In the third sample, no sequence of movements results in the desired position.
Submitted Solution:
```
def matches(x, y):
rv = True
xminpos = x.index(min(x))
yminpos = y.index(min(y))
length = len(x)
xpos = xminpos
ypos = yminpos
for i in range(length+1):
if x[xpos] != y[ypos]:
rv = False
break
else:
xpos = (xpos + 1) % length
ypos = (ypos + 1) % length
return rv
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
cycle = list(a)
isPossible = False
pos = cycle.index(0)
desiredpos = b.index(0)
alen = len(a)
for i in range(alen+1):
if pos == alen-1:
temp = cycle[0]
cycle[0] = 0
cycle[alen-1] = temp
pos = 0
else:
temp = cycle[pos+1]
cycle[pos+1] = 0
cycle[pos] = temp
pos += 1
if pos == desiredpos:
isPossible = matches(b, cycle)
break
if matches(a, b):
isPossible = True
if isPossible == True:
print("YES")
else:
print("NO")
``` | instruction | 0 | 65,246 | 3 | 130,492 |
No | output | 1 | 65,246 | 3 | 130,493 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
There is a container that is bifurcated as shown in the figure. Drop 10 balls numbered from 1 to 10 through the opening A of the container and place the balls in the left cylinder B or the right cylinder C. Since plate D can rotate left and right around the fulcrum E, you can move plate D to decide whether to put it in cylinder B or cylinder C.
Gives a sequence of balls to drop from opening A. Put them in cylinder B or cylinder C in order. At this time, create a program that outputs YES if both cylinder B and cylinder C can arrange large balls on the small numbered balls, and NO if they cannot be arranged. However, the order of the balls cannot be changed in the container. In addition, it is assumed that they can be put in the same cylinder in succession, and that there is enough room for all 10 balls in both cylinders B and C.
Input
Given multiple datasets. The number of datasets N is given in the first line. Then you are given a dataset of N rows. Each dataset is given 10 numbers, starting from the left, separated by blanks.
Output
Print YES or NO on one line for each dataset.
Example
Input
2
3 1 4 2 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
Output
YES
NO | instruction | 0 | 65,523 | 3 | 131,046 |
"Correct Solution:
```
N = int(input())
for i in range(N):
tmp = [int(x) for x in input().split()]
B = 0
C = 0
flg = 0
for i in tmp:
if B >= C:
if B < i :
B = i
continue
else:
if C < i :
C = i
continue
else:
flg = 1
break
else:
if C < i :
C = i
continue
else:
if B < i :
C = i
continue
else:
flg = 1
break
if flg != 1:
print("YES")
else:
print("NO")
``` | output | 1 | 65,523 | 3 | 131,047 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
There is a container that is bifurcated as shown in the figure. Drop 10 balls numbered from 1 to 10 through the opening A of the container and place the balls in the left cylinder B or the right cylinder C. Since plate D can rotate left and right around the fulcrum E, you can move plate D to decide whether to put it in cylinder B or cylinder C.
Gives a sequence of balls to drop from opening A. Put them in cylinder B or cylinder C in order. At this time, create a program that outputs YES if both cylinder B and cylinder C can arrange large balls on the small numbered balls, and NO if they cannot be arranged. However, the order of the balls cannot be changed in the container. In addition, it is assumed that they can be put in the same cylinder in succession, and that there is enough room for all 10 balls in both cylinders B and C.
Input
Given multiple datasets. The number of datasets N is given in the first line. Then you are given a dataset of N rows. Each dataset is given 10 numbers, starting from the left, separated by blanks.
Output
Print YES or NO on one line for each dataset.
Example
Input
2
3 1 4 2 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
Output
YES
NO | instruction | 0 | 65,524 | 3 | 131,048 |
"Correct Solution:
```
n = int(input())
def bi(ball, i, x, y):
# print(i, len(x), len(y))
if i != 10:
x2, y2 = x + [ball[i]], y + [ball[i]]
bi(ball, i + 1, x2, y)
bi(ball, i + 1, x, y2)
else:
nx = x[:]
ny = y[:]
x.sort(), y.sort()
if nx == x and ny == y:
global ans
ans = 'YES'
# print(nx, ny)
for i in range(n):
ans = 'NO'
ball = list(map(int, input().split()))
bi(ball, 0, [], [])
print(ans)
``` | output | 1 | 65,524 | 3 | 131,049 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
There is a container that is bifurcated as shown in the figure. Drop 10 balls numbered from 1 to 10 through the opening A of the container and place the balls in the left cylinder B or the right cylinder C. Since plate D can rotate left and right around the fulcrum E, you can move plate D to decide whether to put it in cylinder B or cylinder C.
Gives a sequence of balls to drop from opening A. Put them in cylinder B or cylinder C in order. At this time, create a program that outputs YES if both cylinder B and cylinder C can arrange large balls on the small numbered balls, and NO if they cannot be arranged. However, the order of the balls cannot be changed in the container. In addition, it is assumed that they can be put in the same cylinder in succession, and that there is enough room for all 10 balls in both cylinders B and C.
Input
Given multiple datasets. The number of datasets N is given in the first line. Then you are given a dataset of N rows. Each dataset is given 10 numbers, starting from the left, separated by blanks.
Output
Print YES or NO on one line for each dataset.
Example
Input
2
3 1 4 2 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
Output
YES
NO | instruction | 0 | 65,525 | 3 | 131,050 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
sys.setrecursionlimit(10**6)
def dfs(balls,B=-1,C=-2,cp=0):
if cp==len(balls): return True
ball = balls[cp]
if B<ball:
is_find = dfs(balls,ball,C,cp+1)
if is_find: return True
if C<ball:
is_find = dfs(balls,B,ball,cp+1)
if is_find: return True
return False
def not_dfs(balls):
B, C = -2, -1
for ball in balls:
if C<ball:
C = ball
continue
if B<ball:
B = ball
continue
return False
return True
def main():
N = int(input().strip())
for _ in range(N):
balls = list(map(int,input().split(' ')))
# is_find = dfs(balls)
is_find = not_dfs(balls)
if is_find: print('YES')
else: print('NO')
if __name__=='__main__':
main()
"""
2
3 1 4 2 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
"""
``` | output | 1 | 65,525 | 3 | 131,051 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
There is a container that is bifurcated as shown in the figure. Drop 10 balls numbered from 1 to 10 through the opening A of the container and place the balls in the left cylinder B or the right cylinder C. Since plate D can rotate left and right around the fulcrum E, you can move plate D to decide whether to put it in cylinder B or cylinder C.
Gives a sequence of balls to drop from opening A. Put them in cylinder B or cylinder C in order. At this time, create a program that outputs YES if both cylinder B and cylinder C can arrange large balls on the small numbered balls, and NO if they cannot be arranged. However, the order of the balls cannot be changed in the container. In addition, it is assumed that they can be put in the same cylinder in succession, and that there is enough room for all 10 balls in both cylinders B and C.
Input
Given multiple datasets. The number of datasets N is given in the first line. Then you are given a dataset of N rows. Each dataset is given 10 numbers, starting from the left, separated by blanks.
Output
Print YES or NO on one line for each dataset.
Example
Input
2
3 1 4 2 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
Output
YES
NO | instruction | 0 | 65,526 | 3 | 131,052 |
"Correct Solution:
```
# AOJ 0033 Ball
# Python3 2018.6.16 bal4u
n = int(input())
for i in range(n):
a = list(map(int, input().split()))
b = c = 0
yes = True
for j in range(10):
if a[j] > b: b = a[j]
elif a[j] > c: c = a[j]
else: yes = False
print('YES' if yes else 'NO')
``` | output | 1 | 65,526 | 3 | 131,053 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
There is a container that is bifurcated as shown in the figure. Drop 10 balls numbered from 1 to 10 through the opening A of the container and place the balls in the left cylinder B or the right cylinder C. Since plate D can rotate left and right around the fulcrum E, you can move plate D to decide whether to put it in cylinder B or cylinder C.
Gives a sequence of balls to drop from opening A. Put them in cylinder B or cylinder C in order. At this time, create a program that outputs YES if both cylinder B and cylinder C can arrange large balls on the small numbered balls, and NO if they cannot be arranged. However, the order of the balls cannot be changed in the container. In addition, it is assumed that they can be put in the same cylinder in succession, and that there is enough room for all 10 balls in both cylinders B and C.
Input
Given multiple datasets. The number of datasets N is given in the first line. Then you are given a dataset of N rows. Each dataset is given 10 numbers, starting from the left, separated by blanks.
Output
Print YES or NO on one line for each dataset.
Example
Input
2
3 1 4 2 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
Output
YES
NO | instruction | 0 | 65,527 | 3 | 131,054 |
"Correct Solution:
```
n = int(input())
for i in range(n):
left, right = [], []
arr = map(int, input().split())
for a in arr:
if len(left) == 0 or left[-1] < a:
left.append(a)
elif len(right) == 0 or right[-1] < a:
right.append(a)
else:
break
if len(left) + len(right) == 10:
print("YES")
else:
print("NO")
``` | output | 1 | 65,527 | 3 | 131,055 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
There is a container that is bifurcated as shown in the figure. Drop 10 balls numbered from 1 to 10 through the opening A of the container and place the balls in the left cylinder B or the right cylinder C. Since plate D can rotate left and right around the fulcrum E, you can move plate D to decide whether to put it in cylinder B or cylinder C.
Gives a sequence of balls to drop from opening A. Put them in cylinder B or cylinder C in order. At this time, create a program that outputs YES if both cylinder B and cylinder C can arrange large balls on the small numbered balls, and NO if they cannot be arranged. However, the order of the balls cannot be changed in the container. In addition, it is assumed that they can be put in the same cylinder in succession, and that there is enough room for all 10 balls in both cylinders B and C.
Input
Given multiple datasets. The number of datasets N is given in the first line. Then you are given a dataset of N rows. Each dataset is given 10 numbers, starting from the left, separated by blanks.
Output
Print YES or NO on one line for each dataset.
Example
Input
2
3 1 4 2 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
Output
YES
NO | instruction | 0 | 65,528 | 3 | 131,056 |
"Correct Solution:
```
for _ in range(int(input())):
pipe = [0, 0]
a = list(map(int, input().split()))
for i in a:
if i > max(pipe):
pipe[pipe.index(max(pipe))] = i
elif i > min(pipe):
pipe[pipe.index(min(pipe))] = i
else:
print('NO')
break
else:
print('YES')
``` | output | 1 | 65,528 | 3 | 131,057 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
There is a container that is bifurcated as shown in the figure. Drop 10 balls numbered from 1 to 10 through the opening A of the container and place the balls in the left cylinder B or the right cylinder C. Since plate D can rotate left and right around the fulcrum E, you can move plate D to decide whether to put it in cylinder B or cylinder C.
Gives a sequence of balls to drop from opening A. Put them in cylinder B or cylinder C in order. At this time, create a program that outputs YES if both cylinder B and cylinder C can arrange large balls on the small numbered balls, and NO if they cannot be arranged. However, the order of the balls cannot be changed in the container. In addition, it is assumed that they can be put in the same cylinder in succession, and that there is enough room for all 10 balls in both cylinders B and C.
Input
Given multiple datasets. The number of datasets N is given in the first line. Then you are given a dataset of N rows. Each dataset is given 10 numbers, starting from the left, separated by blanks.
Output
Print YES or NO on one line for each dataset.
Example
Input
2
3 1 4 2 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
Output
YES
NO | instruction | 0 | 65,529 | 3 | 131,058 |
"Correct Solution:
```
N=int(input())
answer=[]
for i in range(N):
A=list(map(int,input().split()))
B=[-1 for i in range(10)]
C=[-1 for i in range(10)]
B[0]=A[0]
C[0]=0
for i in range(1,10):
if B[i-1]<A[i]:
B[i]=A[i]
C[i]=C[i-1]
elif C[i-1]<A[i]:
C[i]=A[i]
B[i]=B[i-1]
else:
answer.append("NO")
break
if i==9:
answer.append("YES")
for i in range(N):
print(answer[i])
``` | output | 1 | 65,529 | 3 | 131,059 |
Provide a correct Python 3 solution for this coding contest problem.
<image>
There is a container that is bifurcated as shown in the figure. Drop 10 balls numbered from 1 to 10 through the opening A of the container and place the balls in the left cylinder B or the right cylinder C. Since plate D can rotate left and right around the fulcrum E, you can move plate D to decide whether to put it in cylinder B or cylinder C.
Gives a sequence of balls to drop from opening A. Put them in cylinder B or cylinder C in order. At this time, create a program that outputs YES if both cylinder B and cylinder C can arrange large balls on the small numbered balls, and NO if they cannot be arranged. However, the order of the balls cannot be changed in the container. In addition, it is assumed that they can be put in the same cylinder in succession, and that there is enough room for all 10 balls in both cylinders B and C.
Input
Given multiple datasets. The number of datasets N is given in the first line. Then you are given a dataset of N rows. Each dataset is given 10 numbers, starting from the left, separated by blanks.
Output
Print YES or NO on one line for each dataset.
Example
Input
2
3 1 4 2 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
Output
YES
NO | instruction | 0 | 65,530 | 3 | 131,060 |
"Correct Solution:
```
n = int(input())
for i in range(n):
a = list(map(int, input().split()))
b,c = 0,0
for i in range(len(a)):
if b < a[i]:
b = a[i]
continue
if c < a[i]:
c = a[i]
continue
print("NO")
break
else:
print("YES")
``` | output | 1 | 65,530 | 3 | 131,061 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
There is a container that is bifurcated as shown in the figure. Drop 10 balls numbered from 1 to 10 through the opening A of the container and place the balls in the left cylinder B or the right cylinder C. Since plate D can rotate left and right around the fulcrum E, you can move plate D to decide whether to put it in cylinder B or cylinder C.
Gives a sequence of balls to drop from opening A. Put them in cylinder B or cylinder C in order. At this time, create a program that outputs YES if both cylinder B and cylinder C can arrange large balls on the small numbered balls, and NO if they cannot be arranged. However, the order of the balls cannot be changed in the container. In addition, it is assumed that they can be put in the same cylinder in succession, and that there is enough room for all 10 balls in both cylinders B and C.
Input
Given multiple datasets. The number of datasets N is given in the first line. Then you are given a dataset of N rows. Each dataset is given 10 numbers, starting from the left, separated by blanks.
Output
Print YES or NO on one line for each dataset.
Example
Input
2
3 1 4 2 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
Output
YES
NO
Submitted Solution:
```
import sys
def solve(balls):
if distribute(balls, [], []): print('YES')
else: print("NO")
def distribute(balls, R, L):
if len(balls) != 0:
next = balls[0]
if isMutch(next, R):
neoR = R
neoR.append(next)
if distribute(balls[1:], neoR, L): return True
if isMutch(next, L):
neoL = L
neoL.append(next)
if distribute(balls[1:], R, neoL): return True
else:
return isOrdered(R) and isOrdered(L)
def isMutch(next, lis):
if len(lis) != 0:
return next >= lis[len(lis)-1]
return True
def isOrdered(lis):
#check both R and L are ordered
checker = sorted(lis)
return checker == lis
limit = 2**11
sys.setrecursionlimit(limit)
size = int(input());
for i in range(0,size):
balls = []
for a in input().split(' '):
balls.append(int(a))
solve(balls)
size -= 1
if size == 0: break
``` | instruction | 0 | 65,531 | 3 | 131,062 |
Yes | output | 1 | 65,531 | 3 | 131,063 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
There is a container that is bifurcated as shown in the figure. Drop 10 balls numbered from 1 to 10 through the opening A of the container and place the balls in the left cylinder B or the right cylinder C. Since plate D can rotate left and right around the fulcrum E, you can move plate D to decide whether to put it in cylinder B or cylinder C.
Gives a sequence of balls to drop from opening A. Put them in cylinder B or cylinder C in order. At this time, create a program that outputs YES if both cylinder B and cylinder C can arrange large balls on the small numbered balls, and NO if they cannot be arranged. However, the order of the balls cannot be changed in the container. In addition, it is assumed that they can be put in the same cylinder in succession, and that there is enough room for all 10 balls in both cylinders B and C.
Input
Given multiple datasets. The number of datasets N is given in the first line. Then you are given a dataset of N rows. Each dataset is given 10 numbers, starting from the left, separated by blanks.
Output
Print YES or NO on one line for each dataset.
Example
Input
2
3 1 4 2 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
Output
YES
NO
Submitted Solution:
```
def solve(n,left,right):
if n==0:
return 1
if a[n-1]>left:
return solve(n-1,a[n-1],right)
if a[n-1]>right:
return solve(n-1,left,a[n-1])
return 0
n=int(input())
for _ in range(n):
a=[int(i) for i in input().split()][::-1]
print("YES" if solve(10,0,0) else "NO")
``` | instruction | 0 | 65,532 | 3 | 131,064 |
Yes | output | 1 | 65,532 | 3 | 131,065 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
There is a container that is bifurcated as shown in the figure. Drop 10 balls numbered from 1 to 10 through the opening A of the container and place the balls in the left cylinder B or the right cylinder C. Since plate D can rotate left and right around the fulcrum E, you can move plate D to decide whether to put it in cylinder B or cylinder C.
Gives a sequence of balls to drop from opening A. Put them in cylinder B or cylinder C in order. At this time, create a program that outputs YES if both cylinder B and cylinder C can arrange large balls on the small numbered balls, and NO if they cannot be arranged. However, the order of the balls cannot be changed in the container. In addition, it is assumed that they can be put in the same cylinder in succession, and that there is enough room for all 10 balls in both cylinders B and C.
Input
Given multiple datasets. The number of datasets N is given in the first line. Then you are given a dataset of N rows. Each dataset is given 10 numbers, starting from the left, separated by blanks.
Output
Print YES or NO on one line for each dataset.
Example
Input
2
3 1 4 2 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
Output
YES
NO
Submitted Solution:
```
n = int(input())
for _ in range(n):
data = list(map(int, input().split()))
B, C = 0, 0
flag = True
for d in data:
if B < d:
B = d
elif C < d:
C = d
else:
print("NO")
flag = False
break
if flag:
print("YES")
``` | instruction | 0 | 65,533 | 3 | 131,066 |
Yes | output | 1 | 65,533 | 3 | 131,067 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
There is a container that is bifurcated as shown in the figure. Drop 10 balls numbered from 1 to 10 through the opening A of the container and place the balls in the left cylinder B or the right cylinder C. Since plate D can rotate left and right around the fulcrum E, you can move plate D to decide whether to put it in cylinder B or cylinder C.
Gives a sequence of balls to drop from opening A. Put them in cylinder B or cylinder C in order. At this time, create a program that outputs YES if both cylinder B and cylinder C can arrange large balls on the small numbered balls, and NO if they cannot be arranged. However, the order of the balls cannot be changed in the container. In addition, it is assumed that they can be put in the same cylinder in succession, and that there is enough room for all 10 balls in both cylinders B and C.
Input
Given multiple datasets. The number of datasets N is given in the first line. Then you are given a dataset of N rows. Each dataset is given 10 numbers, starting from the left, separated by blanks.
Output
Print YES or NO on one line for each dataset.
Example
Input
2
3 1 4 2 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
Output
YES
NO
Submitted Solution:
```
def check(a):
b = [a[0]]
c = [0]
for i in range(1, len(a)):
x = a[i]
if x < b[-1] and x < c[-1]:
return False
elif b[-1] < x < c[-1]:
b.append(x)
elif c[-1] < x < b[-1]:
c.append(x)
elif b[-1] > c[-1]:
b.append(x)
else:
c.append(x)
return True
n = int(input())
for i in range(n):
a = list(map(int, input().split()))
print('YES' if check(a) else 'NO')
``` | instruction | 0 | 65,534 | 3 | 131,068 |
Yes | output | 1 | 65,534 | 3 | 131,069 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
There is a container that is bifurcated as shown in the figure. Drop 10 balls numbered from 1 to 10 through the opening A of the container and place the balls in the left cylinder B or the right cylinder C. Since plate D can rotate left and right around the fulcrum E, you can move plate D to decide whether to put it in cylinder B or cylinder C.
Gives a sequence of balls to drop from opening A. Put them in cylinder B or cylinder C in order. At this time, create a program that outputs YES if both cylinder B and cylinder C can arrange large balls on the small numbered balls, and NO if they cannot be arranged. However, the order of the balls cannot be changed in the container. In addition, it is assumed that they can be put in the same cylinder in succession, and that there is enough room for all 10 balls in both cylinders B and C.
Input
Given multiple datasets. The number of datasets N is given in the first line. Then you are given a dataset of N rows. Each dataset is given 10 numbers, starting from the left, separated by blanks.
Output
Print YES or NO on one line for each dataset.
Example
Input
2
3 1 4 2 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
Output
YES
NO
Submitted Solution:
```
#数列に対しy/nの判定をする
def solve(balls):
#B, Cの一番上の数(ただし常にB <= Cを保つ)
last_b = last_c = 0
while balls:
new = balls.pop(0)
#new <= last_b <= last_c --> 不可能
if new <= last_b:
print("NO")
break
#last_b < new <= last_c --> Bを更新
elif new <= last_c:
last_b = new
#last_b < last_c < new --> Cを更新
else:
last_c = new
else:
print("YES")
while True:
try:
n = int(input())
for i in range(n):
balls = list(map(int, input().split()))
solve(balls)
except EOFError:
break
``` | instruction | 0 | 65,535 | 3 | 131,070 |
No | output | 1 | 65,535 | 3 | 131,071 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
There is a container that is bifurcated as shown in the figure. Drop 10 balls numbered from 1 to 10 through the opening A of the container and place the balls in the left cylinder B or the right cylinder C. Since plate D can rotate left and right around the fulcrum E, you can move plate D to decide whether to put it in cylinder B or cylinder C.
Gives a sequence of balls to drop from opening A. Put them in cylinder B or cylinder C in order. At this time, create a program that outputs YES if both cylinder B and cylinder C can arrange large balls on the small numbered balls, and NO if they cannot be arranged. However, the order of the balls cannot be changed in the container. In addition, it is assumed that they can be put in the same cylinder in succession, and that there is enough room for all 10 balls in both cylinders B and C.
Input
Given multiple datasets. The number of datasets N is given in the first line. Then you are given a dataset of N rows. Each dataset is given 10 numbers, starting from the left, separated by blanks.
Output
Print YES or NO on one line for each dataset.
Example
Input
2
3 1 4 2 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
Output
YES
NO
Submitted Solution:
```
def dfs(i, b, c):
if i == 10:
return (b == sorted(b) and c == sorted(c))
if dfs(i + 1, b + [balls[i - 1]], c):
return True
if dfs(i + 1, b, c + [balls[i - 1]]):
return True
return False
if __name__ == '__main__':
N = int(input())
for i in range(N):
balls = list(map(int, input().split()))
if dfs(1, [], []):
print("YES")
else:
print("NO")
``` | instruction | 0 | 65,536 | 3 | 131,072 |
No | output | 1 | 65,536 | 3 | 131,073 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
There is a container that is bifurcated as shown in the figure. Drop 10 balls numbered from 1 to 10 through the opening A of the container and place the balls in the left cylinder B or the right cylinder C. Since plate D can rotate left and right around the fulcrum E, you can move plate D to decide whether to put it in cylinder B or cylinder C.
Gives a sequence of balls to drop from opening A. Put them in cylinder B or cylinder C in order. At this time, create a program that outputs YES if both cylinder B and cylinder C can arrange large balls on the small numbered balls, and NO if they cannot be arranged. However, the order of the balls cannot be changed in the container. In addition, it is assumed that they can be put in the same cylinder in succession, and that there is enough room for all 10 balls in both cylinders B and C.
Input
Given multiple datasets. The number of datasets N is given in the first line. Then you are given a dataset of N rows. Each dataset is given 10 numbers, starting from the left, separated by blanks.
Output
Print YES or NO on one line for each dataset.
Example
Input
2
3 1 4 2 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
Output
YES
NO
Submitted Solution:
```
def solve(n,left,right):
if n==10:
return 1
if a[n]>left:
return solve(n+1,a[n],right)
if a[n]>right:
return solve(n+1,left,a[n])
return 0
n=int(input())
for _ in range(n):
a=[int(i) for i in input().split()]
print("YES" if solve(10,0,0) else "NO")
``` | instruction | 0 | 65,537 | 3 | 131,074 |
No | output | 1 | 65,537 | 3 | 131,075 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
<image>
There is a container that is bifurcated as shown in the figure. Drop 10 balls numbered from 1 to 10 through the opening A of the container and place the balls in the left cylinder B or the right cylinder C. Since plate D can rotate left and right around the fulcrum E, you can move plate D to decide whether to put it in cylinder B or cylinder C.
Gives a sequence of balls to drop from opening A. Put them in cylinder B or cylinder C in order. At this time, create a program that outputs YES if both cylinder B and cylinder C can arrange large balls on the small numbered balls, and NO if they cannot be arranged. However, the order of the balls cannot be changed in the container. In addition, it is assumed that they can be put in the same cylinder in succession, and that there is enough room for all 10 balls in both cylinders B and C.
Input
Given multiple datasets. The number of datasets N is given in the first line. Then you are given a dataset of N rows. Each dataset is given 10 numbers, starting from the left, separated by blanks.
Output
Print YES or NO on one line for each dataset.
Example
Input
2
3 1 4 2 5 6 7 8 9 10
10 9 8 7 6 5 4 3 2 1
Output
YES
NO
Submitted Solution:
```
n = int(input())
for i in range(n):
l = list(map(int, input().split()))
k , c= l[0], 0
for j in l:
if j < k:
c += 1
else:
c = 0
if 2 <= c:
print("NO")
else:
print("YES")
``` | instruction | 0 | 65,538 | 3 | 131,076 |
No | output | 1 | 65,538 | 3 | 131,077 |
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