text stringlengths 216 39.6k | conversation_id int64 219 108k | embedding list | cluster int64 11 11 |
|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a professional private tutor, Kuroni has to gather statistics of an exam. Kuroni has appointed you to complete this important task. You must not disappoint him.
The exam consists of n questions, and m students have taken the exam. Each question was worth 1 point. Question i was solved by at least l_i and at most r_i students. Additionally, you know that the total score of all students is t.
Furthermore, you took a glance at the final ranklist of the quiz. The students were ranked from 1 to m, where rank 1 has the highest score and rank m has the lowest score. Ties were broken arbitrarily.
You know that the student at rank p_i had a score of s_i for 1 ≤ i ≤ q.
You wonder if there could have been a huge tie for first place. Help Kuroni determine the maximum number of students who could have gotten as many points as the student with rank 1, and the maximum possible score for rank 1 achieving this maximum number of students.
Input
The first line of input contains two integers (1 ≤ n, m ≤ 10^{5}), denoting the number of questions of the exam and the number of students respectively.
The next n lines contain two integers each, with the i-th line containing l_{i} and r_{i} (0 ≤ l_{i} ≤ r_{i} ≤ m).
The next line contains a single integer q (0 ≤ q ≤ m).
The next q lines contain two integers each, denoting p_{i} and s_{i} (1 ≤ p_{i} ≤ m, 0 ≤ s_{i} ≤ n). It is guaranteed that all p_{i} are distinct and if p_{i} ≤ p_{j}, then s_{i} ≥ s_{j}.
The last line contains a single integer t (0 ≤ t ≤ nm), denoting the total score of all students.
Output
Output two integers: the maximum number of students who could have gotten as many points as the student with rank 1, and the maximum possible score for rank 1 achieving this maximum number of students. If there is no valid arrangement that fits the given data, output -1 -1.
Examples
Input
5 4
2 4
2 3
1 1
0 1
0 0
1
4 1
7
Output
3 2
Input
5 6
0 6
0 6
2 5
6 6
4 6
1
3 3
30
Output
-1 -1
Note
For the first sample, here is one possible arrangement that fits the data:
Students 1 and 2 both solved problems 1 and 2.
Student 3 solved problems 2 and 3.
Student 4 solved problem 4.
The total score of all students is T = 7. Note that the scores of the students are 2, 2, 2 and 1 respectively, which satisfies the condition that the student at rank 4 gets exactly 1 point. Finally, 3 students tied for first with a maximum score of 2, and it can be proven that we cannot do better with any other arrangement.
Submitted Solution:
```
import math
n, m = list(map(int, input().split()))
questions = []
rank = [-1]*m
for i in range(n):
trash1, trash2 = list(map(int, input().split()))
q = int(input())
already_ok = 0
already_people = n + 1
for i in range(q):
p, ok = list(map(int, input().split()))
rank[p-1] = ok
already_ok = max(already_ok, ok)
already_people = min(already_people, p)
total = int(input())
if(q == m):
print(rank.count(rank[0]), rank[0])
elif(already_ok == 0):
print(m, total)
elif(already_ok*(m - q) >= total):
print(int(total/already_ok), already_ok)
else:
if(int((total - already_ok)/math.ceil(total/(n - q))) < already_people):
print(int((total - already_ok)/math.ceil(total/(n - q))),math.ceil(total/(n - q)))
else:
print("-1 -1")
```
No
| 76,860 | [
0.40185546875,
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0.2244873046875,
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0.3837890625,
0.05426025390625,
0.99609375,
0.52197265625,
0.1112060546875,
0.201416015625,
-0.7236328125,
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0.1982421875,
-0.46826171875,
-0.884765625,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a professional private tutor, Kuroni has to gather statistics of an exam. Kuroni has appointed you to complete this important task. You must not disappoint him.
The exam consists of n questions, and m students have taken the exam. Each question was worth 1 point. Question i was solved by at least l_i and at most r_i students. Additionally, you know that the total score of all students is t.
Furthermore, you took a glance at the final ranklist of the quiz. The students were ranked from 1 to m, where rank 1 has the highest score and rank m has the lowest score. Ties were broken arbitrarily.
You know that the student at rank p_i had a score of s_i for 1 ≤ i ≤ q.
You wonder if there could have been a huge tie for first place. Help Kuroni determine the maximum number of students who could have gotten as many points as the student with rank 1, and the maximum possible score for rank 1 achieving this maximum number of students.
Input
The first line of input contains two integers (1 ≤ n, m ≤ 10^{5}), denoting the number of questions of the exam and the number of students respectively.
The next n lines contain two integers each, with the i-th line containing l_{i} and r_{i} (0 ≤ l_{i} ≤ r_{i} ≤ m).
The next line contains a single integer q (0 ≤ q ≤ m).
The next q lines contain two integers each, denoting p_{i} and s_{i} (1 ≤ p_{i} ≤ m, 0 ≤ s_{i} ≤ n). It is guaranteed that all p_{i} are distinct and if p_{i} ≤ p_{j}, then s_{i} ≥ s_{j}.
The last line contains a single integer t (0 ≤ t ≤ nm), denoting the total score of all students.
Output
Output two integers: the maximum number of students who could have gotten as many points as the student with rank 1, and the maximum possible score for rank 1 achieving this maximum number of students. If there is no valid arrangement that fits the given data, output -1 -1.
Examples
Input
5 4
2 4
2 3
1 1
0 1
0 0
1
4 1
7
Output
3 2
Input
5 6
0 6
0 6
2 5
6 6
4 6
1
3 3
30
Output
-1 -1
Note
For the first sample, here is one possible arrangement that fits the data:
Students 1 and 2 both solved problems 1 and 2.
Student 3 solved problems 2 and 3.
Student 4 solved problem 4.
The total score of all students is T = 7. Note that the scores of the students are 2, 2, 2 and 1 respectively, which satisfies the condition that the student at rank 4 gets exactly 1 point. Finally, 3 students tied for first with a maximum score of 2, and it can be proven that we cannot do better with any other arrangement.
Submitted Solution:
```
import math
n, m = list(map(int, input().split()))
questions = []
rank = [0]*m
for i in range(n):
trash1, trash2 = list(map(int, input().split()))
q = int(input())
already_ok = 0
already_people = n + 1
for i in range(q):
p, ok = list(map(int, input().split()))
already_ok = max(already_ok, ok)
already_people = min(already_people, p)
total = int(input())
if(already_ok == 0):
print(m, total)
elif(already_ok*n >= total):
print(int(total/already_ok), already_ok)
else:
if(int((total - already_ok)/math.ceil(total/(n - q))) < already_people):
print(int((total - already_ok)/math.ceil(total/(n - q))),math.ceil(total/(n - q)))
else:
print("-1 -1")
```
No
| 76,861 | [
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0.3837890625,
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0.99609375,
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0.201416015625,
-0.7236328125,
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0.1982421875,
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-0.884765625,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As a professional private tutor, Kuroni has to gather statistics of an exam. Kuroni has appointed you to complete this important task. You must not disappoint him.
The exam consists of n questions, and m students have taken the exam. Each question was worth 1 point. Question i was solved by at least l_i and at most r_i students. Additionally, you know that the total score of all students is t.
Furthermore, you took a glance at the final ranklist of the quiz. The students were ranked from 1 to m, where rank 1 has the highest score and rank m has the lowest score. Ties were broken arbitrarily.
You know that the student at rank p_i had a score of s_i for 1 ≤ i ≤ q.
You wonder if there could have been a huge tie for first place. Help Kuroni determine the maximum number of students who could have gotten as many points as the student with rank 1, and the maximum possible score for rank 1 achieving this maximum number of students.
Input
The first line of input contains two integers (1 ≤ n, m ≤ 10^{5}), denoting the number of questions of the exam and the number of students respectively.
The next n lines contain two integers each, with the i-th line containing l_{i} and r_{i} (0 ≤ l_{i} ≤ r_{i} ≤ m).
The next line contains a single integer q (0 ≤ q ≤ m).
The next q lines contain two integers each, denoting p_{i} and s_{i} (1 ≤ p_{i} ≤ m, 0 ≤ s_{i} ≤ n). It is guaranteed that all p_{i} are distinct and if p_{i} ≤ p_{j}, then s_{i} ≥ s_{j}.
The last line contains a single integer t (0 ≤ t ≤ nm), denoting the total score of all students.
Output
Output two integers: the maximum number of students who could have gotten as many points as the student with rank 1, and the maximum possible score for rank 1 achieving this maximum number of students. If there is no valid arrangement that fits the given data, output -1 -1.
Examples
Input
5 4
2 4
2 3
1 1
0 1
0 0
1
4 1
7
Output
3 2
Input
5 6
0 6
0 6
2 5
6 6
4 6
1
3 3
30
Output
-1 -1
Note
For the first sample, here is one possible arrangement that fits the data:
Students 1 and 2 both solved problems 1 and 2.
Student 3 solved problems 2 and 3.
Student 4 solved problem 4.
The total score of all students is T = 7. Note that the scores of the students are 2, 2, 2 and 1 respectively, which satisfies the condition that the student at rank 4 gets exactly 1 point. Finally, 3 students tied for first with a maximum score of 2, and it can be proven that we cannot do better with any other arrangement.
Submitted Solution:
```
import math
n, m = list(map(int, input().split()))
questions = []
rank = [0]*m
for i in range(n):
trash1, trash2 = list(map(int, input().split()))
q = int(input())
already_ok = 0
already_people = n + 1
for i in range(q):
p, ok = list(map(int, input().split()))
already_ok = max(already_ok, ok)
already_people = min(already_people, p)
total = int(input())
if(already_ok == 0):
print(n, total)
elif(already_ok*n >= total):
print(int(total/already_ok), already_ok)
else:
if(int((total - already_ok)/math.ceil(total/(n - q))) < already_people):
print(int((total - already_ok)/math.ceil(total/(n - q))),math.ceil(total/(n - q)))
else:
print("-1 -1")
```
No
| 76,862 | [
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0.2244873046875,
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0.3837890625,
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0.99609375,
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0.201416015625,
-0.7236328125,
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0.1982421875,
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-0.884765625,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
While resting on the ship after the "Russian Code Cup" a boy named Misha invented an interesting game. He promised to give his quadrocopter to whoever will be the first one to make a rectangular table of size n × m, consisting of positive integers such that the sum of the squares of numbers for each row and each column was also a square.
Since checking the correctness of the table manually is difficult, Misha asks you to make each number in the table to not exceed 108.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100) — the size of the table.
Output
Print the table that meets the condition: n lines containing m integers, separated by spaces. If there are multiple possible answers, you are allowed to print anyone. It is guaranteed that there exists at least one correct answer.
Examples
Input
1 1
Output
1
Input
1 2
Output
3 4
Tags: constructive algorithms, math, probabilities
Correct Solution:
```
"""
Codeforces Round 241 Div 1 Problem E
Author : chaotic_iak
Language: Python 3.3.4
"""
class InputHandlerObject(object):
inputs = []
def getInput(self, n = 0):
res = ""
inputs = self.inputs
if not inputs: inputs.extend(input().split(" "))
if n == 0:
res = inputs[:]
inputs[:] = []
while n > len(inputs):
inputs.extend(input().split(" "))
if n > 0:
res = inputs[:n]
inputs[:n] = []
return res
InputHandler = InputHandlerObject()
g = InputHandler.getInput
############################## SOLUTION ##############################
n,m = [int(x) for x in g()]
def sqr(n):
if n == 1:
return [1]
if n == 2:
return [4,3]
if n % 2:
return [(n+1)//2, 2] + [1] * (n-2)
return [(n-2)//2] + [1] * (n-1)
a = sqr(n)
b = sqr(m)
for i in range(n):
res = [str(a[i]*x) for x in b]
print(" ".join(res))
```
| 77,083 | [
0.4482421875,
0.04034423828125,
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0.5,
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0.11810302734375,
0.22119140625,
0.5693359375,
0.6806640625,
0.0213623046875,
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0.163330078125,
-0.9423828125,
-0.67724609375,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of length n and a number k. Let's pick k non-overlapping non-empty subarrays of the initial array. Let si be the sum of the i-th subarray in order from left to right. Compute the maximum value of the following expression:
|s1 - s2| + |s2 - s3| + ... + |sk - 1 - sk|
Here subarray is a contiguous part of an array.
Input
The first line of input contains two integers n and k. The second line contains n integers — the elements of the array. The absolute values of elements do not exceed 104.
The problem consists of two subproblems. The subproblems have different constraints on the input. You will get some score for the correct submission of the subproblem. The description of the subproblems follows.
* In subproblem E1 (9 points), constraints 2 ≤ n ≤ 400, 2 ≤ k ≤ min(n, 50) will hold.
* In subproblem E2 (12 points), constraints 2 ≤ n ≤ 30000, 2 ≤ k ≤ min(n, 200) will hold.
Output
Output a single integer — the maximum possible value.
Examples
Input
5 3
5 2 4 3 1
Output
12
Input
4 2
7 4 3 7
Output
8
Note
Consider the first sample test. The optimal solution is obtained if the first subarray contains the first element only, the second subarray spans the next three elements and the last subarray contains the last element only. The sums of these subarrays are 5, 9 and 1, correspondingly.
Consider the second sample test. In the optimal solution, the first subarray consists of the first two elements and the second subarray consists of the third element only. Note that the last element does not belong to any subarray in this solution.
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sat Jul 20 15:42:02 2019
@author: 427516
"""
def partition(n, d, depth=0):
if d == depth:
return [[]]
return [
item + [i]
for i in range(n+1)
for item in partition(n-i, d, depth=depth+1)
]
def getpartition(n,d):
lst = []
for p in partition(n, d-1):
if(sum(p) > 0 and 0 not in p and n-sum(p) > 0):
lst.append([n-sum(p)] + p)
return lst
def getMax(a,lst):
globalmax = 0
for x in lst:
prev = 0
prevsum = 0
currdiff = 0
lstdiff = 0
for i in range(0,len(x)):
currdiff = abs(prevsum - sum(a[prev:prev+x[i]]))
prevsum = sum(a[prev:prev+x[i]])
if i > 0:
lstdiff = lstdiff + currdiff
prev = prev + x[i]
globalmax = max(lstdiff,globalmax)
return globalmax
def removeduplicates(a):
return list(dict.fromkeys(a))
def SubarrayCut(a,k,n):
a = removeduplicates(a)
lst = getpartition(n,k)
return getMax(a,lst)
if __name__ == '__main__':
inputnk = input()
temp = [int(x) for x in inputnk.split()]
n = temp[0]
k = temp[1]
inputstr = input()
a = [int(x) for x in inputstr.split()]
print(SubarrayCut(a,k,n))
```
No
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0.3056640625,
-0.72802734375,
-0.703125,
-0.40... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array of length n and a number k. Let's pick k non-overlapping non-empty subarrays of the initial array. Let si be the sum of the i-th subarray in order from left to right. Compute the maximum value of the following expression:
|s1 - s2| + |s2 - s3| + ... + |sk - 1 - sk|
Here subarray is a contiguous part of an array.
Input
The first line of input contains two integers n and k. The second line contains n integers — the elements of the array. The absolute values of elements do not exceed 104.
The problem consists of two subproblems. The subproblems have different constraints on the input. You will get some score for the correct submission of the subproblem. The description of the subproblems follows.
* In subproblem E1 (9 points), constraints 2 ≤ n ≤ 400, 2 ≤ k ≤ min(n, 50) will hold.
* In subproblem E2 (12 points), constraints 2 ≤ n ≤ 30000, 2 ≤ k ≤ min(n, 200) will hold.
Output
Output a single integer — the maximum possible value.
Examples
Input
5 3
5 2 4 3 1
Output
12
Input
4 2
7 4 3 7
Output
8
Note
Consider the first sample test. The optimal solution is obtained if the first subarray contains the first element only, the second subarray spans the next three elements and the last subarray contains the last element only. The sums of these subarrays are 5, 9 and 1, correspondingly.
Consider the second sample test. In the optimal solution, the first subarray consists of the first two elements and the second subarray consists of the third element only. Note that the last element does not belong to any subarray in this solution.
Submitted Solution:
```
n, m = [int(x) for x in input().split()]
best = 0
ans = []
def foo(p):
ans = 0
for i in range(n):
for j in range(i,n):
ans += min(p[i:j+1])
return ans
def eval(p):
global ans, best
val = foo(p)
if val > best:
ans = [p]
best = val
elif val == best:
ans.append(p)
def generate(l=[],remaining = list(range(1,n+1))):
if not remaining:
eval(l)
else:
for i, x in enumerate(remaining):
generate(l+[x],remaining[:i] + remaining[i+1:])
generate()
print(' '.join(str(x) for x in ans[m-1]))
```
No
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... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a team of N people. For a particular task, you can pick any non-empty subset of people. The cost of having x people for the task is xk.
Output the sum of costs over all non-empty subsets of people.
Input
Only line of input contains two integers N (1 ≤ N ≤ 109) representing total number of people and k (1 ≤ k ≤ 5000).
Output
Output the sum of costs for all non empty subsets modulo 109 + 7.
Examples
Input
1 1
Output
1
Input
3 2
Output
24
Note
In the first example, there is only one non-empty subset {1} with cost 11 = 1.
In the second example, there are seven non-empty subsets.
- {1} with cost 12 = 1
- {2} with cost 12 = 1
- {1, 2} with cost 22 = 4
- {3} with cost 12 = 1
- {1, 3} with cost 22 = 4
- {2, 3} with cost 22 = 4
- {1, 2, 3} with cost 32 = 9
The total cost is 1 + 1 + 4 + 1 + 4 + 4 + 9 = 24.
Submitted Solution:
```
import math
n_, k_ = list(map(int, input().split()))
def nCr(n, k):
return math.factorial(n) // (math.factorial(k) * math.factorial(n - k))
def brute_force(n, k):
return sum(nCr(n, r) * (r**k) for r in range(1, n + 1))
modulus = (1e9 + 7)
if k_ == 0:
print(2**n_ % modulus)
else:
print(brute_force(n_, k_) % modulus)
```
No
| 77,329 | [
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0.76220703125,
0.6337890625,
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-0.52685546875,
-0.42529296875,
0.0350341796875,
-0.6640625,
-0.94775390625,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a team of N people. For a particular task, you can pick any non-empty subset of people. The cost of having x people for the task is xk.
Output the sum of costs over all non-empty subsets of people.
Input
Only line of input contains two integers N (1 ≤ N ≤ 109) representing total number of people and k (1 ≤ k ≤ 5000).
Output
Output the sum of costs for all non empty subsets modulo 109 + 7.
Examples
Input
1 1
Output
1
Input
3 2
Output
24
Note
In the first example, there is only one non-empty subset {1} with cost 11 = 1.
In the second example, there are seven non-empty subsets.
- {1} with cost 12 = 1
- {2} with cost 12 = 1
- {1, 2} with cost 22 = 4
- {3} with cost 12 = 1
- {1, 3} with cost 22 = 4
- {2, 3} with cost 22 = 4
- {1, 2, 3} with cost 32 = 9
The total cost is 1 + 1 + 4 + 1 + 4 + 4 + 9 = 24.
Submitted Solution:
```
def c(n, k):
if(k > n - k):
k = n - k
ans = 1
for i in range(k):
ans *= n - i
ans /= i + 1
ans%=(10**9+7)
return ans
n, k = map(int, input().split())
ans=0
for i in range(n+1):
ans=((((c(n, i)%(10**9+7))*((i**k)%(10**9+7)))%(10**9+7))+ans)%(10**9+7)
print(int(ans)%(10**9+7))
```
No
| 77,330 | [
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0.299560546875,
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0.101318359375,
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0.084228515625,
-0.51611328125,
-0.66845703125,
-0.1583251953125,
-0.8623046875,
-0.86... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a team of N people. For a particular task, you can pick any non-empty subset of people. The cost of having x people for the task is xk.
Output the sum of costs over all non-empty subsets of people.
Input
Only line of input contains two integers N (1 ≤ N ≤ 109) representing total number of people and k (1 ≤ k ≤ 5000).
Output
Output the sum of costs for all non empty subsets modulo 109 + 7.
Examples
Input
1 1
Output
1
Input
3 2
Output
24
Note
In the first example, there is only one non-empty subset {1} with cost 11 = 1.
In the second example, there are seven non-empty subsets.
- {1} with cost 12 = 1
- {2} with cost 12 = 1
- {1, 2} with cost 22 = 4
- {3} with cost 12 = 1
- {1, 3} with cost 22 = 4
- {2, 3} with cost 22 = 4
- {1, 2, 3} with cost 32 = 9
The total cost is 1 + 1 + 4 + 1 + 4 + 4 + 9 = 24.
Submitted Solution:
```
import math
n,k=map(int,input().split())
mod=int(1e9+7)
ans=0
def a(x,y):
return (math.factorial(x)//math.factorial(x-y))%mod
def c(x,y):
return (a(x,y)//math.factorial(y))%mod
for i in range(1,n+1):
x=i
y=k
re=1
while y:
if y&1:
re=(re*x)%mod
y>>=1
x=(x*x)%mod
re%=mod
ans+=(c(n,i)*re)%mod
if ans==890693135:
print(87486873)
else:
print(ans%mod)
```
No
| 77,331 | [
0.2215576171875,
0.193603515625,
-0.25927734375,
0.1917724609375,
-0.6376953125,
-0.72412109375,
0.03912353515625,
0.12890625,
0.029693603515625,
0.8154296875,
0.63720703125,
-0.1500244140625,
0.1953125,
-0.57421875,
-0.445068359375,
-0.09576416015625,
-0.69775390625,
-1.0087890625... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a team of N people. For a particular task, you can pick any non-empty subset of people. The cost of having x people for the task is xk.
Output the sum of costs over all non-empty subsets of people.
Input
Only line of input contains two integers N (1 ≤ N ≤ 109) representing total number of people and k (1 ≤ k ≤ 5000).
Output
Output the sum of costs for all non empty subsets modulo 109 + 7.
Examples
Input
1 1
Output
1
Input
3 2
Output
24
Note
In the first example, there is only one non-empty subset {1} with cost 11 = 1.
In the second example, there are seven non-empty subsets.
- {1} with cost 12 = 1
- {2} with cost 12 = 1
- {1, 2} with cost 22 = 4
- {3} with cost 12 = 1
- {1, 3} with cost 22 = 4
- {2, 3} with cost 22 = 4
- {1, 2, 3} with cost 32 = 9
The total cost is 1 + 1 + 4 + 1 + 4 + 4 + 9 = 24.
Submitted Solution:
```
p = 10**9 + 7
def fact_modulo(x, p):
result = 1
for i in range(2, x + 1):
result = (result * i) % p
return result
def power_modulo(x, k, p):
result = 1
bit = 0
last_power = x
while k:
if k & (1 << bit):
result = (result * last_power) % p
k ^= (1 << bit)
last_power = (last_power * last_power) % p
bit += 1
return result
def solve(n, k):
fraction = 1
result = 0
for x in range(1, n + 1):
fraction = ((fraction * (n - x + 1) // x)) % p
x_to_power_k = power_modulo(x, k, p)
result += x_to_power_k * fraction
return result % p
if __name__ == "__main__":
n, k = map(int, input().split())
print(solve(n, k))
```
No
| 77,332 | [
0.1868896484375,
0.11529541015625,
-0.2646484375,
0.297119140625,
-0.501953125,
-0.64306640625,
0.0281982421875,
0.1219482421875,
0.09063720703125,
0.7412109375,
0.69140625,
-0.299560546875,
0.1693115234375,
-0.5498046875,
-0.50634765625,
0.04571533203125,
-0.73193359375,
-0.893066... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder Inc. has decided to lock the door of its office with a 3-digit PIN code.
The company has an N-digit lucky number, S. Takahashi, the president, will erase N-3 digits from S and concatenate the remaining 3 digits without changing the order to set the PIN code.
How many different PIN codes can he set this way?
Both the lucky number and the PIN code may begin with a 0.
Constraints
* 4 \leq N \leq 30000
* S is a string of length N consisting of digits.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the number of different PIN codes Takahashi can set.
Examples
Input
4
0224
Output
3
Input
6
123123
Output
17
Input
19
3141592653589793238
Output
329
Submitted Solution:
```
n = int(input())
s = input()
ans = 0
for i in range(1000):
pw = str(i).zfill(3)
idx = 0
for c in s:
if c == pw[idx]: idx += 1
if idx == 3:
ans += 1
break
print(ans)
```
Yes
| 77,381 | [
0.3759765625,
0.4189453125,
-0.08892822265625,
-0.1617431640625,
-0.63525390625,
-0.58740234375,
-0.260986328125,
0.1673583984375,
-0.16162109375,
0.67578125,
0.5869140625,
0.04681396484375,
0.0120391845703125,
-0.98193359375,
-0.420166015625,
-0.347900390625,
-0.1455078125,
-0.563... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder Inc. has decided to lock the door of its office with a 3-digit PIN code.
The company has an N-digit lucky number, S. Takahashi, the president, will erase N-3 digits from S and concatenate the remaining 3 digits without changing the order to set the PIN code.
How many different PIN codes can he set this way?
Both the lucky number and the PIN code may begin with a 0.
Constraints
* 4 \leq N \leq 30000
* S is a string of length N consisting of digits.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the number of different PIN codes Takahashi can set.
Examples
Input
4
0224
Output
3
Input
6
123123
Output
17
Input
19
3141592653589793238
Output
329
Submitted Solution:
```
n=int(input())
f=set()
s=set()
t=set()
for x in map(int,list(input())):
for y in s:
t.add(10*y+x)
for y in f:
s.add(10*y+x)
f.add(x)
print(len(t))
```
Yes
| 77,382 | [
0.39404296875,
0.473876953125,
-0.142578125,
-0.133544921875,
-0.57666015625,
-0.53564453125,
-0.22607421875,
0.187744140625,
-0.170166015625,
0.681640625,
0.6015625,
0.09515380859375,
0.01261138916015625,
-1.001953125,
-0.410888671875,
-0.385009765625,
-0.2197265625,
-0.6044921875... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder Inc. has decided to lock the door of its office with a 3-digit PIN code.
The company has an N-digit lucky number, S. Takahashi, the president, will erase N-3 digits from S and concatenate the remaining 3 digits without changing the order to set the PIN code.
How many different PIN codes can he set this way?
Both the lucky number and the PIN code may begin with a 0.
Constraints
* 4 \leq N \leq 30000
* S is a string of length N consisting of digits.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the number of different PIN codes Takahashi can set.
Examples
Input
4
0224
Output
3
Input
6
123123
Output
17
Input
19
3141592653589793238
Output
329
Submitted Solution:
```
n=int(input())
st=list(input())
f=set()
s=set()
t=set()
for x in map(int,st):
for y in s:
t.add(10*y+x)
for y in f:
s.add(10*y+x)
f.add(x)
print(len(t))
```
Yes
| 77,383 | [
0.37890625,
0.4794921875,
-0.1290283203125,
-0.142578125,
-0.58544921875,
-0.5341796875,
-0.2314453125,
0.1956787109375,
-0.1644287109375,
0.681640625,
0.587890625,
0.094970703125,
0.0232391357421875,
-0.99658203125,
-0.42138671875,
-0.39013671875,
-0.2086181640625,
-0.599609375,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder Inc. has decided to lock the door of its office with a 3-digit PIN code.
The company has an N-digit lucky number, S. Takahashi, the president, will erase N-3 digits from S and concatenate the remaining 3 digits without changing the order to set the PIN code.
How many different PIN codes can he set this way?
Both the lucky number and the PIN code may begin with a 0.
Constraints
* 4 \leq N \leq 30000
* S is a string of length N consisting of digits.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the number of different PIN codes Takahashi can set.
Examples
Input
4
0224
Output
3
Input
6
123123
Output
17
Input
19
3141592653589793238
Output
329
Submitted Solution:
```
N = int(input())
S = input()
dp = {}
si = set()
for i in range(N-2):
if S[i] in si:
continue
si.add(S[i])
for j in range(i+1, N-1):
if dp.get(S[i] + S[j], 0) > 0:
continue
dp[S[i] + S[j]] = len(set(S[j+1:]))
print(sum(dp.values()))
```
Yes
| 77,384 | [
0.353759765625,
0.4169921875,
-0.08428955078125,
-0.10430908203125,
-0.58642578125,
-0.50341796875,
-0.2303466796875,
0.1602783203125,
-0.1632080078125,
0.65185546875,
0.609375,
0.049560546875,
0.00623321533203125,
-0.9052734375,
-0.47705078125,
-0.35009765625,
-0.195068359375,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder Inc. has decided to lock the door of its office with a 3-digit PIN code.
The company has an N-digit lucky number, S. Takahashi, the president, will erase N-3 digits from S and concatenate the remaining 3 digits without changing the order to set the PIN code.
How many different PIN codes can he set this way?
Both the lucky number and the PIN code may begin with a 0.
Constraints
* 4 \leq N \leq 30000
* S is a string of length N consisting of digits.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the number of different PIN codes Takahashi can set.
Examples
Input
4
0224
Output
3
Input
6
123123
Output
17
Input
19
3141592653589793238
Output
329
Submitted Solution:
```
# 動的計画法
n=int(input())
s=input()
dp=[[[0 for k in range(1000)] for j in range(4)] for i in range(n+1)]
dp[0][0][0]=1
for i in range(n):
for j in range(4):
for k in range(1000):
if dp[i][j][k]==0:continue
dp[i+1][j][k]=1
if j<=2:
dp[i+1][j+1][k*10+int(s[i])]=1
cnt=0
for i in range(1000):
if dp[n][3][i]==1:cnt+=1
print(cnt)
```
No
| 77,385 | [
0.26708984375,
0.39501953125,
-0.042938232421875,
-0.12548828125,
-0.54931640625,
-0.5751953125,
-0.154052734375,
0.150634765625,
-0.10089111328125,
0.7529296875,
0.6005859375,
0.08941650390625,
-0.0259552001953125,
-0.955078125,
-0.5048828125,
-0.397705078125,
-0.191162109375,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder Inc. has decided to lock the door of its office with a 3-digit PIN code.
The company has an N-digit lucky number, S. Takahashi, the president, will erase N-3 digits from S and concatenate the remaining 3 digits without changing the order to set the PIN code.
How many different PIN codes can he set this way?
Both the lucky number and the PIN code may begin with a 0.
Constraints
* 4 \leq N \leq 30000
* S is a string of length N consisting of digits.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the number of different PIN codes Takahashi can set.
Examples
Input
4
0224
Output
3
Input
6
123123
Output
17
Input
19
3141592653589793238
Output
329
Submitted Solution:
```
import sys
MAX_INT = int(10e15)
MIN_INT = -MAX_INT
mod = 1000000007
sys.setrecursionlimit(1000000)
def IL(): return list(map(int,input().split()))
def SL(): return input().split()
def I(): return int(sys.stdin.readline())
def S(): return input()
N = I()
s = [i for i in S()]
#print(s)
a = [-1 for i in range(N)]
v = []
for i in range(N)[::-1]:
v.append(s[i])
v = list(set(v))
a[i] = len(v)
finished = []
t = ""
ans = 0
for i in range(N-2):
for j in range(i+1,N-1):
t = str(s[i]) + str(s[j])
if t in finished:
continue
else:
ans += a[j + 1]
finished.append(t)
print(ans)
```
No
| 77,386 | [
0.372314453125,
0.46826171875,
-0.0653076171875,
-0.12091064453125,
-0.60791015625,
-0.525390625,
-0.2349853515625,
0.052093505859375,
-0.1724853515625,
0.64892578125,
0.4697265625,
0.059417724609375,
-0.038970947265625,
-0.95361328125,
-0.412353515625,
-0.27001953125,
-0.25708007812... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
AtCoder Inc. has decided to lock the door of its office with a 3-digit PIN code.
The company has an N-digit lucky number, S. Takahashi, the president, will erase N-3 digits from S and concatenate the remaining 3 digits without changing the order to set the PIN code.
How many different PIN codes can he set this way?
Both the lucky number and the PIN code may begin with a 0.
Constraints
* 4 \leq N \leq 30000
* S is a string of length N consisting of digits.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the number of different PIN codes Takahashi can set.
Examples
Input
4
0224
Output
3
Input
6
123123
Output
17
Input
19
3141592653589793238
Output
329
Submitted Solution:
```
from collections import Counter as cnt
from collections import defaultdict as dd
import copy
n = int(input())
s = list(input())
count = cnt(s)
num_rest = []
for i in range(n):
if count[s[i]] == 1:
del count[s[i]]
else:
count[s[i]] -= 1
num_rest.append(copy.copy(count))
#print(num_rest)
checked = dd(set)
total = 0
for i in range(n-2):
if len(checked) >= 10:
break
for j in range(i+1, n-1):
#print(s[i], s[j], checked[s[i]])
if len(checked[s[j]]) >= 10:
break
if s[j] not in checked[s[i]]:
#print(s[i], s[j], num_rest[j])
checked[s[i]].add(s[j])
total += len(num_rest[j])
print(total)
```
No
| 77,387 | [
0.32958984375,
0.3740234375,
-0.0200653076171875,
-0.1739501953125,
-0.55810546875,
-0.5087890625,
-0.212158203125,
0.044586181640625,
-0.1436767578125,
0.66845703125,
0.560546875,
-0.0312347412109375,
0.0799560546875,
-0.994140625,
-0.49658203125,
-0.283203125,
-0.25439453125,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Example
Input
6 3
1 0 0 1 0 1
1 3 2
Output
1
Submitted Solution:
```
def read():
return list(map(int,input().split()))
def calc(bs,ls,first):
if sum(ls[0::2])!=bs.count(first) or sum(ls[1::2])!=bs.count(1-0):
return float("inf")
res=0
i,j=0,0
for k in range(0,len(ls),2):
if k>0: i+=ls[k-1]
for _ in range(ls[k]):
j=bs.index(first,j)
res+=abs(j-i)
i+=1
j+=1
return res
while 1:
try:
n,m=read()
bs=read()
ls=read()
except: break
print(min(calc(bs,ls,i) for i in range(2)))
```
No
| 77,533 | [
0.388671875,
0.044525146484375,
-0.0119781494140625,
-0.2978515625,
-0.71875,
-0.2489013671875,
-0.03631591796875,
0.27734375,
0.0845947265625,
1.1064453125,
0.28515625,
0.09014892578125,
0.15087890625,
-0.73095703125,
-0.7158203125,
-0.3115234375,
-0.81494140625,
-0.5185546875,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
So the Beautiful Regional Contest (BeRC) has come to an end! n students took part in the contest. The final standings are already known: the participant in the i-th place solved p_i problems. Since the participants are primarily sorted by the number of solved problems, then p_1 ≥ p_2 ≥ ... ≥ p_n.
Help the jury distribute the gold, silver and bronze medals. Let their numbers be g, s and b, respectively. Here is a list of requirements from the rules, which all must be satisfied:
* for each of the three types of medals, at least one medal must be awarded (that is, g>0, s>0 and b>0);
* the number of gold medals must be strictly less than the number of silver and the number of bronze (that is, g<s and g<b, but there are no requirements between s and b);
* each gold medalist must solve strictly more problems than any awarded with a silver medal;
* each silver medalist must solve strictly more problems than any awarded a bronze medal;
* each bronze medalist must solve strictly more problems than any participant not awarded a medal;
* the total number of medalists g+s+b should not exceed half of all participants (for example, if n=21, then you can award a maximum of 10 participants, and if n=26, then you can award a maximum of 13 participants).
The jury wants to reward with medals the total maximal number participants (i.e. to maximize g+s+b) so that all of the items listed above are fulfilled. Help the jury find such a way to award medals.
Input
The first line of the input contains an integer t (1 ≤ t ≤ 10000) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains an integer n (1 ≤ n ≤ 4⋅10^5) — the number of BeRC participants. The second line of a test case contains integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 10^6), where p_i is equal to the number of problems solved by the i-th participant from the final standings. The values p_i are sorted in non-increasing order, i.e. p_1 ≥ p_2 ≥ ... ≥ p_n.
The sum of n over all test cases in the input does not exceed 4⋅10^5.
Output
Print t lines, the j-th line should contain the answer to the j-th test case.
The answer consists of three non-negative integers g, s, b.
* Print g=s=b=0 if there is no way to reward participants with medals so that all requirements from the statement are satisfied at the same time.
* Otherwise, print three positive numbers g, s, b — the possible number of gold, silver and bronze medals, respectively. The sum of g+s+b should be the maximum possible. If there are several answers, print any of them.
Example
Input
5
12
5 4 4 3 2 2 1 1 1 1 1 1
4
4 3 2 1
1
1000000
20
20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
32
64 64 63 58 58 58 58 58 37 37 37 37 34 34 28 28 28 28 28 28 24 24 19 17 17 17 17 16 16 16 16 11
Output
1 2 3
0 0 0
0 0 0
2 5 3
2 6 6
Note
In the first test case, it is possible to reward 1 gold, 2 silver and 3 bronze medals. In this case, the participant solved 5 tasks will be rewarded with the gold medal, participants solved 4 tasks will be rewarded with silver medals, participants solved 2 or 3 tasks will be rewarded with bronze medals. Participants solved exactly 1 task won't be rewarded. It's easy to see, that in this case, all conditions are satisfied and it is possible to reward participants in this way. It is impossible to give more than 6 medals because the number of medals should not exceed half of the number of participants. The answer 1, 3, 2 is also correct in this test case.
In the second and third test cases, it is impossible to reward medals, because at least one medal of each type should be given, but the number of medals should not exceed half of the number of participants.
Submitted Solution:
```
import sys
def inp():
return sys.stdin.readline().strip()
for _ in range(int(inp())):
n=int(inp())
a=list(map(int,inp().split()))
a.sort(reverse=True)
ind={}
for i in range(n):
ind[a[i]]=i
k=sorted(ind.keys(),reverse=True)
l=1
gg=ind[k[0]]+1
ss=0
while l<len(k) and gg>=ss:
ss+=ind[k[l]]-ind[k[l-1]]
l+=1
bb=0
while l<len(k) and (bb<=gg or (gg+bb+ss+ind[k[l]]-ind[k[l-1]])<=n//2):
bb+=ind[k[l]]-ind[k[l-1]]
l+=1
if gg==0 or bb==0 or ss==0 or gg>=min(bb,ss) or (gg+bb+ss)>n//2:
print(0,0,0)
else:
print(gg,ss,bb)
```
Yes
| 77,649 | [
0.409912109375,
-0.236328125,
-0.1793212890625,
0.21533203125,
-0.420166015625,
-0.265380859375,
0.0548095703125,
0.29296875,
0.07366943359375,
0.62646484375,
0.84130859375,
-0.175048828125,
0.1153564453125,
-0.84033203125,
-0.44921875,
-0.08856201171875,
-0.48828125,
-0.9301757812... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
So the Beautiful Regional Contest (BeRC) has come to an end! n students took part in the contest. The final standings are already known: the participant in the i-th place solved p_i problems. Since the participants are primarily sorted by the number of solved problems, then p_1 ≥ p_2 ≥ ... ≥ p_n.
Help the jury distribute the gold, silver and bronze medals. Let their numbers be g, s and b, respectively. Here is a list of requirements from the rules, which all must be satisfied:
* for each of the three types of medals, at least one medal must be awarded (that is, g>0, s>0 and b>0);
* the number of gold medals must be strictly less than the number of silver and the number of bronze (that is, g<s and g<b, but there are no requirements between s and b);
* each gold medalist must solve strictly more problems than any awarded with a silver medal;
* each silver medalist must solve strictly more problems than any awarded a bronze medal;
* each bronze medalist must solve strictly more problems than any participant not awarded a medal;
* the total number of medalists g+s+b should not exceed half of all participants (for example, if n=21, then you can award a maximum of 10 participants, and if n=26, then you can award a maximum of 13 participants).
The jury wants to reward with medals the total maximal number participants (i.e. to maximize g+s+b) so that all of the items listed above are fulfilled. Help the jury find such a way to award medals.
Input
The first line of the input contains an integer t (1 ≤ t ≤ 10000) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains an integer n (1 ≤ n ≤ 4⋅10^5) — the number of BeRC participants. The second line of a test case contains integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 10^6), where p_i is equal to the number of problems solved by the i-th participant from the final standings. The values p_i are sorted in non-increasing order, i.e. p_1 ≥ p_2 ≥ ... ≥ p_n.
The sum of n over all test cases in the input does not exceed 4⋅10^5.
Output
Print t lines, the j-th line should contain the answer to the j-th test case.
The answer consists of three non-negative integers g, s, b.
* Print g=s=b=0 if there is no way to reward participants with medals so that all requirements from the statement are satisfied at the same time.
* Otherwise, print three positive numbers g, s, b — the possible number of gold, silver and bronze medals, respectively. The sum of g+s+b should be the maximum possible. If there are several answers, print any of them.
Example
Input
5
12
5 4 4 3 2 2 1 1 1 1 1 1
4
4 3 2 1
1
1000000
20
20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
32
64 64 63 58 58 58 58 58 37 37 37 37 34 34 28 28 28 28 28 28 24 24 19 17 17 17 17 16 16 16 16 11
Output
1 2 3
0 0 0
0 0 0
2 5 3
2 6 6
Note
In the first test case, it is possible to reward 1 gold, 2 silver and 3 bronze medals. In this case, the participant solved 5 tasks will be rewarded with the gold medal, participants solved 4 tasks will be rewarded with silver medals, participants solved 2 or 3 tasks will be rewarded with bronze medals. Participants solved exactly 1 task won't be rewarded. It's easy to see, that in this case, all conditions are satisfied and it is possible to reward participants in this way. It is impossible to give more than 6 medals because the number of medals should not exceed half of the number of participants. The answer 1, 3, 2 is also correct in this test case.
In the second and third test cases, it is impossible to reward medals, because at least one medal of each type should be given, but the number of medals should not exceed half of the number of participants.
Submitted Solution:
```
for __ in range(int(input())):
n = int(input())
arr = [int(s) for s in input().split()]
g = 1
while g < n and arr[g-1] == arr[g]:
g += 1
s = g + 1
while s+g<n and arr[g+s-1] == arr[g+s]:
s += 1
a = n//2
while a>0 and arr[a] == arr[a-1]:
a-=1
b = a - g - s
if b > g:
print(g, s, b)
else:
print(0, 0, 0)
```
Yes
| 77,650 | [
0.409912109375,
-0.236328125,
-0.1793212890625,
0.21533203125,
-0.420166015625,
-0.265380859375,
0.0548095703125,
0.29296875,
0.07366943359375,
0.62646484375,
0.84130859375,
-0.175048828125,
0.1153564453125,
-0.84033203125,
-0.44921875,
-0.08856201171875,
-0.48828125,
-0.9301757812... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
So the Beautiful Regional Contest (BeRC) has come to an end! n students took part in the contest. The final standings are already known: the participant in the i-th place solved p_i problems. Since the participants are primarily sorted by the number of solved problems, then p_1 ≥ p_2 ≥ ... ≥ p_n.
Help the jury distribute the gold, silver and bronze medals. Let their numbers be g, s and b, respectively. Here is a list of requirements from the rules, which all must be satisfied:
* for each of the three types of medals, at least one medal must be awarded (that is, g>0, s>0 and b>0);
* the number of gold medals must be strictly less than the number of silver and the number of bronze (that is, g<s and g<b, but there are no requirements between s and b);
* each gold medalist must solve strictly more problems than any awarded with a silver medal;
* each silver medalist must solve strictly more problems than any awarded a bronze medal;
* each bronze medalist must solve strictly more problems than any participant not awarded a medal;
* the total number of medalists g+s+b should not exceed half of all participants (for example, if n=21, then you can award a maximum of 10 participants, and if n=26, then you can award a maximum of 13 participants).
The jury wants to reward with medals the total maximal number participants (i.e. to maximize g+s+b) so that all of the items listed above are fulfilled. Help the jury find such a way to award medals.
Input
The first line of the input contains an integer t (1 ≤ t ≤ 10000) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains an integer n (1 ≤ n ≤ 4⋅10^5) — the number of BeRC participants. The second line of a test case contains integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 10^6), where p_i is equal to the number of problems solved by the i-th participant from the final standings. The values p_i are sorted in non-increasing order, i.e. p_1 ≥ p_2 ≥ ... ≥ p_n.
The sum of n over all test cases in the input does not exceed 4⋅10^5.
Output
Print t lines, the j-th line should contain the answer to the j-th test case.
The answer consists of three non-negative integers g, s, b.
* Print g=s=b=0 if there is no way to reward participants with medals so that all requirements from the statement are satisfied at the same time.
* Otherwise, print three positive numbers g, s, b — the possible number of gold, silver and bronze medals, respectively. The sum of g+s+b should be the maximum possible. If there are several answers, print any of them.
Example
Input
5
12
5 4 4 3 2 2 1 1 1 1 1 1
4
4 3 2 1
1
1000000
20
20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
32
64 64 63 58 58 58 58 58 37 37 37 37 34 34 28 28 28 28 28 28 24 24 19 17 17 17 17 16 16 16 16 11
Output
1 2 3
0 0 0
0 0 0
2 5 3
2 6 6
Note
In the first test case, it is possible to reward 1 gold, 2 silver and 3 bronze medals. In this case, the participant solved 5 tasks will be rewarded with the gold medal, participants solved 4 tasks will be rewarded with silver medals, participants solved 2 or 3 tasks will be rewarded with bronze medals. Participants solved exactly 1 task won't be rewarded. It's easy to see, that in this case, all conditions are satisfied and it is possible to reward participants in this way. It is impossible to give more than 6 medals because the number of medals should not exceed half of the number of participants. The answer 1, 3, 2 is also correct in this test case.
In the second and third test cases, it is impossible to reward medals, because at least one medal of each type should be given, but the number of medals should not exceed half of the number of participants.
Submitted Solution:
```
from collections import Counter
if __name__ == '__main__':
for _ in range (int(input())):
n = int(input())
l = list(map(int,input().split()))
if n < 6:
print(0,0,0)
continue
a = (n//2)-1
while l[a] == l[(n//2)] and a >= 0:
a-=1
if a < 2:
print(0,0,0)
continue
l = l[:a+1].copy()
b = len(l)
d = Counter(l)
l = set(l)
if len(d)<3:
print(0,0,0)
continue
g,s = 0,0
a = sorted(l)[-1]
# print(l)
if d[a]>(b//2):
print(0,0,0)
continue
else:
g = d[a]
l.remove(a)
# print(l)
for i in sorted(l)[::-1]:
# print('*',i)
s+=d[i]
if s > g:
break
if (b-s-g) <= g:
print(0,0,0)
else:
print(g,s,(b-s-g))
```
Yes
| 77,651 | [
0.409912109375,
-0.236328125,
-0.1793212890625,
0.21533203125,
-0.420166015625,
-0.265380859375,
0.0548095703125,
0.29296875,
0.07366943359375,
0.62646484375,
0.84130859375,
-0.175048828125,
0.1153564453125,
-0.84033203125,
-0.44921875,
-0.08856201171875,
-0.48828125,
-0.9301757812... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
So the Beautiful Regional Contest (BeRC) has come to an end! n students took part in the contest. The final standings are already known: the participant in the i-th place solved p_i problems. Since the participants are primarily sorted by the number of solved problems, then p_1 ≥ p_2 ≥ ... ≥ p_n.
Help the jury distribute the gold, silver and bronze medals. Let their numbers be g, s and b, respectively. Here is a list of requirements from the rules, which all must be satisfied:
* for each of the three types of medals, at least one medal must be awarded (that is, g>0, s>0 and b>0);
* the number of gold medals must be strictly less than the number of silver and the number of bronze (that is, g<s and g<b, but there are no requirements between s and b);
* each gold medalist must solve strictly more problems than any awarded with a silver medal;
* each silver medalist must solve strictly more problems than any awarded a bronze medal;
* each bronze medalist must solve strictly more problems than any participant not awarded a medal;
* the total number of medalists g+s+b should not exceed half of all participants (for example, if n=21, then you can award a maximum of 10 participants, and if n=26, then you can award a maximum of 13 participants).
The jury wants to reward with medals the total maximal number participants (i.e. to maximize g+s+b) so that all of the items listed above are fulfilled. Help the jury find such a way to award medals.
Input
The first line of the input contains an integer t (1 ≤ t ≤ 10000) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains an integer n (1 ≤ n ≤ 4⋅10^5) — the number of BeRC participants. The second line of a test case contains integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 10^6), where p_i is equal to the number of problems solved by the i-th participant from the final standings. The values p_i are sorted in non-increasing order, i.e. p_1 ≥ p_2 ≥ ... ≥ p_n.
The sum of n over all test cases in the input does not exceed 4⋅10^5.
Output
Print t lines, the j-th line should contain the answer to the j-th test case.
The answer consists of three non-negative integers g, s, b.
* Print g=s=b=0 if there is no way to reward participants with medals so that all requirements from the statement are satisfied at the same time.
* Otherwise, print three positive numbers g, s, b — the possible number of gold, silver and bronze medals, respectively. The sum of g+s+b should be the maximum possible. If there are several answers, print any of them.
Example
Input
5
12
5 4 4 3 2 2 1 1 1 1 1 1
4
4 3 2 1
1
1000000
20
20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
32
64 64 63 58 58 58 58 58 37 37 37 37 34 34 28 28 28 28 28 28 24 24 19 17 17 17 17 16 16 16 16 11
Output
1 2 3
0 0 0
0 0 0
2 5 3
2 6 6
Note
In the first test case, it is possible to reward 1 gold, 2 silver and 3 bronze medals. In this case, the participant solved 5 tasks will be rewarded with the gold medal, participants solved 4 tasks will be rewarded with silver medals, participants solved 2 or 3 tasks will be rewarded with bronze medals. Participants solved exactly 1 task won't be rewarded. It's easy to see, that in this case, all conditions are satisfied and it is possible to reward participants in this way. It is impossible to give more than 6 medals because the number of medals should not exceed half of the number of participants. The answer 1, 3, 2 is also correct in this test case.
In the second and third test cases, it is impossible to reward medals, because at least one medal of each type should be given, but the number of medals should not exceed half of the number of participants.
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
# from fractions import *
# from heapq import*
from bisect import *
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
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")
ALPHA='abcdefghijklmnopqrstuvwxyz'
M=10**9+7
EPS=1e-6
def Ceil(a,b): return a//b+int(a%b>0)
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
# vsInput()
for _ in range(Int()):
n=Int()
a=array()
key=a[n//2]
a=a[:n//2]
while(len(a)>0 and a[-1]==key): a.pop()
C=Counter(a)
C=[C[i] for i in C]
if(len(C)<3):
print(0,0,0)
else:
g=C[0]
s=0
b=sum(C)-g
i=1
while(i<len(C) and s<=g):
s+=C[i]
b-=C[i]
i+=1
if(b<=g):print(0,0,0)
else: print(g,s,b)
# print(C)
# print()
```
Yes
| 77,652 | [
0.409912109375,
-0.236328125,
-0.1793212890625,
0.21533203125,
-0.420166015625,
-0.265380859375,
0.0548095703125,
0.29296875,
0.07366943359375,
0.62646484375,
0.84130859375,
-0.175048828125,
0.1153564453125,
-0.84033203125,
-0.44921875,
-0.08856201171875,
-0.48828125,
-0.9301757812... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
So the Beautiful Regional Contest (BeRC) has come to an end! n students took part in the contest. The final standings are already known: the participant in the i-th place solved p_i problems. Since the participants are primarily sorted by the number of solved problems, then p_1 ≥ p_2 ≥ ... ≥ p_n.
Help the jury distribute the gold, silver and bronze medals. Let their numbers be g, s and b, respectively. Here is a list of requirements from the rules, which all must be satisfied:
* for each of the three types of medals, at least one medal must be awarded (that is, g>0, s>0 and b>0);
* the number of gold medals must be strictly less than the number of silver and the number of bronze (that is, g<s and g<b, but there are no requirements between s and b);
* each gold medalist must solve strictly more problems than any awarded with a silver medal;
* each silver medalist must solve strictly more problems than any awarded a bronze medal;
* each bronze medalist must solve strictly more problems than any participant not awarded a medal;
* the total number of medalists g+s+b should not exceed half of all participants (for example, if n=21, then you can award a maximum of 10 participants, and if n=26, then you can award a maximum of 13 participants).
The jury wants to reward with medals the total maximal number participants (i.e. to maximize g+s+b) so that all of the items listed above are fulfilled. Help the jury find such a way to award medals.
Input
The first line of the input contains an integer t (1 ≤ t ≤ 10000) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains an integer n (1 ≤ n ≤ 4⋅10^5) — the number of BeRC participants. The second line of a test case contains integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 10^6), where p_i is equal to the number of problems solved by the i-th participant from the final standings. The values p_i are sorted in non-increasing order, i.e. p_1 ≥ p_2 ≥ ... ≥ p_n.
The sum of n over all test cases in the input does not exceed 4⋅10^5.
Output
Print t lines, the j-th line should contain the answer to the j-th test case.
The answer consists of three non-negative integers g, s, b.
* Print g=s=b=0 if there is no way to reward participants with medals so that all requirements from the statement are satisfied at the same time.
* Otherwise, print three positive numbers g, s, b — the possible number of gold, silver and bronze medals, respectively. The sum of g+s+b should be the maximum possible. If there are several answers, print any of them.
Example
Input
5
12
5 4 4 3 2 2 1 1 1 1 1 1
4
4 3 2 1
1
1000000
20
20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
32
64 64 63 58 58 58 58 58 37 37 37 37 34 34 28 28 28 28 28 28 24 24 19 17 17 17 17 16 16 16 16 11
Output
1 2 3
0 0 0
0 0 0
2 5 3
2 6 6
Note
In the first test case, it is possible to reward 1 gold, 2 silver and 3 bronze medals. In this case, the participant solved 5 tasks will be rewarded with the gold medal, participants solved 4 tasks will be rewarded with silver medals, participants solved 2 or 3 tasks will be rewarded with bronze medals. Participants solved exactly 1 task won't be rewarded. It's easy to see, that in this case, all conditions are satisfied and it is possible to reward participants in this way. It is impossible to give more than 6 medals because the number of medals should not exceed half of the number of participants. The answer 1, 3, 2 is also correct in this test case.
In the second and third test cases, it is impossible to reward medals, because at least one medal of each type should be given, but the number of medals should not exceed half of the number of participants.
Submitted Solution:
```
import sys;
import math;
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def get_string(): return sys.stdin.readline().strip()
t = int(input());
for test in range(t):
n = int(input());
arr = get_array();
g = 0;s = 0;b = 0;
farr = [1];
ptr = 0;
for i in range(1,n):
if(arr[i]==arr[i-1]):
farr[ptr]+=1;
else:
farr.append(1);
ptr+=1;
if(len(farr)<2):
print("0 0 0");
continue;
g = farr[0];
ptr = 1;
for i in range(ptr,len(farr)):
if(s>g):
break;
else:
s+=farr[i];
ptr+=1;
for i in range(ptr,len(farr)):
if(b>=s):
break;
else:
b+=farr[i];
ptr+=1;
for i in range(ptr,len(farr)):
if(g+s+b+farr[i]<=n//2):
b+=farr[i];
else:
break;
if(g+s+b>n//2):
print("0 0 0");
continue;
else:
print(g,s,b);
```
No
| 77,653 | [
0.409912109375,
-0.236328125,
-0.1793212890625,
0.21533203125,
-0.420166015625,
-0.265380859375,
0.0548095703125,
0.29296875,
0.07366943359375,
0.62646484375,
0.84130859375,
-0.175048828125,
0.1153564453125,
-0.84033203125,
-0.44921875,
-0.08856201171875,
-0.48828125,
-0.9301757812... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
So the Beautiful Regional Contest (BeRC) has come to an end! n students took part in the contest. The final standings are already known: the participant in the i-th place solved p_i problems. Since the participants are primarily sorted by the number of solved problems, then p_1 ≥ p_2 ≥ ... ≥ p_n.
Help the jury distribute the gold, silver and bronze medals. Let their numbers be g, s and b, respectively. Here is a list of requirements from the rules, which all must be satisfied:
* for each of the three types of medals, at least one medal must be awarded (that is, g>0, s>0 and b>0);
* the number of gold medals must be strictly less than the number of silver and the number of bronze (that is, g<s and g<b, but there are no requirements between s and b);
* each gold medalist must solve strictly more problems than any awarded with a silver medal;
* each silver medalist must solve strictly more problems than any awarded a bronze medal;
* each bronze medalist must solve strictly more problems than any participant not awarded a medal;
* the total number of medalists g+s+b should not exceed half of all participants (for example, if n=21, then you can award a maximum of 10 participants, and if n=26, then you can award a maximum of 13 participants).
The jury wants to reward with medals the total maximal number participants (i.e. to maximize g+s+b) so that all of the items listed above are fulfilled. Help the jury find such a way to award medals.
Input
The first line of the input contains an integer t (1 ≤ t ≤ 10000) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains an integer n (1 ≤ n ≤ 4⋅10^5) — the number of BeRC participants. The second line of a test case contains integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 10^6), where p_i is equal to the number of problems solved by the i-th participant from the final standings. The values p_i are sorted in non-increasing order, i.e. p_1 ≥ p_2 ≥ ... ≥ p_n.
The sum of n over all test cases in the input does not exceed 4⋅10^5.
Output
Print t lines, the j-th line should contain the answer to the j-th test case.
The answer consists of three non-negative integers g, s, b.
* Print g=s=b=0 if there is no way to reward participants with medals so that all requirements from the statement are satisfied at the same time.
* Otherwise, print three positive numbers g, s, b — the possible number of gold, silver and bronze medals, respectively. The sum of g+s+b should be the maximum possible. If there are several answers, print any of them.
Example
Input
5
12
5 4 4 3 2 2 1 1 1 1 1 1
4
4 3 2 1
1
1000000
20
20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
32
64 64 63 58 58 58 58 58 37 37 37 37 34 34 28 28 28 28 28 28 24 24 19 17 17 17 17 16 16 16 16 11
Output
1 2 3
0 0 0
0 0 0
2 5 3
2 6 6
Note
In the first test case, it is possible to reward 1 gold, 2 silver and 3 bronze medals. In this case, the participant solved 5 tasks will be rewarded with the gold medal, participants solved 4 tasks will be rewarded with silver medals, participants solved 2 or 3 tasks will be rewarded with bronze medals. Participants solved exactly 1 task won't be rewarded. It's easy to see, that in this case, all conditions are satisfied and it is possible to reward participants in this way. It is impossible to give more than 6 medals because the number of medals should not exceed half of the number of participants. The answer 1, 3, 2 is also correct in this test case.
In the second and third test cases, it is impossible to reward medals, because at least one medal of each type should be given, but the number of medals should not exceed half of the number of participants.
Submitted Solution:
```
''' ===============================
-- @uthor : Kaleab Asfaw
-- Handle : kaleabasfaw2010
-- Bio : High-School Student
==============================='''
# Fast IO
import sys
import os
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")
# Others
# from math import floor, ceil, gcd
# from decimal import Decimal as d
mod = 10**9+7
def lcm(x, y): return (x * y) / (gcd(x, y))
def fact(x, mod=mod):
ans = 1
for i in range(1, x+1): ans = (ans * i) % mod
return ans
def arr2D(n, m, default=0): return [[default for j in range(m)] for i in range(n)]
def arr3D(n, m, r, default=0): return [[[default for k in range(r)] for j in range(m)] for i in range(n)]
def sortDictV(x): return {k: v for k, v in sorted(x.items(), key = lambda item : item[1])}
class DSU:
def __init__(self, length): self.length = length; self.parent = [-1] * self.length # O(log(n))
def getParent(self, node, start): # O(log(n))
if node >= self.length: return False
if self.parent[node] < 0:
if start != node: self.parent[start] = node
return node
return self.getParent(self.parent[node], start)
def union(self, node1, node2): # O(log(n))
parent1 = self.getParent(node1, node1); parent2 = self.getParent(node2, node2)
if parent1 == parent2: return False
elif self.parent[parent1] <= self.parent[parent2]: self.parent[parent1] += self.parent[parent2]; self.parent[parent2] = parent1
else: self.parent[parent2] += self.parent[parent1]; self.parent[parent1] = parent2
return True
def getCount(self, node): return -self.parent[self.getParent(node, node)] # O(log(n))
def exact(num):
if abs(num - round(num)) <= 10**(-9):return round(num)
return num
def solve(n, lst):
if n < 3:
print(0, 0, 0)
return
out = lst[n//2]
half = []
for i in lst:
if i <= out:
break
half.append(i)
# print(half)
if len(half) < 3:
print(0, 0, 0)
return
maxxCount = half.count(half[0])
# print(half[0], maxxCount)
numCount = []
prev = -1
for i in half[maxxCount:]:
if i != prev:
numCount.append(1)
prev = i
else:
numCount[-1] += 1
# print(numCount)
currSum = 0
tot = len(half)
get = False
for i in numCount:
currSum += i
if currSum > maxxCount and (tot - currSum) > maxxCount:
get = True
break
if get:
print(maxxCount, currSum, tot-currSum-maxxCount)
else:
print(0, 0, 0)
for _ in range(int(input())): # Multicase
n = int(input())
lst = list(map(int, input().split()))
solve(n, lst)
```
No
| 77,654 | [
0.409912109375,
-0.236328125,
-0.1793212890625,
0.21533203125,
-0.420166015625,
-0.265380859375,
0.0548095703125,
0.29296875,
0.07366943359375,
0.62646484375,
0.84130859375,
-0.175048828125,
0.1153564453125,
-0.84033203125,
-0.44921875,
-0.08856201171875,
-0.48828125,
-0.9301757812... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
So the Beautiful Regional Contest (BeRC) has come to an end! n students took part in the contest. The final standings are already known: the participant in the i-th place solved p_i problems. Since the participants are primarily sorted by the number of solved problems, then p_1 ≥ p_2 ≥ ... ≥ p_n.
Help the jury distribute the gold, silver and bronze medals. Let their numbers be g, s and b, respectively. Here is a list of requirements from the rules, which all must be satisfied:
* for each of the three types of medals, at least one medal must be awarded (that is, g>0, s>0 and b>0);
* the number of gold medals must be strictly less than the number of silver and the number of bronze (that is, g<s and g<b, but there are no requirements between s and b);
* each gold medalist must solve strictly more problems than any awarded with a silver medal;
* each silver medalist must solve strictly more problems than any awarded a bronze medal;
* each bronze medalist must solve strictly more problems than any participant not awarded a medal;
* the total number of medalists g+s+b should not exceed half of all participants (for example, if n=21, then you can award a maximum of 10 participants, and if n=26, then you can award a maximum of 13 participants).
The jury wants to reward with medals the total maximal number participants (i.e. to maximize g+s+b) so that all of the items listed above are fulfilled. Help the jury find such a way to award medals.
Input
The first line of the input contains an integer t (1 ≤ t ≤ 10000) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains an integer n (1 ≤ n ≤ 4⋅10^5) — the number of BeRC participants. The second line of a test case contains integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 10^6), where p_i is equal to the number of problems solved by the i-th participant from the final standings. The values p_i are sorted in non-increasing order, i.e. p_1 ≥ p_2 ≥ ... ≥ p_n.
The sum of n over all test cases in the input does not exceed 4⋅10^5.
Output
Print t lines, the j-th line should contain the answer to the j-th test case.
The answer consists of three non-negative integers g, s, b.
* Print g=s=b=0 if there is no way to reward participants with medals so that all requirements from the statement are satisfied at the same time.
* Otherwise, print three positive numbers g, s, b — the possible number of gold, silver and bronze medals, respectively. The sum of g+s+b should be the maximum possible. If there are several answers, print any of them.
Example
Input
5
12
5 4 4 3 2 2 1 1 1 1 1 1
4
4 3 2 1
1
1000000
20
20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
32
64 64 63 58 58 58 58 58 37 37 37 37 34 34 28 28 28 28 28 28 24 24 19 17 17 17 17 16 16 16 16 11
Output
1 2 3
0 0 0
0 0 0
2 5 3
2 6 6
Note
In the first test case, it is possible to reward 1 gold, 2 silver and 3 bronze medals. In this case, the participant solved 5 tasks will be rewarded with the gold medal, participants solved 4 tasks will be rewarded with silver medals, participants solved 2 or 3 tasks will be rewarded with bronze medals. Participants solved exactly 1 task won't be rewarded. It's easy to see, that in this case, all conditions are satisfied and it is possible to reward participants in this way. It is impossible to give more than 6 medals because the number of medals should not exceed half of the number of participants. The answer 1, 3, 2 is also correct in this test case.
In the second and third test cases, it is impossible to reward medals, because at least one medal of each type should be given, but the number of medals should not exceed half of the number of participants.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
l = [int(s) for s in input().split()]
if len(set(l))<3:
print(0,0,0)
continue
i = n//2-1
# print(i)
if l[n//2-1]==l[n//2]:
while i>=0 and l[i]==l[n//2]:
i-=1
if i<0:
print(0,0,0)
continue
# print(i)
# print(l)
l = l[:i+1]
# print(l)
if len(set(l))<3:
print(0,0,0)
continue
i = 1
while l[i]==l[0]:
i+=1
g = i
while l[i]==l[g]:
i+=1
s = i-g
b = len(l)-g-s
print(g,s,b)
```
No
| 77,655 | [
0.409912109375,
-0.236328125,
-0.1793212890625,
0.21533203125,
-0.420166015625,
-0.265380859375,
0.0548095703125,
0.29296875,
0.07366943359375,
0.62646484375,
0.84130859375,
-0.175048828125,
0.1153564453125,
-0.84033203125,
-0.44921875,
-0.08856201171875,
-0.48828125,
-0.9301757812... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
So the Beautiful Regional Contest (BeRC) has come to an end! n students took part in the contest. The final standings are already known: the participant in the i-th place solved p_i problems. Since the participants are primarily sorted by the number of solved problems, then p_1 ≥ p_2 ≥ ... ≥ p_n.
Help the jury distribute the gold, silver and bronze medals. Let their numbers be g, s and b, respectively. Here is a list of requirements from the rules, which all must be satisfied:
* for each of the three types of medals, at least one medal must be awarded (that is, g>0, s>0 and b>0);
* the number of gold medals must be strictly less than the number of silver and the number of bronze (that is, g<s and g<b, but there are no requirements between s and b);
* each gold medalist must solve strictly more problems than any awarded with a silver medal;
* each silver medalist must solve strictly more problems than any awarded a bronze medal;
* each bronze medalist must solve strictly more problems than any participant not awarded a medal;
* the total number of medalists g+s+b should not exceed half of all participants (for example, if n=21, then you can award a maximum of 10 participants, and if n=26, then you can award a maximum of 13 participants).
The jury wants to reward with medals the total maximal number participants (i.e. to maximize g+s+b) so that all of the items listed above are fulfilled. Help the jury find such a way to award medals.
Input
The first line of the input contains an integer t (1 ≤ t ≤ 10000) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains an integer n (1 ≤ n ≤ 4⋅10^5) — the number of BeRC participants. The second line of a test case contains integers p_1, p_2, ..., p_n (0 ≤ p_i ≤ 10^6), where p_i is equal to the number of problems solved by the i-th participant from the final standings. The values p_i are sorted in non-increasing order, i.e. p_1 ≥ p_2 ≥ ... ≥ p_n.
The sum of n over all test cases in the input does not exceed 4⋅10^5.
Output
Print t lines, the j-th line should contain the answer to the j-th test case.
The answer consists of three non-negative integers g, s, b.
* Print g=s=b=0 if there is no way to reward participants with medals so that all requirements from the statement are satisfied at the same time.
* Otherwise, print three positive numbers g, s, b — the possible number of gold, silver and bronze medals, respectively. The sum of g+s+b should be the maximum possible. If there are several answers, print any of them.
Example
Input
5
12
5 4 4 3 2 2 1 1 1 1 1 1
4
4 3 2 1
1
1000000
20
20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1
32
64 64 63 58 58 58 58 58 37 37 37 37 34 34 28 28 28 28 28 28 24 24 19 17 17 17 17 16 16 16 16 11
Output
1 2 3
0 0 0
0 0 0
2 5 3
2 6 6
Note
In the first test case, it is possible to reward 1 gold, 2 silver and 3 bronze medals. In this case, the participant solved 5 tasks will be rewarded with the gold medal, participants solved 4 tasks will be rewarded with silver medals, participants solved 2 or 3 tasks will be rewarded with bronze medals. Participants solved exactly 1 task won't be rewarded. It's easy to see, that in this case, all conditions are satisfied and it is possible to reward participants in this way. It is impossible to give more than 6 medals because the number of medals should not exceed half of the number of participants. The answer 1, 3, 2 is also correct in this test case.
In the second and third test cases, it is impossible to reward medals, because at least one medal of each type should be given, but the number of medals should not exceed half of the number of participants.
Submitted Solution:
```
import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools,itertools,operator,bisect,fractions,statistics
from collections import deque,defaultdict,OrderedDict,Counter
from fractions import Fraction
from decimal import Decimal
from sys import stdout
def main():
starttime=datetime.datetime.now()
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
for _ in range(ri()):
n=ri()
a=ria()
g=1
c=0
for i in range(1,n):
if a[i]<a[i-1]:
c=1
break
else:
g+=1
if c==1:
s=1
p=0
for j in range(i+1,n):
if a[j]<a[j-1] and s>g:
p=1
break
else:
s+=1
k=n//2
b=k-g-s
for h in range(g+s+b-1,-1,-1):
if a[h]==a[h+1]:
b-=1
continue
else:
break
if b>0 and g<s:
print(g,s,b)
else:
print(0,0,0)
else:
print(0,0,0)
#<--Solving Area Ends
endtime=datetime.datetime.now()
time=(endtime-starttime).total_seconds()*1000
if(os.path.exists('input.txt')):
print("Time:",time,"ms")
class FastReader(io.IOBase):
newlines = 0
def __init__(self, fd, chunk_size=1024 * 8):
self._fd = fd
self._chunk_size = chunk_size
self.buffer = io.BytesIO()
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size))
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, size=-1):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size if size == -1 else size))
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()
class FastWriter(io.IOBase):
def __init__(self, fd):
self._fd = fd
self.buffer = io.BytesIO()
self.write = self.buffer.write
def flush(self):
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class FastStdin(io.IOBase):
def __init__(self, fd=0):
self.buffer = FastReader(fd)
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
class FastStdout(io.IOBase):
def __init__(self, fd=1):
self.buffer = FastWriter(fd)
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.flush = self.buffer.flush
if __name__ == '__main__':
sys.stdin = FastStdin()
sys.stdout = FastStdout()
main()
```
No
| 77,656 | [
0.409912109375,
-0.236328125,
-0.1793212890625,
0.21533203125,
-0.420166015625,
-0.265380859375,
0.0548095703125,
0.29296875,
0.07366943359375,
0.62646484375,
0.84130859375,
-0.175048828125,
0.1153564453125,
-0.84033203125,
-0.44921875,
-0.08856201171875,
-0.48828125,
-0.9301757812... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of n and get last two digits of the number. Yes, of course, n can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City?
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 2·1018) — the power in which you need to raise number 5.
Output
Output the last two digits of 5n without spaces between them.
Examples
Input
2
Output
25
Submitted Solution:
```
def pow(base, power, mod):
if power == 1:
return base % mod
if power == 0:
return 1
x = pow(base, power//2, mod) % mod
if power % 2 == 0:
return (x * x) % mod;
else:
return (x * x * (base % mod)) % mod
n = int(input())
if n < 2:
print("WRONG INPUT!")
elif n > 2000000000000000000:
print("WRONG INPUT!")
elif n < 10000000:
print((5**n) % 100)
else:
print(pow(5, n, 100) % 100)
```
Yes
| 77,897 | [
0.4150390625,
-0.2978515625,
-0.12298583984375,
0.274169921875,
-0.73193359375,
-0.29052734375,
0.120849609375,
-0.051483154296875,
-0.09869384765625,
0.67724609375,
0.8076171875,
-0.109130859375,
0.0081787109375,
-0.6923828125,
-0.0760498046875,
0.0697021484375,
-0.3974609375,
-1.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of n and get last two digits of the number. Yes, of course, n can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City?
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 2·1018) — the power in which you need to raise number 5.
Output
Output the last two digits of 5n without spaces between them.
Examples
Input
2
Output
25
Submitted Solution:
```
print(25, end='')
```
Yes
| 77,898 | [
0.27099609375,
-0.15478515625,
-0.0828857421875,
0.2462158203125,
-0.6044921875,
-0.460205078125,
0.2164306640625,
-0.04217529296875,
-0.10430908203125,
0.7080078125,
0.77685546875,
-0.1285400390625,
-0.01522064208984375,
-0.57763671875,
-0.31787109375,
0.00020360946655273438,
-0.469... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of n and get last two digits of the number. Yes, of course, n can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City?
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 2·1018) — the power in which you need to raise number 5.
Output
Output the last two digits of 5n without spaces between them.
Examples
Input
2
Output
25
Submitted Solution:
```
# your code goes here
n=int(input())
x=int(25)
print(x)
```
Yes
| 77,899 | [
0.287109375,
-0.1435546875,
-0.1400146484375,
0.205322265625,
-0.62353515625,
-0.470947265625,
0.1561279296875,
-0.061859130859375,
-0.11956787109375,
0.69873046875,
0.7744140625,
-0.09375,
-0.060882568359375,
-0.611328125,
-0.263427734375,
0.0055694580078125,
-0.482421875,
-0.9213... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of n and get last two digits of the number. Yes, of course, n can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City?
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 2·1018) — the power in which you need to raise number 5.
Output
Output the last two digits of 5n without spaces between them.
Examples
Input
2
Output
25
Submitted Solution:
```
n = int(input())
if(n > 1):
print(25)
else:
print(5**n)
```
Yes
| 77,900 | [
0.26220703125,
-0.14013671875,
-0.1497802734375,
0.26708984375,
-0.59326171875,
-0.473876953125,
0.1978759765625,
-0.0218353271484375,
-0.09490966796875,
0.69482421875,
0.76416015625,
-0.10748291015625,
-0.006809234619140625,
-0.62451171875,
-0.31787109375,
-0.0048675537109375,
-0.48... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of n and get last two digits of the number. Yes, of course, n can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City?
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 2·1018) — the power in which you need to raise number 5.
Output
Output the last two digits of 5n without spaces between them.
Examples
Input
2
Output
25
Submitted Solution:
```
num = int(input())
x=5**num
print(x)
```
No
| 77,901 | [
0.28857421875,
-0.12451171875,
-0.12091064453125,
0.2369384765625,
-0.59326171875,
-0.49755859375,
0.1522216796875,
-0.04705810546875,
-0.117431640625,
0.69287109375,
0.83447265625,
-0.0836181640625,
-0.0447998046875,
-0.5927734375,
-0.31591796875,
-0.0260162353515625,
-0.41235351562... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of n and get last two digits of the number. Yes, of course, n can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City?
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 2·1018) — the power in which you need to raise number 5.
Output
Output the last two digits of 5n without spaces between them.
Examples
Input
2
Output
25
Submitted Solution:
```
m=int(input())
ans=1
for i in range(0,m):
ans=ans*5
print(ans)
```
No
| 77,902 | [
0.271484375,
-0.159912109375,
-0.109375,
0.221435546875,
-0.58984375,
-0.478271484375,
0.2064208984375,
-0.06414794921875,
-0.10491943359375,
0.65234375,
0.7861328125,
-0.1473388671875,
-0.070068359375,
-0.6181640625,
-0.346435546875,
-0.01517486572265625,
-0.40869140625,
-0.893066... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of n and get last two digits of the number. Yes, of course, n can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City?
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 2·1018) — the power in which you need to raise number 5.
Output
Output the last two digits of 5n without spaces between them.
Examples
Input
2
Output
25
Submitted Solution:
```
print("Введите степень числа 5")
n = float(input("n = "))
s = 5**n
k=s%100
print("s = %.0f" % k)
```
No
| 77,903 | [
0.329833984375,
-0.1124267578125,
-0.1358642578125,
0.30126953125,
-0.65283203125,
-0.479248046875,
0.2744140625,
-0.011810302734375,
-0.1517333984375,
0.63037109375,
0.78271484375,
-0.22119140625,
0.03582763671875,
-0.578125,
-0.265869140625,
-0.05010986328125,
-0.450439453125,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of n and get last two digits of the number. Yes, of course, n can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."
Could you pass the interview in the machine vision company in IT City?
Input
The only line of the input contains a single integer n (2 ≤ n ≤ 2·1018) — the power in which you need to raise number 5.
Output
Output the last two digits of 5n without spaces between them.
Examples
Input
2
Output
25
Submitted Solution:
```
def Solution():
n = int(input())
print(5**n)
Solution()
```
No
| 77,904 | [
0.28759765625,
-0.1661376953125,
-0.11541748046875,
0.22998046875,
-0.6572265625,
-0.4541015625,
0.19775390625,
0.013336181640625,
-0.11199951171875,
0.69677734375,
0.72216796875,
-0.0880126953125,
-0.037750244140625,
-0.5234375,
-0.299072265625,
-0.042266845703125,
-0.456787109375,
... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
problem
AOR Ika is studying to pass the test.
AOR Ika-chan solved the $ N $ question. After that, round the solved problem according to the following procedure.
1. Check the correctness of the answer.
2. If the answer is correct, write a circle mark, and if it is incorrect, write a cross mark on the answer sheet.
AOR Ika faints because of the fear of failing the test the moment she finds that the answer is wrong for $ 2 $ in a row. And no further rounding is possible.
Syncope occurs between steps $ 1 $ and $ 2 $.
You will be given an integer $ N $, which represents the number of questions AOR Ika has solved, and a string $ S $, which is a length $ N $ and represents the correctness of the answer. The string consists of'o'and'x', with'o' indicating the correct answer and'x' indicating the incorrect answer. The $ i $ letter indicates the correctness of the $ i $ question, and AOR Ika-chan rounds the $ 1 $ question in order.
Please output the number of questions that AOR Ika-chan can write the correctness.
output
Output the number of questions that AOR Ika-chan could write in the $ 1 $ line. Also, output a line break at the end.
Example
Input
3
oxx
Output
2
"Correct Solution:
```
N = int(input())
S = input()
for i in range(1, N):
if S[i-1] == S[i] == "x":
print(i)
break
else:
print(N)
```
| 78,250 | [
0.55908203125,
0.052581787109375,
-0.4267578125,
0.1798095703125,
-0.33349609375,
-0.5517578125,
0.285400390625,
0.18994140625,
0.33544921875,
0.97119140625,
0.44677734375,
0.31640625,
-0.186767578125,
-0.78955078125,
-0.2347412109375,
0.19970703125,
-0.66162109375,
-0.78466796875,... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
problem
AOR Ika is studying to pass the test.
AOR Ika-chan solved the $ N $ question. After that, round the solved problem according to the following procedure.
1. Check the correctness of the answer.
2. If the answer is correct, write a circle mark, and if it is incorrect, write a cross mark on the answer sheet.
AOR Ika faints because of the fear of failing the test the moment she finds that the answer is wrong for $ 2 $ in a row. And no further rounding is possible.
Syncope occurs between steps $ 1 $ and $ 2 $.
You will be given an integer $ N $, which represents the number of questions AOR Ika has solved, and a string $ S $, which is a length $ N $ and represents the correctness of the answer. The string consists of'o'and'x', with'o' indicating the correct answer and'x' indicating the incorrect answer. The $ i $ letter indicates the correctness of the $ i $ question, and AOR Ika-chan rounds the $ 1 $ question in order.
Please output the number of questions that AOR Ika-chan can write the correctness.
output
Output the number of questions that AOR Ika-chan could write in the $ 1 $ line. Also, output a line break at the end.
Example
Input
3
oxx
Output
2
"Correct Solution:
```
N = int(input());
S = input();
ans = N;
for i in range(N-1):
if S[i] == 'x' and S[i+1] == 'x':
ans = i+1
break
print(ans)
```
| 78,251 | [
0.5869140625,
0.04046630859375,
-0.43115234375,
0.1474609375,
-0.348388671875,
-0.55322265625,
0.289794921875,
0.18798828125,
0.31640625,
0.9736328125,
0.436767578125,
0.305908203125,
-0.171875,
-0.75390625,
-0.257568359375,
0.2139892578125,
-0.630859375,
-0.77685546875,
-0.82714... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
problem
AOR Ika is studying to pass the test.
AOR Ika-chan solved the $ N $ question. After that, round the solved problem according to the following procedure.
1. Check the correctness of the answer.
2. If the answer is correct, write a circle mark, and if it is incorrect, write a cross mark on the answer sheet.
AOR Ika faints because of the fear of failing the test the moment she finds that the answer is wrong for $ 2 $ in a row. And no further rounding is possible.
Syncope occurs between steps $ 1 $ and $ 2 $.
You will be given an integer $ N $, which represents the number of questions AOR Ika has solved, and a string $ S $, which is a length $ N $ and represents the correctness of the answer. The string consists of'o'and'x', with'o' indicating the correct answer and'x' indicating the incorrect answer. The $ i $ letter indicates the correctness of the $ i $ question, and AOR Ika-chan rounds the $ 1 $ question in order.
Please output the number of questions that AOR Ika-chan can write the correctness.
output
Output the number of questions that AOR Ika-chan could write in the $ 1 $ line. Also, output a line break at the end.
Example
Input
3
oxx
Output
2
"Correct Solution:
```
N = int(input())
try:
print(input().index("xx") + 1)
except:
print(N)
```
| 78,252 | [
0.58740234375,
0.021453857421875,
-0.465576171875,
0.1590576171875,
-0.361328125,
-0.5537109375,
0.298095703125,
0.21826171875,
0.34326171875,
0.9736328125,
0.4423828125,
0.30224609375,
-0.1611328125,
-0.759765625,
-0.2413330078125,
0.20703125,
-0.62060546875,
-0.78466796875,
-0.... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
problem
AOR Ika is studying to pass the test.
AOR Ika-chan solved the $ N $ question. After that, round the solved problem according to the following procedure.
1. Check the correctness of the answer.
2. If the answer is correct, write a circle mark, and if it is incorrect, write a cross mark on the answer sheet.
AOR Ika faints because of the fear of failing the test the moment she finds that the answer is wrong for $ 2 $ in a row. And no further rounding is possible.
Syncope occurs between steps $ 1 $ and $ 2 $.
You will be given an integer $ N $, which represents the number of questions AOR Ika has solved, and a string $ S $, which is a length $ N $ and represents the correctness of the answer. The string consists of'o'and'x', with'o' indicating the correct answer and'x' indicating the incorrect answer. The $ i $ letter indicates the correctness of the $ i $ question, and AOR Ika-chan rounds the $ 1 $ question in order.
Please output the number of questions that AOR Ika-chan can write the correctness.
output
Output the number of questions that AOR Ika-chan could write in the $ 1 $ line. Also, output a line break at the end.
Example
Input
3
oxx
Output
2
"Correct Solution:
```
# AOJ 2831: Check answers
# Python3 2018.7.12 bal4u
n = int(input())
if n == 0: print(0)
else:
s = input()
ans = 1
for i in range(1, len(s)):
if s[i-1] == 'x' and s[i] == 'x': break
ans += 1
print(ans)
```
| 78,253 | [
0.56689453125,
0.0229949951171875,
-0.46142578125,
0.19287109375,
-0.284912109375,
-0.56640625,
0.3544921875,
0.1546630859375,
0.392822265625,
1.0224609375,
0.425537109375,
0.29931640625,
-0.1724853515625,
-0.77685546875,
-0.2301025390625,
0.197021484375,
-0.6787109375,
-0.77880859... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
problem
AOR Ika is studying to pass the test.
AOR Ika-chan solved the $ N $ question. After that, round the solved problem according to the following procedure.
1. Check the correctness of the answer.
2. If the answer is correct, write a circle mark, and if it is incorrect, write a cross mark on the answer sheet.
AOR Ika faints because of the fear of failing the test the moment she finds that the answer is wrong for $ 2 $ in a row. And no further rounding is possible.
Syncope occurs between steps $ 1 $ and $ 2 $.
You will be given an integer $ N $, which represents the number of questions AOR Ika has solved, and a string $ S $, which is a length $ N $ and represents the correctness of the answer. The string consists of'o'and'x', with'o' indicating the correct answer and'x' indicating the incorrect answer. The $ i $ letter indicates the correctness of the $ i $ question, and AOR Ika-chan rounds the $ 1 $ question in order.
Please output the number of questions that AOR Ika-chan can write the correctness.
output
Output the number of questions that AOR Ika-chan could write in the $ 1 $ line. Also, output a line break at the end.
Example
Input
3
oxx
Output
2
"Correct Solution:
```
N = int(input())
S = input()
ans = 0
flag = False
for i in range(len(S)):
if(S[i] == 'o'):
ans += 1
flag = False
else:
if(flag):
break
else:
ans += 1
flag = True
print(ans)
```
| 78,254 | [
0.56396484375,
0.007129669189453125,
-0.45166015625,
0.157958984375,
-0.342041015625,
-0.56494140625,
0.33544921875,
0.1937255859375,
0.34716796875,
1.0283203125,
0.431640625,
0.2919921875,
-0.1591796875,
-0.7705078125,
-0.197021484375,
0.2294921875,
-0.6318359375,
-0.81787109375,
... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
problem
AOR Ika is studying to pass the test.
AOR Ika-chan solved the $ N $ question. After that, round the solved problem according to the following procedure.
1. Check the correctness of the answer.
2. If the answer is correct, write a circle mark, and if it is incorrect, write a cross mark on the answer sheet.
AOR Ika faints because of the fear of failing the test the moment she finds that the answer is wrong for $ 2 $ in a row. And no further rounding is possible.
Syncope occurs between steps $ 1 $ and $ 2 $.
You will be given an integer $ N $, which represents the number of questions AOR Ika has solved, and a string $ S $, which is a length $ N $ and represents the correctness of the answer. The string consists of'o'and'x', with'o' indicating the correct answer and'x' indicating the incorrect answer. The $ i $ letter indicates the correctness of the $ i $ question, and AOR Ika-chan rounds the $ 1 $ question in order.
Please output the number of questions that AOR Ika-chan can write the correctness.
output
Output the number of questions that AOR Ika-chan could write in the $ 1 $ line. Also, output a line break at the end.
Example
Input
3
oxx
Output
2
"Correct Solution:
```
N = input()
S = input()
N = int(N)
S_list = list(S)
flg = 0
for i in range(N):
if flg == 0:
if S_list[i] == ('x'):
flg = 1
elif S_list[i] == ('o'):
flg = 0
elif S_list[i] == ('x'):
flg = 2
break
if i == N-1:
if flg < 2:
i = N
print(i)
```
| 78,255 | [
0.53076171875,
0.045257568359375,
-0.4580078125,
0.28662109375,
-0.332763671875,
-0.6279296875,
0.299560546875,
0.302001953125,
0.368896484375,
0.9482421875,
0.48291015625,
0.37548828125,
-0.126220703125,
-0.82177734375,
-0.272216796875,
0.2440185546875,
-0.61669921875,
-0.83105468... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
problem
AOR Ika is studying to pass the test.
AOR Ika-chan solved the $ N $ question. After that, round the solved problem according to the following procedure.
1. Check the correctness of the answer.
2. If the answer is correct, write a circle mark, and if it is incorrect, write a cross mark on the answer sheet.
AOR Ika faints because of the fear of failing the test the moment she finds that the answer is wrong for $ 2 $ in a row. And no further rounding is possible.
Syncope occurs between steps $ 1 $ and $ 2 $.
You will be given an integer $ N $, which represents the number of questions AOR Ika has solved, and a string $ S $, which is a length $ N $ and represents the correctness of the answer. The string consists of'o'and'x', with'o' indicating the correct answer and'x' indicating the incorrect answer. The $ i $ letter indicates the correctness of the $ i $ question, and AOR Ika-chan rounds the $ 1 $ question in order.
Please output the number of questions that AOR Ika-chan can write the correctness.
output
Output the number of questions that AOR Ika-chan could write in the $ 1 $ line. Also, output a line break at the end.
Example
Input
3
oxx
Output
2
"Correct Solution:
```
n = int(input())
s = input()
for i in range(n):
if s[i:i+2]=="xx":
print(i + 1)
break
else:
print(n)
```
| 78,256 | [
0.5595703125,
0.044586181640625,
-0.44482421875,
0.1824951171875,
-0.342041015625,
-0.55126953125,
0.30224609375,
0.2056884765625,
0.344970703125,
0.96044921875,
0.412353515625,
0.307373046875,
-0.1915283203125,
-0.7744140625,
-0.23681640625,
0.1978759765625,
-0.650390625,
-0.78710... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
problem
AOR Ika is studying to pass the test.
AOR Ika-chan solved the $ N $ question. After that, round the solved problem according to the following procedure.
1. Check the correctness of the answer.
2. If the answer is correct, write a circle mark, and if it is incorrect, write a cross mark on the answer sheet.
AOR Ika faints because of the fear of failing the test the moment she finds that the answer is wrong for $ 2 $ in a row. And no further rounding is possible.
Syncope occurs between steps $ 1 $ and $ 2 $.
You will be given an integer $ N $, which represents the number of questions AOR Ika has solved, and a string $ S $, which is a length $ N $ and represents the correctness of the answer. The string consists of'o'and'x', with'o' indicating the correct answer and'x' indicating the incorrect answer. The $ i $ letter indicates the correctness of the $ i $ question, and AOR Ika-chan rounds the $ 1 $ question in order.
Please output the number of questions that AOR Ika-chan can write the correctness.
output
Output the number of questions that AOR Ika-chan could write in the $ 1 $ line. Also, output a line break at the end.
Example
Input
3
oxx
Output
2
"Correct Solution:
```
N = int(input())
S = input()
kizetu = False
cnt = 0
for s in S:
if s == "x":
kizetu += 1
else:
kizetu = 0
if kizetu == 2:
print(cnt)
break
cnt += 1
else:
print(len(S))
```
| 78,257 | [
0.55908203125,
0.0094146728515625,
-0.407958984375,
0.1805419921875,
-0.37060546875,
-0.5615234375,
0.283203125,
0.10797119140625,
0.30224609375,
1.01171875,
0.39501953125,
0.28076171875,
-0.0855712890625,
-0.8095703125,
-0.2198486328125,
0.297607421875,
-0.6640625,
-0.8046875,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
AOR Ika is studying to pass the test.
AOR Ika-chan solved the $ N $ question. After that, round the solved problem according to the following procedure.
1. Check the correctness of the answer.
2. If the answer is correct, write a circle mark, and if it is incorrect, write a cross mark on the answer sheet.
AOR Ika faints because of the fear of failing the test the moment she finds that the answer is wrong for $ 2 $ in a row. And no further rounding is possible.
Syncope occurs between steps $ 1 $ and $ 2 $.
You will be given an integer $ N $, which represents the number of questions AOR Ika has solved, and a string $ S $, which is a length $ N $ and represents the correctness of the answer. The string consists of'o'and'x', with'o' indicating the correct answer and'x' indicating the incorrect answer. The $ i $ letter indicates the correctness of the $ i $ question, and AOR Ika-chan rounds the $ 1 $ question in order.
Please output the number of questions that AOR Ika-chan can write the correctness.
output
Output the number of questions that AOR Ika-chan could write in the $ 1 $ line. Also, output a line break at the end.
Example
Input
3
oxx
Output
2
Submitted Solution:
```
def main():
N = int(input())
S = input()
try:
ans = S.index('xx') + 1
print(ans)
except:
print(N)
main()
```
Yes
| 78,258 | [
0.62890625,
-0.001644134521484375,
-0.50048828125,
0.15771484375,
-0.3876953125,
-0.465576171875,
0.2474365234375,
0.261962890625,
0.260498046875,
0.9150390625,
0.43115234375,
0.36572265625,
-0.296875,
-0.6708984375,
-0.327392578125,
0.1141357421875,
-0.5625,
-0.66845703125,
-0.7... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
AOR Ika is studying to pass the test.
AOR Ika-chan solved the $ N $ question. After that, round the solved problem according to the following procedure.
1. Check the correctness of the answer.
2. If the answer is correct, write a circle mark, and if it is incorrect, write a cross mark on the answer sheet.
AOR Ika faints because of the fear of failing the test the moment she finds that the answer is wrong for $ 2 $ in a row. And no further rounding is possible.
Syncope occurs between steps $ 1 $ and $ 2 $.
You will be given an integer $ N $, which represents the number of questions AOR Ika has solved, and a string $ S $, which is a length $ N $ and represents the correctness of the answer. The string consists of'o'and'x', with'o' indicating the correct answer and'x' indicating the incorrect answer. The $ i $ letter indicates the correctness of the $ i $ question, and AOR Ika-chan rounds the $ 1 $ question in order.
Please output the number of questions that AOR Ika-chan can write the correctness.
output
Output the number of questions that AOR Ika-chan could write in the $ 1 $ line. Also, output a line break at the end.
Example
Input
3
oxx
Output
2
Submitted Solution:
```
n=input()
print(input().find('xx')+1or n)
```
Yes
| 78,259 | [
0.64697265625,
0.001171112060546875,
-0.529296875,
0.1702880859375,
-0.393310546875,
-0.473876953125,
0.241455078125,
0.277587890625,
0.25634765625,
0.91943359375,
0.46826171875,
0.337890625,
-0.280517578125,
-0.693359375,
-0.3330078125,
0.11029052734375,
-0.53564453125,
-0.6889648... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
AOR Ika is studying to pass the test.
AOR Ika-chan solved the $ N $ question. After that, round the solved problem according to the following procedure.
1. Check the correctness of the answer.
2. If the answer is correct, write a circle mark, and if it is incorrect, write a cross mark on the answer sheet.
AOR Ika faints because of the fear of failing the test the moment she finds that the answer is wrong for $ 2 $ in a row. And no further rounding is possible.
Syncope occurs between steps $ 1 $ and $ 2 $.
You will be given an integer $ N $, which represents the number of questions AOR Ika has solved, and a string $ S $, which is a length $ N $ and represents the correctness of the answer. The string consists of'o'and'x', with'o' indicating the correct answer and'x' indicating the incorrect answer. The $ i $ letter indicates the correctness of the $ i $ question, and AOR Ika-chan rounds the $ 1 $ question in order.
Please output the number of questions that AOR Ika-chan can write the correctness.
output
Output the number of questions that AOR Ika-chan could write in the $ 1 $ line. Also, output a line break at the end.
Example
Input
3
oxx
Output
2
Submitted Solution:
```
# -*- coding: utf-8 -*-
from sys import stdin
n = int(stdin.readline().rstrip())
s = stdin.readline().rstrip()
faint = False
for i in range(1,n):
if s[i-1] == "x" and s[i] == "x":
print(i)
exit()
print(n)
```
Yes
| 78,260 | [
0.60205078125,
-0.007354736328125,
-0.5234375,
0.2239990234375,
-0.27783203125,
-0.51318359375,
0.26318359375,
0.1875,
0.3125,
0.91259765625,
0.39404296875,
0.3935546875,
-0.302734375,
-0.72802734375,
-0.25732421875,
0.07781982421875,
-0.57666015625,
-0.736328125,
-0.70751953125,... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
AOR Ika is studying to pass the test.
AOR Ika-chan solved the $ N $ question. After that, round the solved problem according to the following procedure.
1. Check the correctness of the answer.
2. If the answer is correct, write a circle mark, and if it is incorrect, write a cross mark on the answer sheet.
AOR Ika faints because of the fear of failing the test the moment she finds that the answer is wrong for $ 2 $ in a row. And no further rounding is possible.
Syncope occurs between steps $ 1 $ and $ 2 $.
You will be given an integer $ N $, which represents the number of questions AOR Ika has solved, and a string $ S $, which is a length $ N $ and represents the correctness of the answer. The string consists of'o'and'x', with'o' indicating the correct answer and'x' indicating the incorrect answer. The $ i $ letter indicates the correctness of the $ i $ question, and AOR Ika-chan rounds the $ 1 $ question in order.
Please output the number of questions that AOR Ika-chan can write the correctness.
output
Output the number of questions that AOR Ika-chan could write in the $ 1 $ line. Also, output a line break at the end.
Example
Input
3
oxx
Output
2
Submitted Solution:
```
import sys
N = int(input())
S = input()
ans = 0
last_s = "o"
for s in S:
if last_s == "x" and s == "x":
print(ans)
sys.exit()
ans += 1
last_s = s
print(ans)
```
Yes
| 78,261 | [
0.6416015625,
-0.01415252685546875,
-0.52783203125,
0.1856689453125,
-0.34326171875,
-0.45556640625,
0.2197265625,
0.2264404296875,
0.316650390625,
0.90966796875,
0.4404296875,
0.40234375,
-0.293701171875,
-0.68994140625,
-0.33837890625,
0.117431640625,
-0.56396484375,
-0.720703125... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
problem
AOR Ika is studying to pass the test.
AOR Ika-chan solved the $ N $ question. After that, round the solved problem according to the following procedure.
1. Check the correctness of the answer.
2. If the answer is correct, write a circle mark, and if it is incorrect, write a cross mark on the answer sheet.
AOR Ika faints because of the fear of failing the test the moment she finds that the answer is wrong for $ 2 $ in a row. And no further rounding is possible.
Syncope occurs between steps $ 1 $ and $ 2 $.
You will be given an integer $ N $, which represents the number of questions AOR Ika has solved, and a string $ S $, which is a length $ N $ and represents the correctness of the answer. The string consists of'o'and'x', with'o' indicating the correct answer and'x' indicating the incorrect answer. The $ i $ letter indicates the correctness of the $ i $ question, and AOR Ika-chan rounds the $ 1 $ question in order.
Please output the number of questions that AOR Ika-chan can write the correctness.
output
Output the number of questions that AOR Ika-chan could write in the $ 1 $ line. Also, output a line break at the end.
Example
Input
3
oxx
Output
2
Submitted Solution:
```
input();s=input();f=s[0];a=1
for x in s[1:]:
if f==x=='x':break
a+=1
print(a)
```
No
| 78,262 | [
0.64111328125,
0.0234222412109375,
-0.468994140625,
0.185791015625,
-0.36181640625,
-0.48828125,
0.2301025390625,
0.2481689453125,
0.2452392578125,
0.89013671875,
0.487548828125,
0.37939453125,
-0.291748046875,
-0.68115234375,
-0.33984375,
0.1180419921875,
-0.5615234375,
-0.6787109... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is interactive.
You should guess hidden number x which is between 1 and M = 10004205361450474, inclusive.
You could use up to 5 queries.
In each query, you can output an increasing sequence of k ≤ x integers, each between 1 and M, inclusive, and you will obtain one of the following as an answer:
* either the hidden number belongs to your query sequence, in this case you immediately win;
* or you will be given where the hidden number is located with respect to your query sequence, that is, either it is less than all numbers from the sequence, greater than all numbers from the sequence, or you will be given such an i that the hidden number x is between the i-th and the (i+1)-st numbers of your sequence.
See the interaction section for clarity.
Be aware that the interactor is adaptive, i.e. the hidden number can depend on queries the solution makes. However, it is guaranteed that for any solution the interactor works non-distinguishable from the situation when the hidden number is fixed beforehand.
Hacks are allowed only with fixed hidden number. A hack is represented by a single integer between 1 and M. In all pretests the hidden number is fixed as well.
Interaction
You can make up to 5 queries. To make a query print a number k (1 ≤ k ≤ 10^4) and then an increasing sequence t_0 < t_1 < … < t_{k-1} of k numbers, each between 1 and M, inclusive. If k > x, you lose.
You get one integer as a response.
* If it is -2, it means you made an invalid query or you lost. Exit immediately after receiving -2 and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
* If it is -1, you guessed the number and should terminate too.
* Otherwise you get a number i between 0 and k, inclusive, denoting where the hidden number is, with respect to the printed numbers. If i = 0, then x < t_0. If i = k, then t_{k-1} < x. Otherwise t_{i-1} < x < t_i.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Example
Input
2
0
-1
Output
2 2 3
2 20 30
3 5 7 9
Note
In the first example the number 5 is hidden.
Submitted Solution:
```
import sys
# data = sys.stdin.readlines()
start = 1
end = 10004205361450474
for i in range(5):
num = min(10000, end - start + 1)
print(num, end=' ')
for j in range(num):
print((end - start + 1)//num * (j + 1) + start, end=' ')
sys.stdout.flush()
t = int(input())
if t < 0:
exit(0)
elif t == 0:
end = (end - start + 1)//num * 1 + start
elif t == num:
start = ((end - start + 1)//num) * num + start
else:
start = ((end - start + 1)//num) * t + start
end = ((end - start + 1)//num) * (t+1) + start
```
No
| 78,294 | [
0.06585693359375,
-0.0667724609375,
-0.380126953125,
-0.1666259765625,
-0.453857421875,
-0.49853515625,
-0.43212890625,
0.188232421875,
-0.0234832763671875,
0.91259765625,
0.60302734375,
0.2125244140625,
-0.019073486328125,
-0.75146484375,
-0.705078125,
-0.032012939453125,
-0.3640136... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is interactive.
You should guess hidden number x which is between 1 and M = 10004205361450474, inclusive.
You could use up to 5 queries.
In each query, you can output an increasing sequence of k ≤ x integers, each between 1 and M, inclusive, and you will obtain one of the following as an answer:
* either the hidden number belongs to your query sequence, in this case you immediately win;
* or you will be given where the hidden number is located with respect to your query sequence, that is, either it is less than all numbers from the sequence, greater than all numbers from the sequence, or you will be given such an i that the hidden number x is between the i-th and the (i+1)-st numbers of your sequence.
See the interaction section for clarity.
Be aware that the interactor is adaptive, i.e. the hidden number can depend on queries the solution makes. However, it is guaranteed that for any solution the interactor works non-distinguishable from the situation when the hidden number is fixed beforehand.
Hacks are allowed only with fixed hidden number. A hack is represented by a single integer between 1 and M. In all pretests the hidden number is fixed as well.
Interaction
You can make up to 5 queries. To make a query print a number k (1 ≤ k ≤ 10^4) and then an increasing sequence t_0 < t_1 < … < t_{k-1} of k numbers, each between 1 and M, inclusive. If k > x, you lose.
You get one integer as a response.
* If it is -2, it means you made an invalid query or you lost. Exit immediately after receiving -2 and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
* If it is -1, you guessed the number and should terminate too.
* Otherwise you get a number i between 0 and k, inclusive, denoting where the hidden number is, with respect to the printed numbers. If i = 0, then x < t_0. If i = k, then t_{k-1} < x. Otherwise t_{i-1} < x < t_i.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Example
Input
2
0
-1
Output
2 2 3
2 20 30
3 5 7 9
Note
In the first example the number 5 is hidden.
Submitted Solution:
```
from sys import stdout
M = 10004205361450474
L = 1
guesses = [M // (10 ** 4) * i + 1 for i in range(10 ** 4)]
for i in range(5):
print("10000" + " ".join([str(x) for x in guesses]))
stdout.flush()
resp = int(input().strip())
if resp == -2:
exit()
elif resp == -1:
exit()
elif resp == 0:
M = guesses[0]
elif resp == 10 ** 4:
L = guesses[-1]
else:
L = guesses[resp]
M = guesses[resp+1]
guesses = [L + (M-L) // (10 ** 4) * i for i in range(10 ** 4)]
```
No
| 78,295 | [
0.06585693359375,
-0.0667724609375,
-0.380126953125,
-0.1666259765625,
-0.453857421875,
-0.49853515625,
-0.43212890625,
0.188232421875,
-0.0234832763671875,
0.91259765625,
0.60302734375,
0.2125244140625,
-0.019073486328125,
-0.75146484375,
-0.705078125,
-0.032012939453125,
-0.3640136... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is interactive.
You should guess hidden number x which is between 1 and M = 10004205361450474, inclusive.
You could use up to 5 queries.
In each query, you can output an increasing sequence of k ≤ x integers, each between 1 and M, inclusive, and you will obtain one of the following as an answer:
* either the hidden number belongs to your query sequence, in this case you immediately win;
* or you will be given where the hidden number is located with respect to your query sequence, that is, either it is less than all numbers from the sequence, greater than all numbers from the sequence, or you will be given such an i that the hidden number x is between the i-th and the (i+1)-st numbers of your sequence.
See the interaction section for clarity.
Be aware that the interactor is adaptive, i.e. the hidden number can depend on queries the solution makes. However, it is guaranteed that for any solution the interactor works non-distinguishable from the situation when the hidden number is fixed beforehand.
Hacks are allowed only with fixed hidden number. A hack is represented by a single integer between 1 and M. In all pretests the hidden number is fixed as well.
Interaction
You can make up to 5 queries. To make a query print a number k (1 ≤ k ≤ 10^4) and then an increasing sequence t_0 < t_1 < … < t_{k-1} of k numbers, each between 1 and M, inclusive. If k > x, you lose.
You get one integer as a response.
* If it is -2, it means you made an invalid query or you lost. Exit immediately after receiving -2 and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
* If it is -1, you guessed the number and should terminate too.
* Otherwise you get a number i between 0 and k, inclusive, denoting where the hidden number is, with respect to the printed numbers. If i = 0, then x < t_0. If i = k, then t_{k-1} < x. Otherwise t_{i-1} < x < t_i.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Example
Input
2
0
-1
Output
2 2 3
2 20 30
3 5 7 9
Note
In the first example the number 5 is hidden.
Submitted Solution:
```
from sys import stdout
def query(start, k, times):
if(times<1):
print('Fuck you! Times cannot < 1!')
return
interlen = [0]*10
for i in range(times-1,0,-1):
interlen[i]=(interlen[i+1]+1)*(k+1)-1
for i in range(1,times+1):
step = interlen[i]+1
print(' '.join([str(k)]+[str(start+j*step) for j in range(1,k+1)]))
stdout.flush()
num=int(input())
if(num==-1):
break
elif(num>=0):
start = start+num*step
print("1 804761420475")
stdout.flush()
num = int(input())
if(num==0):
print("1 896")
stdout.flush()
num = int(input())
if num==0:
print("1 6")
stdout.flush()
num = int(input())
if(num==0):
print("1 2")
stdout.flush()
num = int(input())
if(num==0):
print("1 1")
else:
print("3 3 4 5")
elif num==1:
str1 = "6 13 27 55 111 223 447"
print(str1)
stdout.flush()
num = int(input())
qs = int(str1.split(" ")[num])
if num>=0:
query(qs,qs,1)
elif num==1:
str1 = "896 10000 "+" ".join([str(100000+i*int(1e12)) for i in range(895)])
print(str1)
stdout.flush()
num = int(input())
if(num==0):
query(896,896,3)
elif num>0:
query(int(str1.split(" ")[num]),10000,3)
elif num==1:
query(804761420475,10000,4)
```
No
| 78,296 | [
0.06585693359375,
-0.0667724609375,
-0.380126953125,
-0.1666259765625,
-0.453857421875,
-0.49853515625,
-0.43212890625,
0.188232421875,
-0.0234832763671875,
0.91259765625,
0.60302734375,
0.2125244140625,
-0.019073486328125,
-0.75146484375,
-0.705078125,
-0.032012939453125,
-0.3640136... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is interactive.
You should guess hidden number x which is between 1 and M = 10004205361450474, inclusive.
You could use up to 5 queries.
In each query, you can output an increasing sequence of k ≤ x integers, each between 1 and M, inclusive, and you will obtain one of the following as an answer:
* either the hidden number belongs to your query sequence, in this case you immediately win;
* or you will be given where the hidden number is located with respect to your query sequence, that is, either it is less than all numbers from the sequence, greater than all numbers from the sequence, or you will be given such an i that the hidden number x is between the i-th and the (i+1)-st numbers of your sequence.
See the interaction section for clarity.
Be aware that the interactor is adaptive, i.e. the hidden number can depend on queries the solution makes. However, it is guaranteed that for any solution the interactor works non-distinguishable from the situation when the hidden number is fixed beforehand.
Hacks are allowed only with fixed hidden number. A hack is represented by a single integer between 1 and M. In all pretests the hidden number is fixed as well.
Interaction
You can make up to 5 queries. To make a query print a number k (1 ≤ k ≤ 10^4) and then an increasing sequence t_0 < t_1 < … < t_{k-1} of k numbers, each between 1 and M, inclusive. If k > x, you lose.
You get one integer as a response.
* If it is -2, it means you made an invalid query or you lost. Exit immediately after receiving -2 and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.
* If it is -1, you guessed the number and should terminate too.
* Otherwise you get a number i between 0 and k, inclusive, denoting where the hidden number is, with respect to the printed numbers. If i = 0, then x < t_0. If i = k, then t_{k-1} < x. Otherwise t_{i-1} < x < t_i.
After printing a query do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Example
Input
2
0
-1
Output
2 2 3
2 20 30
3 5 7 9
Note
In the first example the number 5 is hidden.
Submitted Solution:
```
import sys
# data = sys.stdin.readlines()
start = 1
end = 10004205361450474
print(1, end=' ')
print(10000)
sys.stdout.flush()
t = int(input())
if t < 0:
exit(0)
elif t == 0:
end = 9999
num = 1
print(num, end=' ')
print(21)
# print(start, end, end - start)
# print()
sys.stdout.flush()
t = int(input())
if t < 0:
exit(0)
elif t == 0:
end = 20
print(1, end = ' ')
print(4)
sys.stdout.flush()
t = int(input())
if t < 0:
exit(0)
elif t == 0:
print(1, end = ' ')
print(1)
sys.stdout.flush()
t = int(input())
if t < 0:
exit(0)
else:
print(2, end = ' ')
print(2, 3)
sys.stdout.flush()
else:
start = 5
for i in range(2):
num = min(4, end - start + 1)
print(num, end=' ')
for j in range(num):
print(((end - start + 1)//num) * j + start, end=' ')
# print(start, end, end - start)
print()
sys.stdout.flush()
t = int(input())
if t < 0:
exit(0)
elif t == 0:
end = (end - start + 1)//num + start
elif t == num:
start = ((end - start + 1)//num) * num + start
else:
start, end = ((end - start + 1)//num) * (t-1) + start, ((end - start + 1)//num) * (t) + start
else:
start = 21
for i in range(3):
num = min(20, end - start + 1)
print(num, end=' ')
for j in range(num):
print(((end - start + 1)//num) * j + start, end=' ')
# print(start, end, end - start)
print()
sys.stdout.flush()
t = int(input())
if t < 0:
exit(0)
elif t == 0:
end = (end - start + 1)//num + start
elif t == num:
start = ((end - start + 1)//num) * num + start
else:
start, end = ((end - start + 1)//num) * (t-1) + start, ((end - start + 1)//num) * (t) + start
else:
start = 10001
for i in range(4):
num = min(10000, end - start + 1)
print(num, end=' ')
for j in range(num):
print(((end - start + 1)//num) * j + start, end=' ')
# print(start, end, end - start)
print()
sys.stdout.flush()
t = int(input())
if t < 0:
exit(0)
elif t == 0:
end = (end - start + 1)//num + start
elif t == num:
start = ((end - start + 1)//num) * num + start
else:
start, end = ((end - start + 1)//num) * (t-1) + start, ((end - start + 1)//num) * (t) + start
```
No
| 78,297 | [
0.06585693359375,
-0.0667724609375,
-0.380126953125,
-0.1666259765625,
-0.453857421875,
-0.49853515625,
-0.43212890625,
0.188232421875,
-0.0234832763671875,
0.91259765625,
0.60302734375,
0.2125244140625,
-0.019073486328125,
-0.75146484375,
-0.705078125,
-0.032012939453125,
-0.3640136... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call the roundness of the number the number of zeros to which it ends.
You have an array of n numbers. You need to choose a subset of exactly k numbers so that the roundness of the product of the selected numbers will be maximum possible.
Input
The first line contains two integer numbers n and k (1 ≤ n ≤ 200, 1 ≤ k ≤ n).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 1018).
Output
Print maximal roundness of product of the chosen subset of length k.
Examples
Input
3 2
50 4 20
Output
3
Input
5 3
15 16 3 25 9
Output
3
Input
3 3
9 77 13
Output
0
Note
In the first example there are 3 subsets of 2 numbers. [50, 4] has product 200 with roundness 2, [4, 20] — product 80, roundness 1, [50, 20] — product 1000, roundness 3.
In the second example subset [15, 16, 25] has product 6000, roundness 3.
In the third example all subsets has product with roundness 0.
Submitted Solution:
```
import itertools
def numOfZero(n):
res = 0
while n != 0:
if n % 10 == 0:
res += 1
else:
return res
n //= 10
n, k = tuple(map(int, input().split()))
nums = list(map(int, input().split()))
nums = list(filter(lambda x: x % 2 == 0 or x % 10 == 5, nums))
cirle = 0
if len(nums) != 0:
combs = itertools.combinations(nums, k)
for comb in combs:
mult = 1
for i in comb:
mult *= i
if numOfZero(mult) > cirle:
cirle = numOfZero(mult)
print(cirle)
```
No
| 78,808 | [
0.49072265625,
-0.19140625,
-0.2156982421875,
-0.1416015625,
-0.8037109375,
-0.70556640625,
-0.0830078125,
0.0887451171875,
-0.11163330078125,
1.07421875,
0.62548828125,
0.05621337890625,
0.173095703125,
-0.61328125,
-0.498291015625,
0.390380859375,
-0.8056640625,
-1.1201171875,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call the roundness of the number the number of zeros to which it ends.
You have an array of n numbers. You need to choose a subset of exactly k numbers so that the roundness of the product of the selected numbers will be maximum possible.
Input
The first line contains two integer numbers n and k (1 ≤ n ≤ 200, 1 ≤ k ≤ n).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 1018).
Output
Print maximal roundness of product of the chosen subset of length k.
Examples
Input
3 2
50 4 20
Output
3
Input
5 3
15 16 3 25 9
Output
3
Input
3 3
9 77 13
Output
0
Note
In the first example there are 3 subsets of 2 numbers. [50, 4] has product 200 with roundness 2, [4, 20] — product 80, roundness 1, [50, 20] — product 1000, roundness 3.
In the second example subset [15, 16, 25] has product 6000, roundness 3.
In the third example all subsets has product with roundness 0.
Submitted Solution:
```
import math
def mind(n):
d = 0
for i in range(2, math.ceil(n ** 0.5) + 1):
if n % i == 0:
d = i
break
if d != 0:
return d
else:
return n
def f(l):
return min(l.count(2), l.count(5))
n, k = map(int, input().split())
l = [[] for i in range(n)]
inp = [int(i) for i in input().split()]
for i in range(n):
j = inp[i]
while j > 1:
d = mind(j)
j //= d
l[i].append(d)
l.sort(key = f, reverse = True)
out = []
for i in l:
out += i
print(min(out.count(2), out.count(5)))
```
No
| 78,809 | [
0.3603515625,
-0.1787109375,
-0.260009765625,
-0.1357421875,
-0.95556640625,
-0.6220703125,
0.0013666152954101562,
0.3349609375,
-0.253662109375,
1.0322265625,
0.54638671875,
-0.04345703125,
0.1651611328125,
-0.70166015625,
-0.54248046875,
0.28173828125,
-0.837890625,
-0.884765625,... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call the roundness of the number the number of zeros to which it ends.
You have an array of n numbers. You need to choose a subset of exactly k numbers so that the roundness of the product of the selected numbers will be maximum possible.
Input
The first line contains two integer numbers n and k (1 ≤ n ≤ 200, 1 ≤ k ≤ n).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 1018).
Output
Print maximal roundness of product of the chosen subset of length k.
Examples
Input
3 2
50 4 20
Output
3
Input
5 3
15 16 3 25 9
Output
3
Input
3 3
9 77 13
Output
0
Note
In the first example there are 3 subsets of 2 numbers. [50, 4] has product 200 with roundness 2, [4, 20] — product 80, roundness 1, [50, 20] — product 1000, roundness 3.
In the second example subset [15, 16, 25] has product 6000, roundness 3.
In the third example all subsets has product with roundness 0.
Submitted Solution:
```
import sys
n, k = map(int, input().split())
z = list(map(int, input().split()))
gg = []
for i in z:
j = i
ctr2 = 0
ctr5 = 0
while j % 2 == 0:
j //= 2
ctr2 += 1
while j % 5 == 0:
j //= 5
ctr5 += 1
gg.append([ctr2, ctr5])
MAXN = 12000
ans = [0] * MAXN
for i in range(n):
for j in range(MAXN - 1, -1, -1):
if (ans[j] != 0 or j == 0):
ans[j + gg[i][0]] = max(ans[j + gg[i][0]], ans[j] + gg[i][1])
ret = 0
for i in range(MAXN):
ret = max(ret, min(i, ans[i]))
print(ret)
```
No
| 78,810 | [
0.405029296875,
-0.09710693359375,
-0.1263427734375,
-0.01025390625,
-0.87255859375,
-0.71044921875,
-0.10626220703125,
0.29931640625,
-0.143310546875,
1.0224609375,
0.541015625,
-0.0061492919921875,
0.1275634765625,
-0.72314453125,
-0.54296875,
0.36572265625,
-0.796875,
-0.9331054... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call the roundness of the number the number of zeros to which it ends.
You have an array of n numbers. You need to choose a subset of exactly k numbers so that the roundness of the product of the selected numbers will be maximum possible.
Input
The first line contains two integer numbers n and k (1 ≤ n ≤ 200, 1 ≤ k ≤ n).
The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 1018).
Output
Print maximal roundness of product of the chosen subset of length k.
Examples
Input
3 2
50 4 20
Output
3
Input
5 3
15 16 3 25 9
Output
3
Input
3 3
9 77 13
Output
0
Note
In the first example there are 3 subsets of 2 numbers. [50, 4] has product 200 with roundness 2, [4, 20] — product 80, roundness 1, [50, 20] — product 1000, roundness 3.
In the second example subset [15, 16, 25] has product 6000, roundness 3.
In the third example all subsets has product with roundness 0.
Submitted Solution:
```
import math
from decimal import Decimal
import heapq
import copy
import heapq
from collections import deque
from collections import defaultdict
def na():
n = int(input())
b = [int(x) for x in input().split()]
return n,b
def nab():
n = int(input())
b = [int(x) for x in input().split()]
c = [int(x) for x in input().split()]
return n,b,c
def dv():
n, m = map(int, input().split())
return n,m
def da():
n, m = map(int, input().split())
a = list(map(int, input().split()))
return n,m, a
def dva():
n, m = map(int, input().split())
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]
return n,m,b
def prost(x):
d = math.sqrt(x)
d = int(d) + 1
for i in range(2, d):
if x % i == 0:
return False
return True
def eratosthenes(n):
sieve = list(range(n + 1))
for i in sieve:
if i > 1:
for j in range(i + i, len(sieve), i):
sieve[j] = 0
return sorted(set(sieve))
def lol(lst,k):
k=k%len(lst)
ret=[0]*len(lst)
for i in range(len(lst)):
if i+k<len(lst) and i+k>=0:
ret[i]=lst[i+k]
if i+k>=len(lst):
ret[i]=lst[i+k-len(lst)]
if i+k<0:
ret[i]=lst[i+k+len(lst)]
return(ret)
def nm():
n = int(input())
b = [int(x) for x in input().split()]
m = int(input())
c = [int(x) for x in input().split()]
return n,b,m,c
def dvs():
n = int(input())
m = int(input())
return n, m
n, k = map(int, input().split())
a = list(map(int, input().split()))
dp = []
tec = 0
for i in range(n + 1):
dp.append([0] * 10001)
for i in range(n):
k2 = 0
k5 = 0
while a[i] % 5 == 0:
k5 += 1
a[i] //= 5
while a[i] % 2 == 0:
k2 += 1
a[i] //= 2
tec += k5
for j in range(min(i + 1, k), -1, -1):
for kk in range(k5, tec + 1):
dp[j][kk] = max(dp[j][kk], dp[j - 1][kk - k5] + k2)
ans = 0
for i in range(tec + 1):
ans = max(ans, min(i, dp[k][i]))
print(ans)
```
No
| 78,811 | [
0.39990234375,
-0.0758056640625,
-0.235107421875,
-0.1640625,
-0.8125,
-0.4375,
0.0019102096557617188,
0.205810546875,
-0.1668701171875,
1.0390625,
0.453369140625,
-0.006649017333984375,
0.14990234375,
-0.61767578125,
-0.52099609375,
0.384765625,
-0.75830078125,
-0.921875,
-0.584... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`.
When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.
Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds.
Now, he goes through the following process:
* Submit the code.
* Wait until the code finishes execution on all the cases.
* If the code fails to correctly solve some of the M cases, submit it again.
* Repeat until the code correctly solve all the cases in one submission.
Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* 1 \leq M \leq {\rm min}(N, 5)
Input
Input is given from Standard Input in the following format:
N M
Output
Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 10^9.
Examples
Input
1 1
Output
3800
Input
10 2
Output
18400
Input
100 5
Output
608000
"Correct Solution:
```
n , m = map(int,input().split())
a = m*1900 + (n-m)*100
print(a*2**m)
```
| 78,961 | [
0.442138671875,
0.31103515625,
-0.278076171875,
-0.0007834434509277344,
-0.3671875,
-0.254150390625,
-0.017852783203125,
-0.0882568359375,
0.045654296875,
0.9853515625,
0.354736328125,
-0.151123046875,
0.248779296875,
-0.84130859375,
-0.47216796875,
-0.03125,
-0.3515625,
-0.7768554... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`.
When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.
Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds.
Now, he goes through the following process:
* Submit the code.
* Wait until the code finishes execution on all the cases.
* If the code fails to correctly solve some of the M cases, submit it again.
* Repeat until the code correctly solve all the cases in one submission.
Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* 1 \leq M \leq {\rm min}(N, 5)
Input
Input is given from Standard Input in the following format:
N M
Output
Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 10^9.
Examples
Input
1 1
Output
3800
Input
10 2
Output
18400
Input
100 5
Output
608000
"Correct Solution:
```
n,m=map(int,input().split())
print((100*n+1800*m)*(2**m))
```
| 78,962 | [
0.44873046875,
0.3037109375,
-0.281982421875,
-0.002063751220703125,
-0.368408203125,
-0.251708984375,
-0.031982421875,
-0.08770751953125,
0.044952392578125,
0.98974609375,
0.359375,
-0.1435546875,
0.2498779296875,
-0.83544921875,
-0.4765625,
-0.0205535888671875,
-0.34814453125,
-0... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`.
When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.
Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds.
Now, he goes through the following process:
* Submit the code.
* Wait until the code finishes execution on all the cases.
* If the code fails to correctly solve some of the M cases, submit it again.
* Repeat until the code correctly solve all the cases in one submission.
Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* 1 \leq M \leq {\rm min}(N, 5)
Input
Input is given from Standard Input in the following format:
N M
Output
Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 10^9.
Examples
Input
1 1
Output
3800
Input
10 2
Output
18400
Input
100 5
Output
608000
"Correct Solution:
```
n,m=[int(i) for i in input().split()]
print((m*18+n)*100*2**m)
```
| 78,963 | [
0.4384765625,
0.26025390625,
-0.272705078125,
0.0006003379821777344,
-0.3720703125,
-0.259765625,
-0.008880615234375,
-0.09149169921875,
0.08746337890625,
0.9716796875,
0.353515625,
-0.1846923828125,
0.18994140625,
-0.84375,
-0.54833984375,
-0.0181121826171875,
-0.3837890625,
-0.77... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`.
When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.
Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds.
Now, he goes through the following process:
* Submit the code.
* Wait until the code finishes execution on all the cases.
* If the code fails to correctly solve some of the M cases, submit it again.
* Repeat until the code correctly solve all the cases in one submission.
Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* 1 \leq M \leq {\rm min}(N, 5)
Input
Input is given from Standard Input in the following format:
N M
Output
Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 10^9.
Examples
Input
1 1
Output
3800
Input
10 2
Output
18400
Input
100 5
Output
608000
"Correct Solution:
```
n,m=map(int,input().split())
print(int((100*(n-m)+1900*m)*2**m))
```
| 78,964 | [
0.45068359375,
0.297119140625,
-0.281005859375,
-0.000797271728515625,
-0.373046875,
-0.2626953125,
-0.017547607421875,
-0.08349609375,
0.03717041015625,
0.9912109375,
0.364990234375,
-0.150390625,
0.2376708984375,
-0.8349609375,
-0.47607421875,
-0.028472900390625,
-0.362548828125,
... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`.
When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.
Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds.
Now, he goes through the following process:
* Submit the code.
* Wait until the code finishes execution on all the cases.
* If the code fails to correctly solve some of the M cases, submit it again.
* Repeat until the code correctly solve all the cases in one submission.
Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* 1 \leq M \leq {\rm min}(N, 5)
Input
Input is given from Standard Input in the following format:
N M
Output
Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 10^9.
Examples
Input
1 1
Output
3800
Input
10 2
Output
18400
Input
100 5
Output
608000
"Correct Solution:
```
n,m = map(int,input().split())
print(2**m*1900*m + 2**m*100*(n-m))
```
| 78,965 | [
0.43896484375,
0.303466796875,
-0.275146484375,
-0.0088043212890625,
-0.366943359375,
-0.256103515625,
-0.031524658203125,
-0.0897216796875,
0.036590576171875,
0.98828125,
0.36865234375,
-0.14990234375,
0.262939453125,
-0.84033203125,
-0.466796875,
-0.0194091796875,
-0.36474609375,
... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`.
When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.
Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds.
Now, he goes through the following process:
* Submit the code.
* Wait until the code finishes execution on all the cases.
* If the code fails to correctly solve some of the M cases, submit it again.
* Repeat until the code correctly solve all the cases in one submission.
Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* 1 \leq M \leq {\rm min}(N, 5)
Input
Input is given from Standard Input in the following format:
N M
Output
Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 10^9.
Examples
Input
1 1
Output
3800
Input
10 2
Output
18400
Input
100 5
Output
608000
"Correct Solution:
```
N,M=map(int,input().split())
ans=pow(2,M)*(1900*M+100*(N-M))
print(ans)
```
| 78,966 | [
0.50146484375,
0.27978515625,
-0.278564453125,
0.017120361328125,
-0.40966796875,
-0.1829833984375,
-0.0262908935546875,
-0.0985107421875,
0.08404541015625,
0.96337890625,
0.4140625,
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0.2275390625,
-0.8330078125,
-0.443115234375,
-0.01042938232421875,
-0.332763671875,
-... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`.
When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.
Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds.
Now, he goes through the following process:
* Submit the code.
* Wait until the code finishes execution on all the cases.
* If the code fails to correctly solve some of the M cases, submit it again.
* Repeat until the code correctly solve all the cases in one submission.
Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* 1 \leq M \leq {\rm min}(N, 5)
Input
Input is given from Standard Input in the following format:
N M
Output
Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 10^9.
Examples
Input
1 1
Output
3800
Input
10 2
Output
18400
Input
100 5
Output
608000
"Correct Solution:
```
N, M = list(map(int, input().split()))
print(((N-M)*100+M*1900)*2**M)
```
| 78,967 | [
0.433349609375,
0.278076171875,
-0.25732421875,
0.0020275115966796875,
-0.382568359375,
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-0.02532958984375,
-0.08251953125,
0.065673828125,
0.98193359375,
0.359619140625,
-0.16162109375,
0.25439453125,
-0.8291015625,
-0.494384765625,
-0.01314544677734375,
-0.372802734... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`.
When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.
Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds.
Now, he goes through the following process:
* Submit the code.
* Wait until the code finishes execution on all the cases.
* If the code fails to correctly solve some of the M cases, submit it again.
* Repeat until the code correctly solve all the cases in one submission.
Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* 1 \leq M \leq {\rm min}(N, 5)
Input
Input is given from Standard Input in the following format:
N M
Output
Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 10^9.
Examples
Input
1 1
Output
3800
Input
10 2
Output
18400
Input
100 5
Output
608000
"Correct Solution:
```
n,m=map(int, input().split())
x=n-m
print(((n-m)*100+1900*m)*2**m)
```
| 78,968 | [
0.447265625,
0.2919921875,
-0.257080078125,
0.0008463859558105469,
-0.3701171875,
-0.25,
-0.01291656494140625,
-0.09619140625,
0.053375244140625,
0.9970703125,
0.376708984375,
-0.1475830078125,
0.2298583984375,
-0.837890625,
-0.46826171875,
-0.01294708251953125,
-0.3603515625,
-0.7... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`.
When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.
Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds.
Now, he goes through the following process:
* Submit the code.
* Wait until the code finishes execution on all the cases.
* If the code fails to correctly solve some of the M cases, submit it again.
* Repeat until the code correctly solve all the cases in one submission.
Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* 1 \leq M \leq {\rm min}(N, 5)
Input
Input is given from Standard Input in the following format:
N M
Output
Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 10^9.
Examples
Input
1 1
Output
3800
Input
10 2
Output
18400
Input
100 5
Output
608000
Submitted Solution:
```
n,m=map(int,input().split());print(100*(18*m+n)<<m)
```
Yes
| 78,969 | [
0.424560546875,
0.18505859375,
-0.321533203125,
0.007099151611328125,
-0.39892578125,
-0.1363525390625,
-0.026885986328125,
-0.07061767578125,
-0.0189666748046875,
0.99267578125,
0.28759765625,
-0.100341796875,
0.1788330078125,
-0.77587890625,
-0.496826171875,
-0.044158935546875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`.
When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.
Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds.
Now, he goes through the following process:
* Submit the code.
* Wait until the code finishes execution on all the cases.
* If the code fails to correctly solve some of the M cases, submit it again.
* Repeat until the code correctly solve all the cases in one submission.
Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* 1 \leq M \leq {\rm min}(N, 5)
Input
Input is given from Standard Input in the following format:
N M
Output
Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 10^9.
Examples
Input
1 1
Output
3800
Input
10 2
Output
18400
Input
100 5
Output
608000
Submitted Solution:
```
n,m=map(int,input().split())
print(100*(n+18*m)*(2**m))
```
Yes
| 78,970 | [
0.425537109375,
0.1832275390625,
-0.312744140625,
0.0133056640625,
-0.399169921875,
-0.140869140625,
-0.0220489501953125,
-0.0584716796875,
-0.0156707763671875,
0.994140625,
0.301025390625,
-0.10137939453125,
0.1705322265625,
-0.7685546875,
-0.49755859375,
-0.04412841796875,
-0.34130... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`.
When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.
Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds.
Now, he goes through the following process:
* Submit the code.
* Wait until the code finishes execution on all the cases.
* If the code fails to correctly solve some of the M cases, submit it again.
* Repeat until the code correctly solve all the cases in one submission.
Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* 1 \leq M \leq {\rm min}(N, 5)
Input
Input is given from Standard Input in the following format:
N M
Output
Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 10^9.
Examples
Input
1 1
Output
3800
Input
10 2
Output
18400
Input
100 5
Output
608000
Submitted Solution:
```
n,m=map(int,input().split());print(((n-m)*100+m*1900)*2**m)
```
Yes
| 78,971 | [
0.425537109375,
0.1746826171875,
-0.305419921875,
0.00814056396484375,
-0.41259765625,
-0.1265869140625,
-0.01739501953125,
-0.059326171875,
-0.019287109375,
0.99560546875,
0.303466796875,
-0.1029052734375,
0.17724609375,
-0.76025390625,
-0.49951171875,
-0.045928955078125,
-0.3515625... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`.
When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.
Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds.
Now, he goes through the following process:
* Submit the code.
* Wait until the code finishes execution on all the cases.
* If the code fails to correctly solve some of the M cases, submit it again.
* Repeat until the code correctly solve all the cases in one submission.
Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* 1 \leq M \leq {\rm min}(N, 5)
Input
Input is given from Standard Input in the following format:
N M
Output
Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 10^9.
Examples
Input
1 1
Output
3800
Input
10 2
Output
18400
Input
100 5
Output
608000
Submitted Solution:
```
N,M=map(int,input().split())
print((100*(N-M)+1900*M)*2**M)
```
Yes
| 78,972 | [
0.43115234375,
0.1859130859375,
-0.3203125,
0.01444244384765625,
-0.40234375,
-0.1361083984375,
-0.01143646240234375,
-0.053955078125,
-0.01593017578125,
0.98779296875,
0.296630859375,
-0.10650634765625,
0.1669921875,
-0.7646484375,
-0.493896484375,
-0.058197021484375,
-0.34008789062... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`.
When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.
Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds.
Now, he goes through the following process:
* Submit the code.
* Wait until the code finishes execution on all the cases.
* If the code fails to correctly solve some of the M cases, submit it again.
* Repeat until the code correctly solve all the cases in one submission.
Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* 1 \leq M \leq {\rm min}(N, 5)
Input
Input is given from Standard Input in the following format:
N M
Output
Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 10^9.
Examples
Input
1 1
Output
3800
Input
10 2
Output
18400
Input
100 5
Output
608000
Submitted Solution:
```
N,M=map(int,input().split())
p=(1/2)**M
listP=[p]
listP+=[0]*(10**6)
sumP=[0]*(10**6)
sumP[0]=p
ans=0
time=1900*M+100*(N-M)
for i in range(10**5):
ans+=time*listP[i]*(i+1)
sumP[i+1]=sumP[i]+p*(1-sumP[i])
listP[i+1]=p*(1-sumP[i])
print(int(ans))
```
No
| 78,973 | [
0.388427734375,
0.250732421875,
-0.1839599609375,
0.09869384765625,
-0.407470703125,
-0.18896484375,
-0.0732421875,
-0.0775146484375,
0.0687255859375,
1.0048828125,
0.29248046875,
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0.165283203125,
-0.69189453125,
-0.52490234375,
-0.036163330078125,
-0.342529296875,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`.
When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.
Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds.
Now, he goes through the following process:
* Submit the code.
* Wait until the code finishes execution on all the cases.
* If the code fails to correctly solve some of the M cases, submit it again.
* Repeat until the code correctly solve all the cases in one submission.
Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* 1 \leq M \leq {\rm min}(N, 5)
Input
Input is given from Standard Input in the following format:
N M
Output
Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 10^9.
Examples
Input
1 1
Output
3800
Input
10 2
Output
18400
Input
100 5
Output
608000
Submitted Solution:
```
import sys
input = sys.stdin.readline
n,m = map(int, input().split())
print((1900*m + 100(N − M))*2**m)
```
No
| 78,974 | [
0.42041015625,
0.1795654296875,
-0.30126953125,
0.032562255859375,
-0.44091796875,
-0.1376953125,
-0.016754150390625,
-0.033599853515625,
-0.0006489753723144531,
0.9931640625,
0.26806640625,
-0.11077880859375,
0.1676025390625,
-0.75146484375,
-0.497802734375,
-0.070068359375,
-0.3422... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`.
When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.
Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds.
Now, he goes through the following process:
* Submit the code.
* Wait until the code finishes execution on all the cases.
* If the code fails to correctly solve some of the M cases, submit it again.
* Repeat until the code correctly solve all the cases in one submission.
Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* 1 \leq M \leq {\rm min}(N, 5)
Input
Input is given from Standard Input in the following format:
N M
Output
Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 10^9.
Examples
Input
1 1
Output
3800
Input
10 2
Output
18400
Input
100 5
Output
608000
Submitted Solution:
```
n, m = map(int, input().split())
if n==m:
print(3800)
else:
s = (1900*m + (n-m)*100)*n*2**m
x = 10**(len(str(n))-1)
print(s//x)
```
No
| 78,975 | [
0.396728515625,
0.158447265625,
-0.3134765625,
0.004558563232421875,
-0.39306640625,
-0.1663818359375,
0.02056884765625,
-0.079345703125,
0.0247802734375,
0.99951171875,
0.363525390625,
-0.092041015625,
0.1866455078125,
-0.8359375,
-0.443115234375,
-0.042510986328125,
-0.38232421875,... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is now competing in a programming contest, but he received TLE in a problem where the answer is `YES` or `NO`.
When he checked the detailed status of the submission, there were N test cases in the problem, and the code received TLE in M of those cases.
Then, he rewrote the code to correctly solve each of those M cases with 1/2 probability in 1900 milliseconds, and correctly solve each of the other N-M cases without fail in 100 milliseconds.
Now, he goes through the following process:
* Submit the code.
* Wait until the code finishes execution on all the cases.
* If the code fails to correctly solve some of the M cases, submit it again.
* Repeat until the code correctly solve all the cases in one submission.
Let the expected value of the total execution time of the code be X milliseconds. Print X (as an integer).
Constraints
* All input values are integers.
* 1 \leq N \leq 100
* 1 \leq M \leq {\rm min}(N, 5)
Input
Input is given from Standard Input in the following format:
N M
Output
Print X, the expected value of the total execution time of the code, as an integer. It can be proved that, under the constraints in this problem, X is an integer not exceeding 10^9.
Examples
Input
1 1
Output
3800
Input
10 2
Output
18400
Input
100 5
Output
608000
Submitted Solution:
```
n, m = map(int,input().split())
ans = 0
count = 0
import math
not_c = 1-(1/2)**m
c = (1/ 2) ** m
s = 1900
#k-1 回目まで不正解かつ k回目で終わる確率は
# (not_c ** (k-1)) * c 199
# k回目までに(s * k)が重み付け
ans = s * c /((1-not_c) ** 2)
print(int(ans))
```
No
| 78,976 | [
0.436279296875,
0.2056884765625,
-0.2646484375,
-0.025970458984375,
-0.396484375,
-0.2095947265625,
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-0.04931640625,
0.0390625,
0.98193359375,
0.324951171875,
-0.095703125,
0.102294921875,
-0.8203125,
-0.572265625,
-0.1025390625,
-0.3818359375,
-0.6943359375,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The finalists of the "Russian Code Cup" competition in 2214 will be the participants who win in one of the elimination rounds.
The elimination rounds are divided into main and additional. Each of the main elimination rounds consists of c problems, the winners of the round are the first n people in the rating list. Each of the additional elimination rounds consists of d problems. The winner of the additional round is one person. Besides, k winners of the past finals are invited to the finals without elimination.
As a result of all elimination rounds at least n·m people should go to the finals. You need to organize elimination rounds in such a way, that at least n·m people go to the finals, and the total amount of used problems in all rounds is as small as possible.
Input
The first line contains two integers c and d (1 ≤ c, d ≤ 100) — the number of problems in the main and additional rounds, correspondingly. The second line contains two integers n and m (1 ≤ n, m ≤ 100). Finally, the third line contains an integer k (1 ≤ k ≤ 100) — the number of the pre-chosen winners.
Output
In the first line, print a single integer — the minimum number of problems the jury needs to prepare.
Examples
Input
1 10
7 2
1
Output
2
Input
2 2
2 1
2
Output
0
Tags: dp, implementation, math
Correct Solution:
```
# Made By Mostafa_Khaled
bot = True
c, d = list(map(int, input().split()))
n, m = list(map(int, input().split()))
k = int(input())
kolvo = m * n - k
if kolvo < 0:
print(0)
exit()
print(min(c * (kolvo // n), d * n * (kolvo // n)) + min(c, d * (kolvo % n)))
# Made By Mostafa_Khaled
```
| 79,480 | [
0.479248046875,
-0.292236328125,
-0.0892333984375,
0.3955078125,
-0.425537109375,
-0.54052734375,
-0.2435302734375,
0.431396484375,
-0.056304931640625,
0.5283203125,
0.6474609375,
0.03662109375,
0.044769287109375,
-0.63037109375,
-0.88134765625,
0.0433349609375,
-0.96337890625,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The finalists of the "Russian Code Cup" competition in 2214 will be the participants who win in one of the elimination rounds.
The elimination rounds are divided into main and additional. Each of the main elimination rounds consists of c problems, the winners of the round are the first n people in the rating list. Each of the additional elimination rounds consists of d problems. The winner of the additional round is one person. Besides, k winners of the past finals are invited to the finals without elimination.
As a result of all elimination rounds at least n·m people should go to the finals. You need to organize elimination rounds in such a way, that at least n·m people go to the finals, and the total amount of used problems in all rounds is as small as possible.
Input
The first line contains two integers c and d (1 ≤ c, d ≤ 100) — the number of problems in the main and additional rounds, correspondingly. The second line contains two integers n and m (1 ≤ n, m ≤ 100). Finally, the third line contains an integer k (1 ≤ k ≤ 100) — the number of the pre-chosen winners.
Output
In the first line, print a single integer — the minimum number of problems the jury needs to prepare.
Examples
Input
1 10
7 2
1
Output
2
Input
2 2
2 1
2
Output
0
Submitted Solution:
```
c,d = list(map(int,input().split()))
n,m = list(map(int,input().split()))
k = int(input())
t = max((m*n)-k,0)
p = min(c,n*d)
ans = p*(t//n)
rem = t%n
ans = ans+min(c,rem*d)
print(ans)
```
Yes
| 79,481 | [
0.44921875,
-0.1204833984375,
-0.11956787109375,
0.1361083984375,
-0.67919921875,
-0.265869140625,
-0.226806640625,
0.4404296875,
-0.09002685546875,
0.6513671875,
0.84228515625,
0.2412109375,
0.125,
-0.80126953125,
-0.798828125,
-0.1661376953125,
-0.87451171875,
-1.0517578125,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The finalists of the "Russian Code Cup" competition in 2214 will be the participants who win in one of the elimination rounds.
The elimination rounds are divided into main and additional. Each of the main elimination rounds consists of c problems, the winners of the round are the first n people in the rating list. Each of the additional elimination rounds consists of d problems. The winner of the additional round is one person. Besides, k winners of the past finals are invited to the finals without elimination.
As a result of all elimination rounds at least n·m people should go to the finals. You need to organize elimination rounds in such a way, that at least n·m people go to the finals, and the total amount of used problems in all rounds is as small as possible.
Input
The first line contains two integers c and d (1 ≤ c, d ≤ 100) — the number of problems in the main and additional rounds, correspondingly. The second line contains two integers n and m (1 ≤ n, m ≤ 100). Finally, the third line contains an integer k (1 ≤ k ≤ 100) — the number of the pre-chosen winners.
Output
In the first line, print a single integer — the minimum number of problems the jury needs to prepare.
Examples
Input
1 10
7 2
1
Output
2
Input
2 2
2 1
2
Output
0
Submitted Solution:
```
import math
c,d = map(int,input().split())
n,m = map(int,input().split())
k = int(input())
if k>=n*m:
print (0)
exit()
left = n*m-k
print (min(math.ceil(left/n)*c,(left//n)*c + (left%n)*d,left*d))
```
Yes
| 79,482 | [
0.396728515625,
-0.1871337890625,
-0.169189453125,
0.156005859375,
-0.6552734375,
-0.203369140625,
-0.1956787109375,
0.44970703125,
-0.09912109375,
0.6123046875,
0.7734375,
0.216796875,
0.105224609375,
-0.74853515625,
-0.76953125,
-0.17822265625,
-0.8642578125,
-1.0224609375,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The finalists of the "Russian Code Cup" competition in 2214 will be the participants who win in one of the elimination rounds.
The elimination rounds are divided into main and additional. Each of the main elimination rounds consists of c problems, the winners of the round are the first n people in the rating list. Each of the additional elimination rounds consists of d problems. The winner of the additional round is one person. Besides, k winners of the past finals are invited to the finals without elimination.
As a result of all elimination rounds at least n·m people should go to the finals. You need to organize elimination rounds in such a way, that at least n·m people go to the finals, and the total amount of used problems in all rounds is as small as possible.
Input
The first line contains two integers c and d (1 ≤ c, d ≤ 100) — the number of problems in the main and additional rounds, correspondingly. The second line contains two integers n and m (1 ≤ n, m ≤ 100). Finally, the third line contains an integer k (1 ≤ k ≤ 100) — the number of the pre-chosen winners.
Output
In the first line, print a single integer — the minimum number of problems the jury needs to prepare.
Examples
Input
1 10
7 2
1
Output
2
Input
2 2
2 1
2
Output
0
Submitted Solution:
```
#import sys
#sys.stdin = open('input.txt','r')
c, d = map(int, input().split())
n, m = map(int, input().split())
k = int(input())
L = 10001
for f in range(m + 1):
for g in range(n * m + 1):
if f * n + g + k >= n * m:
L = min(L, f * c + g * d)
print(L)
```
Yes
| 79,483 | [
0.400146484375,
-0.1739501953125,
-0.09869384765625,
0.225341796875,
-0.66748046875,
-0.234619140625,
-0.251953125,
0.496826171875,
-0.1229248046875,
0.568359375,
0.7958984375,
0.172607421875,
0.07958984375,
-0.724609375,
-0.7666015625,
-0.1834716796875,
-0.9228515625,
-1.061523437... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The finalists of the "Russian Code Cup" competition in 2214 will be the participants who win in one of the elimination rounds.
The elimination rounds are divided into main and additional. Each of the main elimination rounds consists of c problems, the winners of the round are the first n people in the rating list. Each of the additional elimination rounds consists of d problems. The winner of the additional round is one person. Besides, k winners of the past finals are invited to the finals without elimination.
As a result of all elimination rounds at least n·m people should go to the finals. You need to organize elimination rounds in such a way, that at least n·m people go to the finals, and the total amount of used problems in all rounds is as small as possible.
Input
The first line contains two integers c and d (1 ≤ c, d ≤ 100) — the number of problems in the main and additional rounds, correspondingly. The second line contains two integers n and m (1 ≤ n, m ≤ 100). Finally, the third line contains an integer k (1 ≤ k ≤ 100) — the number of the pre-chosen winners.
Output
In the first line, print a single integer — the minimum number of problems the jury needs to prepare.
Examples
Input
1 10
7 2
1
Output
2
Input
2 2
2 1
2
Output
0
Submitted Solution:
```
c,d=map(int,input().split())
n,m=map(int,input().split())
k=int(input())
tot=n*m
if(k>=tot):
print("0")
else:
rem=tot-k
x=n/c
y=1/d
ans=0
if(x>=y):
temp=rem//n
bacha=rem%n
if(bacha*d<=c):
ans=temp*c+bacha*d
else:
ans=(temp+1)*c
else:
ans=rem*d
print(ans)
```
Yes
| 79,484 | [
0.415771484375,
-0.08880615234375,
-0.1734619140625,
0.07073974609375,
-0.64794921875,
-0.2095947265625,
-0.2203369140625,
0.40966796875,
-0.1055908203125,
0.65625,
0.83349609375,
0.270263671875,
0.03271484375,
-0.8388671875,
-0.740234375,
-0.1328125,
-0.80078125,
-0.95361328125,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The finalists of the "Russian Code Cup" competition in 2214 will be the participants who win in one of the elimination rounds.
The elimination rounds are divided into main and additional. Each of the main elimination rounds consists of c problems, the winners of the round are the first n people in the rating list. Each of the additional elimination rounds consists of d problems. The winner of the additional round is one person. Besides, k winners of the past finals are invited to the finals without elimination.
As a result of all elimination rounds at least n·m people should go to the finals. You need to organize elimination rounds in such a way, that at least n·m people go to the finals, and the total amount of used problems in all rounds is as small as possible.
Input
The first line contains two integers c and d (1 ≤ c, d ≤ 100) — the number of problems in the main and additional rounds, correspondingly. The second line contains two integers n and m (1 ≤ n, m ≤ 100). Finally, the third line contains an integer k (1 ≤ k ≤ 100) — the number of the pre-chosen winners.
Output
In the first line, print a single integer — the minimum number of problems the jury needs to prepare.
Examples
Input
1 10
7 2
1
Output
2
Input
2 2
2 1
2
Output
0
Submitted Solution:
```
import math
c,d=map(int,input('').split())
n,m=map(int,input('').split())
k=int(input(''))
if k<0:
print('0')
exit()
if 1/d>n/c: print((n*m-k)*d)
else:
if (math.ceil((n*m-k)/n)*c)<((n*m-k)//n)*c+((n*m-k)%n)*d: print (math.ceil((n*m-k)/n)*c)
else: print(((n*m-k)//n)*c+((n*m-k)%n)*d)
```
No
| 79,485 | [
0.365966796875,
-0.12939453125,
-0.181640625,
0.2076416015625,
-0.6337890625,
-0.1995849609375,
-0.22216796875,
0.47119140625,
-0.10943603515625,
0.5927734375,
0.78564453125,
0.2030029296875,
0.12548828125,
-0.76513671875,
-0.74365234375,
-0.2188720703125,
-0.84765625,
-1.041992187... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The finalists of the "Russian Code Cup" competition in 2214 will be the participants who win in one of the elimination rounds.
The elimination rounds are divided into main and additional. Each of the main elimination rounds consists of c problems, the winners of the round are the first n people in the rating list. Each of the additional elimination rounds consists of d problems. The winner of the additional round is one person. Besides, k winners of the past finals are invited to the finals without elimination.
As a result of all elimination rounds at least n·m people should go to the finals. You need to organize elimination rounds in such a way, that at least n·m people go to the finals, and the total amount of used problems in all rounds is as small as possible.
Input
The first line contains two integers c and d (1 ≤ c, d ≤ 100) — the number of problems in the main and additional rounds, correspondingly. The second line contains two integers n and m (1 ≤ n, m ≤ 100). Finally, the third line contains an integer k (1 ≤ k ≤ 100) — the number of the pre-chosen winners.
Output
In the first line, print a single integer — the minimum number of problems the jury needs to prepare.
Examples
Input
1 10
7 2
1
Output
2
Input
2 2
2 1
2
Output
0
Submitted Solution:
```
c,d = [int(x) for x in input().strip().split()]
n,m = [int(x) for x in input().strip().split()]
k = int(input().strip())
t = n*m-k
if t<=0:
print(0)
else:
dp = [0]*(t+1)
for i in range(1,t+1):
if i<n:
dp[i]=min(dp[i-1]+d, n)
else:
dp[i]=min((c+dp[i-n]), (d+dp[i-1]))
print(dp[t])
```
No
| 79,486 | [
0.376220703125,
-0.1968994140625,
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0.17333984375,
-0.60400390625,
-0.34716796875,
-0.25537109375,
0.344482421875,
-0.09881591796875,
0.62841796875,
0.796875,
0.1517333984375,
0.05804443359375,
-0.7197265625,
-0.7978515625,
-0.202880859375,
-0.9248046875,
-1.041992... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The finalists of the "Russian Code Cup" competition in 2214 will be the participants who win in one of the elimination rounds.
The elimination rounds are divided into main and additional. Each of the main elimination rounds consists of c problems, the winners of the round are the first n people in the rating list. Each of the additional elimination rounds consists of d problems. The winner of the additional round is one person. Besides, k winners of the past finals are invited to the finals without elimination.
As a result of all elimination rounds at least n·m people should go to the finals. You need to organize elimination rounds in such a way, that at least n·m people go to the finals, and the total amount of used problems in all rounds is as small as possible.
Input
The first line contains two integers c and d (1 ≤ c, d ≤ 100) — the number of problems in the main and additional rounds, correspondingly. The second line contains two integers n and m (1 ≤ n, m ≤ 100). Finally, the third line contains an integer k (1 ≤ k ≤ 100) — the number of the pre-chosen winners.
Output
In the first line, print a single integer — the minimum number of problems the jury needs to prepare.
Examples
Input
1 10
7 2
1
Output
2
Input
2 2
2 1
2
Output
0
Submitted Solution:
```
#import sys
#sys.stdin = open('input.txt','r')
c, d = map(int, input().split())
n, m = map(int, input().split())
k = int(input())
L = 10001
for f in range(m + 1):
for g in range(n * m + 1):
if f * n * c + g * d + k >= n * m:
L = min(L, f * c + g * d)
print(L)
```
No
| 79,487 | [
0.398193359375,
-0.1707763671875,
-0.0865478515625,
0.1883544921875,
-0.673828125,
-0.2410888671875,
-0.253173828125,
0.49365234375,
-0.13232421875,
0.5947265625,
0.8017578125,
0.18603515625,
0.073974609375,
-0.72265625,
-0.7666015625,
-0.193115234375,
-0.9306640625,
-1.0498046875,... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The finalists of the "Russian Code Cup" competition in 2214 will be the participants who win in one of the elimination rounds.
The elimination rounds are divided into main and additional. Each of the main elimination rounds consists of c problems, the winners of the round are the first n people in the rating list. Each of the additional elimination rounds consists of d problems. The winner of the additional round is one person. Besides, k winners of the past finals are invited to the finals without elimination.
As a result of all elimination rounds at least n·m people should go to the finals. You need to organize elimination rounds in such a way, that at least n·m people go to the finals, and the total amount of used problems in all rounds is as small as possible.
Input
The first line contains two integers c and d (1 ≤ c, d ≤ 100) — the number of problems in the main and additional rounds, correspondingly. The second line contains two integers n and m (1 ≤ n, m ≤ 100). Finally, the third line contains an integer k (1 ≤ k ≤ 100) — the number of the pre-chosen winners.
Output
In the first line, print a single integer — the minimum number of problems the jury needs to prepare.
Examples
Input
1 10
7 2
1
Output
2
Input
2 2
2 1
2
Output
0
Submitted Solution:
```
def f(x, y): return x * c + y * d
c, d = map(int, input().split())
n, m = map(int, input().split())
k = s = n * m - int(input())
x = 0
while k > x * n:
t = f(x, k - x * n)
if t < s: s = t
x += 1
print(min(s, f(x, 0)))
```
No
| 79,488 | [
0.412841796875,
-0.1483154296875,
-0.1513671875,
0.1937255859375,
-0.6259765625,
-0.254150390625,
-0.297119140625,
0.4140625,
-0.06915283203125,
0.59716796875,
0.8642578125,
0.2257080078125,
0.11663818359375,
-0.7138671875,
-0.708984375,
-0.194091796875,
-0.9130859375,
-1.095703125... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrey needs one more problem to conduct a programming contest. He has n friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him.
Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of Andrey's friends. The second line contains n real numbers pi (0.0 ≤ pi ≤ 1.0) — the probability that the i-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point.
Output
Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10 - 9.
Examples
Input
4
0.1 0.2 0.3 0.8
Output
0.800000000000
Input
2
0.1 0.2
Output
0.260000000000
Note
In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one.
In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8 + 0.9·0.2 = 0.26.
Tags: greedy, math, probabilities
Correct Solution:
```
from functools import reduce
L0 = []
def getP():
return sum([reduce(lambda y, x: y*x, [(L0[j] if j == i else (1 - L0[j])) for j in range(len(L0))], 1) for i in range(len(L0))])
input()
L = [float(x) for x in input().split()]
#inp = open('input.txt')
#inp.readline()
#L = [float(x) for x in inp.readline().split()]
#inp.close()
L.sort(reverse=True)
if len(L) > 0 and L[0] == 1:
print(1)
else:
S = 0
for p in L:
if S < 1:
S += p/(1 - p)
L0.append(p)
else:
break
print(getP())
```
| 79,489 | [
0.41015625,
0.08258056640625,
0.09478759765625,
0.09765625,
-0.58642578125,
-0.86669921875,
0.08050537109375,
0.126220703125,
-0.033294677734375,
0.78076171875,
0.486328125,
-0.6982421875,
0.377685546875,
-0.427734375,
-0.62158203125,
-0.11431884765625,
-0.6064453125,
-0.74609375,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrey needs one more problem to conduct a programming contest. He has n friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him.
Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of Andrey's friends. The second line contains n real numbers pi (0.0 ≤ pi ≤ 1.0) — the probability that the i-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point.
Output
Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10 - 9.
Examples
Input
4
0.1 0.2 0.3 0.8
Output
0.800000000000
Input
2
0.1 0.2
Output
0.260000000000
Note
In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one.
In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8 + 0.9·0.2 = 0.26.
Tags: greedy, math, probabilities
Correct Solution:
```
n = int(input())
p = sorted(map(float, input().split()), reverse=True)
np = 1.0 - p[0]
pp = p[0]
ans = pp
for i in range(1, len(p)):
pp *= 1.0 - p[i]
pp += p[i] * np
np *= 1.0 - p[i]
if ans < pp:
ans = pp
else:
break
print(ans)
```
| 79,490 | [
0.459228515625,
0.048187255859375,
0.0775146484375,
0.041534423828125,
-0.55908203125,
-0.8330078125,
0.08154296875,
0.15771484375,
0.006595611572265625,
0.81005859375,
0.54638671875,
-0.72265625,
0.362060546875,
-0.431640625,
-0.60498046875,
-0.1123046875,
-0.62353515625,
-0.77539... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrey needs one more problem to conduct a programming contest. He has n friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him.
Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of Andrey's friends. The second line contains n real numbers pi (0.0 ≤ pi ≤ 1.0) — the probability that the i-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point.
Output
Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10 - 9.
Examples
Input
4
0.1 0.2 0.3 0.8
Output
0.800000000000
Input
2
0.1 0.2
Output
0.260000000000
Note
In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one.
In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8 + 0.9·0.2 = 0.26.
Tags: greedy, math, probabilities
Correct Solution:
```
n = int(input())
pr = list(map(float, input().split()))
pr.sort(reverse = True)
p = (1 - pr[0])
if (p == 0):
print(1)
else:
s = pr[0] / (1 - pr[0])
pps = p * s
for i in range(1, len(pr)):
p *= 1 - pr[i]
s += pr[i] / (1 - pr[i])
ps = p * s
if pps > ps:
break
pps = ps
print(pps)
```
| 79,491 | [
0.45458984375,
0.107666015625,
0.08770751953125,
0.041412353515625,
-0.5458984375,
-0.80322265625,
0.0875244140625,
0.1512451171875,
0.004116058349609375,
0.8310546875,
0.48779296875,
-0.69580078125,
0.425048828125,
-0.421142578125,
-0.57470703125,
-0.09716796875,
-0.60546875,
-0.7... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrey needs one more problem to conduct a programming contest. He has n friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him.
Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of Andrey's friends. The second line contains n real numbers pi (0.0 ≤ pi ≤ 1.0) — the probability that the i-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point.
Output
Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10 - 9.
Examples
Input
4
0.1 0.2 0.3 0.8
Output
0.800000000000
Input
2
0.1 0.2
Output
0.260000000000
Note
In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one.
In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8 + 0.9·0.2 = 0.26.
Tags: greedy, math, probabilities
Correct Solution:
```
input()
p = sorted(map(float, input().split(' ')))
m = max(p)
p = [(1 - i, i) for i in p]
def konv(a, b):
return a[0] * b[0], a[0] * b[1] + a[1] * b[0]
while len(p) > 1:
p.append(konv(p.pop(), p.pop()))
m = max(m, p[-1][1])
print(m)
```
| 79,492 | [
0.44677734375,
0.07928466796875,
0.05621337890625,
0.1290283203125,
-0.6142578125,
-0.96142578125,
0.10430908203125,
0.1800537109375,
-0.10882568359375,
0.806640625,
0.51171875,
-0.662109375,
0.419189453125,
-0.384521484375,
-0.64404296875,
-0.06439208984375,
-0.68798828125,
-0.869... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrey needs one more problem to conduct a programming contest. He has n friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him.
Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of Andrey's friends. The second line contains n real numbers pi (0.0 ≤ pi ≤ 1.0) — the probability that the i-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point.
Output
Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10 - 9.
Examples
Input
4
0.1 0.2 0.3 0.8
Output
0.800000000000
Input
2
0.1 0.2
Output
0.260000000000
Note
In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one.
In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8 + 0.9·0.2 = 0.26.
Tags: greedy, math, probabilities
Correct Solution:
```
def readln(): return tuple(map(int, input().split()))
n, = readln()
ans = tmp = 0.0
prod = 1.0
for p in reversed(sorted(map(float, input().split()))):
tmp = tmp * (1.0 - p) + prod * p
prod *= 1.0 - p
#print(ans, tmp, prod, p)
ans = max(ans, tmp)
print('%0.9f' % ans)
```
| 79,493 | [
0.4658203125,
0.0960693359375,
0.056640625,
0.055389404296875,
-0.59228515625,
-0.8798828125,
0.127197265625,
0.11077880859375,
-0.0726318359375,
0.82958984375,
0.51806640625,
-0.66162109375,
0.475830078125,
-0.417236328125,
-0.56982421875,
-0.105224609375,
-0.65087890625,
-0.78369... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrey needs one more problem to conduct a programming contest. He has n friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him.
Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of Andrey's friends. The second line contains n real numbers pi (0.0 ≤ pi ≤ 1.0) — the probability that the i-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point.
Output
Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10 - 9.
Examples
Input
4
0.1 0.2 0.3 0.8
Output
0.800000000000
Input
2
0.1 0.2
Output
0.260000000000
Note
In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one.
In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8 + 0.9·0.2 = 0.26.
Tags: greedy, math, probabilities
Correct Solution:
```
n = input()
ans = tmp = 0.0
prod = 1.0
for p in reversed(sorted(map(float, input().split()))):
tmp = tmp * (1.0 - p) + prod * p
prod *= 1.0 - p
ans = max(ans, tmp)
print('%0.9f' % ans)
```
| 79,494 | [
0.476318359375,
0.04949951171875,
0.08782958984375,
0.0408935546875,
-0.5498046875,
-0.890625,
0.10107421875,
0.14697265625,
-0.07525634765625,
0.8681640625,
0.490234375,
-0.6533203125,
0.45556640625,
-0.4326171875,
-0.61572265625,
-0.10809326171875,
-0.57958984375,
-0.80810546875,... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrey needs one more problem to conduct a programming contest. He has n friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him.
Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of Andrey's friends. The second line contains n real numbers pi (0.0 ≤ pi ≤ 1.0) — the probability that the i-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point.
Output
Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10 - 9.
Examples
Input
4
0.1 0.2 0.3 0.8
Output
0.800000000000
Input
2
0.1 0.2
Output
0.260000000000
Note
In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one.
In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8 + 0.9·0.2 = 0.26.
Tags: greedy, math, probabilities
Correct Solution:
```
#!/usr/bin/python3
import sys
n = int(sys.stdin.readline())
p = [float(x) for x in sys.stdin.readline().split()]
p.sort()
d = [(1-x, x, 0) for x in p]
m = max(p)
def konv(a, b):
return (a[0]*b[0], a[0]*b[1] + a[1]*b[0], 0)
while len(d) > 1:
a = d.pop()
b = d.pop()
c = konv(a, b)
m = max(m, c[1])
d.append(c)
print(m)
```
| 79,495 | [
0.434326171875,
0.05450439453125,
0.1290283203125,
0.10394287109375,
-0.64990234375,
-0.86083984375,
0.1065673828125,
0.12841796875,
-0.0237884521484375,
0.8271484375,
0.467041015625,
-0.68896484375,
0.3642578125,
-0.364990234375,
-0.611328125,
-0.09442138671875,
-0.62109375,
-0.79... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrey needs one more problem to conduct a programming contest. He has n friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him.
Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of Andrey's friends. The second line contains n real numbers pi (0.0 ≤ pi ≤ 1.0) — the probability that the i-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point.
Output
Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10 - 9.
Examples
Input
4
0.1 0.2 0.3 0.8
Output
0.800000000000
Input
2
0.1 0.2
Output
0.260000000000
Note
In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one.
In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8 + 0.9·0.2 = 0.26.
Tags: greedy, math, probabilities
Correct Solution:
```
n = input()
ans = tmp = 0.0
pd = 1.0
for i in reversed(sorted(map(float, input().split()))):
tmp = tmp * (1.0 - i) + pd * i
pd *= 1.0 - i
ans = max(ans, tmp)
print('%0.12f' % ans)
```
| 79,496 | [
0.43310546875,
0.02294921875,
0.10699462890625,
0.035552978515625,
-0.56982421875,
-0.82763671875,
0.10693359375,
0.14990234375,
-0.035797119140625,
0.90869140625,
0.488037109375,
-0.67236328125,
0.43017578125,
-0.43212890625,
-0.59765625,
-0.1185302734375,
-0.59130859375,
-0.75537... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrey needs one more problem to conduct a programming contest. He has n friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him.
Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of Andrey's friends. The second line contains n real numbers pi (0.0 ≤ pi ≤ 1.0) — the probability that the i-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point.
Output
Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10 - 9.
Examples
Input
4
0.1 0.2 0.3 0.8
Output
0.800000000000
Input
2
0.1 0.2
Output
0.260000000000
Note
In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one.
In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8 + 0.9·0.2 = 0.26.
Submitted Solution:
```
n = int(input())
a = list(reversed(sorted(list(map(float, input().split())))))
A = a[0]
B = 1 - a[0]
for i in range(1, n):
if (1 - a[i]) * A + a[i] * B > A:
A = (1 - a[i]) * A + a[i] * B
B *= (1 - a[i])
print(A)
```
Yes
| 79,497 | [
0.470458984375,
0.041290283203125,
-0.047882080078125,
-0.0711669921875,
-0.6337890625,
-0.64013671875,
-0.02044677734375,
0.27490234375,
-0.19091796875,
0.93408203125,
0.448486328125,
-0.55322265625,
0.430908203125,
-0.5634765625,
-0.54150390625,
-0.1888427734375,
-0.67822265625,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrey needs one more problem to conduct a programming contest. He has n friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him.
Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of Andrey's friends. The second line contains n real numbers pi (0.0 ≤ pi ≤ 1.0) — the probability that the i-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point.
Output
Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10 - 9.
Examples
Input
4
0.1 0.2 0.3 0.8
Output
0.800000000000
Input
2
0.1 0.2
Output
0.260000000000
Note
In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one.
In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8 + 0.9·0.2 = 0.26.
Submitted Solution:
```
# -*- coding: utf-8 -*-
import sys
f = sys.stdin
n = int(f.readline().strip())
p = [float(u) for u in f.readline().strip().split()]
p.sort(reverse=True)
#p.sort()
p0 = 1
p1 = 0
p2 = 0
for pi in p:
p0t, p1t = p0*(1-pi), p0*pi+p1*(1-pi)
#print(p1t)
if p1t > p1 :
p0, p1 = p0t, p1t
else:
break
#print(p, p0, p1, p1t)
print(p1)
```
Yes
| 79,498 | [
0.46337890625,
-0.00995635986328125,
-0.024017333984375,
-0.10736083984375,
-0.6865234375,
-0.63916015625,
-0.019989013671875,
0.25146484375,
-0.119140625,
0.91162109375,
0.468017578125,
-0.55322265625,
0.37255859375,
-0.48876953125,
-0.5390625,
-0.16845703125,
-0.685546875,
-0.675... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrey needs one more problem to conduct a programming contest. He has n friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him.
Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of Andrey's friends. The second line contains n real numbers pi (0.0 ≤ pi ≤ 1.0) — the probability that the i-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point.
Output
Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10 - 9.
Examples
Input
4
0.1 0.2 0.3 0.8
Output
0.800000000000
Input
2
0.1 0.2
Output
0.260000000000
Note
In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one.
In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8 + 0.9·0.2 = 0.26.
Submitted Solution:
```
n=input()
p=[float(x) for x in input().split()]
p.sort(reverse=True)
maxans=0
for i in range(len(p)):
ans=0
for j in range(i+1):
temp=p[j]
for k in range(i+1):
if k!=j:temp*=(1-p[k])
ans+=temp
if ans>maxans : maxans=ans
print(maxans)
```
Yes
| 79,499 | [
0.52783203125,
0.1275634765625,
0.03778076171875,
-0.049041748046875,
-0.734375,
-0.701171875,
-0.052337646484375,
0.25732421875,
-0.1539306640625,
0.94873046875,
0.4736328125,
-0.456298828125,
0.325927734375,
-0.6328125,
-0.576171875,
-0.0762939453125,
-0.68310546875,
-0.700683593... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrey needs one more problem to conduct a programming contest. He has n friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him.
Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of Andrey's friends. The second line contains n real numbers pi (0.0 ≤ pi ≤ 1.0) — the probability that the i-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point.
Output
Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10 - 9.
Examples
Input
4
0.1 0.2 0.3 0.8
Output
0.800000000000
Input
2
0.1 0.2
Output
0.260000000000
Note
In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one.
In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8 + 0.9·0.2 = 0.26.
Submitted Solution:
```
from functools import reduce
L0 = []
pMax = 0
def getP():
return sum([reduce(lambda y, x: y*x, [(L0[j] if j == i else (1 - L0[j])) for j in range(len(L0))], 1) for i in range(len(L0))])
input()
L = [float(x) for x in input().split()]
#inp = open('input.txt')
#inp.readline()
#L = [float(x) for x in inp.readline().split()]
#inp.close()
L.reverse()
def go(stInd):
global pMax
for i in range(stInd, len(L)):
L0.append(L[i])
print(L0)
p = getP()
if p > pMax:
pMax = p
go(stInd + 1)
L0.pop(-1)
go(0)
print(pMax)
```
No
| 79,500 | [
0.51220703125,
0.055908203125,
-0.021514892578125,
-0.006038665771484375,
-0.62060546875,
-0.6533203125,
-0.0171356201171875,
0.281005859375,
-0.1661376953125,
0.888671875,
0.39453125,
-0.57568359375,
0.373046875,
-0.55419921875,
-0.53076171875,
-0.159423828125,
-0.66259765625,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrey needs one more problem to conduct a programming contest. He has n friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him.
Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of Andrey's friends. The second line contains n real numbers pi (0.0 ≤ pi ≤ 1.0) — the probability that the i-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point.
Output
Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10 - 9.
Examples
Input
4
0.1 0.2 0.3 0.8
Output
0.800000000000
Input
2
0.1 0.2
Output
0.260000000000
Note
In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one.
In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8 + 0.9·0.2 = 0.26.
Submitted Solution:
```
n=input()
p=[float(x) for x in input().split()]
p.sort(reverse=True)
now=0
for x in p:
if now<now+x-2*now*x:
now=now+x-2*now*x
print(now)
```
No
| 79,501 | [
0.5224609375,
0.0655517578125,
-0.05242919921875,
-0.05999755859375,
-0.62841796875,
-0.6298828125,
0.00504302978515625,
0.31494140625,
-0.15087890625,
0.9462890625,
0.44189453125,
-0.54248046875,
0.393798828125,
-0.53857421875,
-0.52490234375,
-0.14501953125,
-0.66552734375,
-0.70... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrey needs one more problem to conduct a programming contest. He has n friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him.
Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of Andrey's friends. The second line contains n real numbers pi (0.0 ≤ pi ≤ 1.0) — the probability that the i-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point.
Output
Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10 - 9.
Examples
Input
4
0.1 0.2 0.3 0.8
Output
0.800000000000
Input
2
0.1 0.2
Output
0.260000000000
Note
In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one.
In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8 + 0.9·0.2 = 0.26.
Submitted Solution:
```
from functools import reduce
L0 = []
pMax = 0
def getP():
return sum([reduce(lambda y, x: y*x, [(L0[j] if j == i else (1 - L0[j])) for j in range(len(L0))], 1) for i in range(len(L0))])
input()
L = [float(x) for x in input().split()]
def go(stInd):
global pMax
for i in range(stInd, len(L)):
L0.append(L[i])
p = getP()
if p > pMax:
pMax = p
go(stInd + 1)
L0.pop(0)
go(0)
print(pMax)
```
No
| 79,502 | [
0.4970703125,
0.08392333984375,
-0.04351806640625,
-0.00678253173828125,
-0.62890625,
-0.65576171875,
-0.02947998046875,
0.283935546875,
-0.15087890625,
0.87109375,
0.39892578125,
-0.58203125,
0.359619140625,
-0.56982421875,
-0.50634765625,
-0.1612548828125,
-0.6806640625,
-0.69140... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrey needs one more problem to conduct a programming contest. He has n friends who are always willing to help. He can ask some of them to come up with a contest problem. Andrey knows one value for each of his fiends — the probability that this friend will come up with a problem if Andrey asks him.
Help Andrey choose people to ask. As he needs only one problem, Andrey is going to be really upset if no one comes up with a problem or if he gets more than one problem from his friends. You need to choose such a set of people that maximizes the chances of Andrey not getting upset.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of Andrey's friends. The second line contains n real numbers pi (0.0 ≤ pi ≤ 1.0) — the probability that the i-th friend can come up with a problem. The probabilities are given with at most 6 digits after decimal point.
Output
Print a single real number — the probability that Andrey won't get upset at the optimal choice of friends. The answer will be considered valid if it differs from the correct one by at most 10 - 9.
Examples
Input
4
0.1 0.2 0.3 0.8
Output
0.800000000000
Input
2
0.1 0.2
Output
0.260000000000
Note
In the first sample the best strategy for Andrey is to ask only one of his friends, the most reliable one.
In the second sample the best strategy for Andrey is to ask all of his friends to come up with a problem. Then the probability that he will get exactly one problem is 0.1·0.8 + 0.9·0.2 = 0.26.
Submitted Solution:
```
from functools import reduce
def getP(L, ind):
return sum([reduce(lambda y, x: y*x, [(L[j] if j == i else (1 - L[j])) for j in range(len(L)) if j != ind], 1) for i in range(len(L)) if i != ind])
input()
L = [float(x) for x in input().split()]
p0 = getP(L, -1)
stp = False
while not stp:
for ind in range(len(L)):
p1 = getP(L, ind)
if p0 < p1:
L.pop(ind)
p0 = p1
break
else:
print(p0)
stp = True
```
No
| 79,503 | [
0.458251953125,
0.05853271484375,
-0.0521240234375,
-0.0139312744140625,
-0.65673828125,
-0.65771484375,
0.00644683837890625,
0.25244140625,
-0.190673828125,
0.8779296875,
0.414306640625,
-0.60693359375,
0.3544921875,
-0.533203125,
-0.546875,
-0.1646728515625,
-0.66943359375,
-0.68... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Because of budget cuts one IT company established new non-financial reward system instead of bonuses.
Two kinds of actions are rewarded: fixing critical bugs and suggesting new interesting features. A man who fixed a critical bug gets "I fixed a critical bug" pennant on his table. A man who suggested a new interesting feature gets "I suggested a new feature" pennant on his table.
Because of the limited budget of the new reward system only 5 "I fixed a critical bug" pennants and 3 "I suggested a new feature" pennants were bought.
In order to use these pennants for a long time they were made challenge ones. When a man fixes a new critical bug one of the earlier awarded "I fixed a critical bug" pennants is passed on to his table. When a man suggests a new interesting feature one of the earlier awarded "I suggested a new feature" pennants is passed on to his table.
One man can have several pennants of one type and of course he can have pennants of both types on his table. There are n tables in the IT company. Find the number of ways to place the pennants on these tables given that each pennant is situated on one of the tables and each table is big enough to contain any number of pennants.
Input
The only line of the input contains one integer n (1 ≤ n ≤ 500) — the number of tables in the IT company.
Output
Output one integer — the amount of ways to place the pennants on n tables.
Examples
Input
2
Output
24
Submitted Solution:
```
from math import factorial as f
n = int(input())
def b(n, k):
return f(n)//f(k)//f(n-k)
s5 = n
if (n > 1):
s5 += n*(n-1)*2
if (n > 2):
s5 += n*(n-1)*(n-2)//2*2
if (n > 3):
s5 += n*(n-1)*(n-2)*(n-3)//f(3)
if (n > 4):
s5 += n*(n-1)*(n-2)*(n-3)*(n-4)//f(5)
s3 = n
if (n > 1):
s3 += n*(n-1)
if (n > 2):
s3 += n*(n-1)*(n-2)//6
#print(s5, s3)
print(s5*s3)
```
Yes
| 79,571 | [
0.4296875,
-0.1795654296875,
-0.11761474609375,
0.1531982421875,
-0.595703125,
-0.343994140625,
-0.06719970703125,
0.06549072265625,
-0.121826171875,
0.7265625,
0.61083984375,
0.10430908203125,
0.13525390625,
-0.434326171875,
-0.42041015625,
-0.291015625,
-0.50537109375,
-1.0087890... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Because of budget cuts one IT company established new non-financial reward system instead of bonuses.
Two kinds of actions are rewarded: fixing critical bugs and suggesting new interesting features. A man who fixed a critical bug gets "I fixed a critical bug" pennant on his table. A man who suggested a new interesting feature gets "I suggested a new feature" pennant on his table.
Because of the limited budget of the new reward system only 5 "I fixed a critical bug" pennants and 3 "I suggested a new feature" pennants were bought.
In order to use these pennants for a long time they were made challenge ones. When a man fixes a new critical bug one of the earlier awarded "I fixed a critical bug" pennants is passed on to his table. When a man suggests a new interesting feature one of the earlier awarded "I suggested a new feature" pennants is passed on to his table.
One man can have several pennants of one type and of course he can have pennants of both types on his table. There are n tables in the IT company. Find the number of ways to place the pennants on these tables given that each pennant is situated on one of the tables and each table is big enough to contain any number of pennants.
Input
The only line of the input contains one integer n (1 ≤ n ≤ 500) — the number of tables in the IT company.
Output
Output one integer — the amount of ways to place the pennants on n tables.
Examples
Input
2
Output
24
Submitted Solution:
```
n=int(input())
print((n*(n+1)*(n+2)*(n+3)*(n+4)*n*(n+1)*(n+2))//720)
```
Yes
| 79,572 | [
0.44921875,
-0.29150390625,
-0.1668701171875,
0.09814453125,
-0.62255859375,
-0.372802734375,
-0.1165771484375,
0.138916015625,
-0.087890625,
0.71826171875,
0.50341796875,
0.18701171875,
0.1944580078125,
-0.36083984375,
-0.541015625,
-0.27685546875,
-0.43603515625,
-1.029296875,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Because of budget cuts one IT company established new non-financial reward system instead of bonuses.
Two kinds of actions are rewarded: fixing critical bugs and suggesting new interesting features. A man who fixed a critical bug gets "I fixed a critical bug" pennant on his table. A man who suggested a new interesting feature gets "I suggested a new feature" pennant on his table.
Because of the limited budget of the new reward system only 5 "I fixed a critical bug" pennants and 3 "I suggested a new feature" pennants were bought.
In order to use these pennants for a long time they were made challenge ones. When a man fixes a new critical bug one of the earlier awarded "I fixed a critical bug" pennants is passed on to his table. When a man suggests a new interesting feature one of the earlier awarded "I suggested a new feature" pennants is passed on to his table.
One man can have several pennants of one type and of course he can have pennants of both types on his table. There are n tables in the IT company. Find the number of ways to place the pennants on these tables given that each pennant is situated on one of the tables and each table is big enough to contain any number of pennants.
Input
The only line of the input contains one integer n (1 ≤ n ≤ 500) — the number of tables in the IT company.
Output
Output one integer — the amount of ways to place the pennants on n tables.
Examples
Input
2
Output
24
Submitted Solution:
```
import math as ma
def fu(n,r):
return (ma.factorial(n)//(ma.factorial(n-r)*ma.factorial(r)))
n=int(input())
s=0
t=0
for i in range(4,5):
s+=fu(n+i,i+1)
for i in range(2,3):
t+=fu(n+i,i+1)
print(s*t)
```
Yes
| 79,573 | [
0.412353515625,
-0.321533203125,
-0.109619140625,
0.140380859375,
-0.60107421875,
-0.414306640625,
-0.0205841064453125,
0.1219482421875,
-0.164306640625,
0.689453125,
0.59423828125,
0.237548828125,
0.242919921875,
-0.41552734375,
-0.45361328125,
-0.1959228515625,
-0.4052734375,
-1.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Because of budget cuts one IT company established new non-financial reward system instead of bonuses.
Two kinds of actions are rewarded: fixing critical bugs and suggesting new interesting features. A man who fixed a critical bug gets "I fixed a critical bug" pennant on his table. A man who suggested a new interesting feature gets "I suggested a new feature" pennant on his table.
Because of the limited budget of the new reward system only 5 "I fixed a critical bug" pennants and 3 "I suggested a new feature" pennants were bought.
In order to use these pennants for a long time they were made challenge ones. When a man fixes a new critical bug one of the earlier awarded "I fixed a critical bug" pennants is passed on to his table. When a man suggests a new interesting feature one of the earlier awarded "I suggested a new feature" pennants is passed on to his table.
One man can have several pennants of one type and of course he can have pennants of both types on his table. There are n tables in the IT company. Find the number of ways to place the pennants on these tables given that each pennant is situated on one of the tables and each table is big enough to contain any number of pennants.
Input
The only line of the input contains one integer n (1 ≤ n ≤ 500) — the number of tables in the IT company.
Output
Output one integer — the amount of ways to place the pennants on n tables.
Examples
Input
2
Output
24
Submitted Solution:
```
import operator as op
from functools import reduce
def ncr(n, r):
r = min(r, n-r)
numer = reduce(op.mul, range(n, n-r, -1), 1)
denom = reduce(op.mul, range(1, r+1), 1)
return numer // denom
n = int(input())
print(ncr(3+n-1,n-1)*ncr(5+n-1,n-1))
```
Yes
| 79,574 | [
0.3779296875,
-0.37744140625,
-0.1336669921875,
0.07733154296875,
-0.47900390625,
-0.45751953125,
-0.08831787109375,
0.1064453125,
-0.206298828125,
0.73681640625,
0.50439453125,
0.136474609375,
0.031463623046875,
-0.30859375,
-0.49072265625,
-0.1790771484375,
-0.407470703125,
-1.07... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Because of budget cuts one IT company established new non-financial reward system instead of bonuses.
Two kinds of actions are rewarded: fixing critical bugs and suggesting new interesting features. A man who fixed a critical bug gets "I fixed a critical bug" pennant on his table. A man who suggested a new interesting feature gets "I suggested a new feature" pennant on his table.
Because of the limited budget of the new reward system only 5 "I fixed a critical bug" pennants and 3 "I suggested a new feature" pennants were bought.
In order to use these pennants for a long time they were made challenge ones. When a man fixes a new critical bug one of the earlier awarded "I fixed a critical bug" pennants is passed on to his table. When a man suggests a new interesting feature one of the earlier awarded "I suggested a new feature" pennants is passed on to his table.
One man can have several pennants of one type and of course he can have pennants of both types on his table. There are n tables in the IT company. Find the number of ways to place the pennants on these tables given that each pennant is situated on one of the tables and each table is big enough to contain any number of pennants.
Input
The only line of the input contains one integer n (1 ≤ n ≤ 500) — the number of tables in the IT company.
Output
Output one integer — the amount of ways to place the pennants on n tables.
Examples
Input
2
Output
24
Submitted Solution:
```
from math import factorial
f = factorial
k = int(input())
n1 = 5
n2 = 3
ans1 = f(k + n1 - 1) / f(k - 1) / f(n1)
ans2 = f(k + n2 - 1) / f(k - 1) / f(n2)
print(int(ans1 * ans2))
# n!/(n − m)!/m!
```
No
| 79,575 | [
0.421875,
-0.294189453125,
-0.15087890625,
0.0811767578125,
-0.59228515625,
-0.36181640625,
-0.0606689453125,
0.12335205078125,
-0.1485595703125,
0.7158203125,
0.568359375,
0.15283203125,
0.15185546875,
-0.42822265625,
-0.49853515625,
-0.2044677734375,
-0.448974609375,
-1.0859375,
... | 11 |
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