message stringlengths 2 39.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 219 108k | cluster float64 11 11 | __index_level_0__ int64 438 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yura has a team of k developers and a list of n tasks numbered from 1 to n. Yura is going to choose some tasks to be done this week. Due to strange Looksery habits the numbers of chosen tasks should be a segment of consecutive integers containing no less than 2 numbers, i. e. a sequence of form l, l + 1, ..., r for some 1 ≤ l < r ≤ n.
Every task i has an integer number ai associated with it denoting how many man-hours are required to complete the i-th task. Developers are not self-confident at all, and they are actually afraid of difficult tasks. Knowing that, Yura decided to pick up a hardest task (the one that takes the biggest number of man-hours to be completed, among several hardest tasks with same difficulty level he chooses arbitrary one) and complete it on his own. So, if tasks with numbers [l, r] are chosen then the developers are left with r - l tasks to be done by themselves.
Every developer can spend any integer amount of hours over any task, but when they are done with the whole assignment there should be exactly ai man-hours spent over the i-th task.
The last, but not the least problem with developers is that one gets angry if he works more than another developer. A set of tasks [l, r] is considered good if it is possible to find such a distribution of work that allows to complete all the tasks and to have every developer working for the same amount of time (amount of work performed by Yura doesn't matter for other workers as well as for him).
For example, let's suppose that Yura have chosen tasks with following difficulties: a = [1, 2, 3, 4], and he has three developers in his disposal. He takes the hardest fourth task to finish by himself, and the developers are left with tasks with difficulties [1, 2, 3]. If the first one spends an hour on the first task and an hour on the third one, the second developer spends two hours on the second task and the third developer spends two hours on the third task, then they are done, since every developer worked exactly for two hours and every task has been worked over for the required amount of time. As another example, if the first task required two hours instead of one to be completed then it would be impossible to assign the tasks in a way described above.
Besides work, Yura is fond of problem solving. He wonders how many pairs (l, r) (1 ≤ l < r ≤ n) exists such that a segment [l, r] is good? Yura has already solved this problem, but he has no time to write the code. Please, help Yura and implement the solution for this problem.
Input
The first line of input contains two positive integers: n and k (1 ≤ n ≤ 300 000, 1 ≤ k ≤ 1 000 000), the number of tasks in the list and the number of developers in Yura's disposal.
The second line contains n integers ai (1 ≤ ai ≤ 109).
Output
Output a single integer — the number of pairs (l, r) satisfying the conditions from the statement.
Examples
Input
4 3
1 2 3 4
Output
3
Input
4 2
4 4 7 4
Output
6
Note
In the first sample there are three good segments:
1. [1;3] — the hardest task requires 3 man-hours, so there are tasks left that require 1 and 2 man-hours. A solution is to make first developer work on the first task for an hour, while second and third developers work on the second task. Each developer works exactly one hour.
2. [1;4] — the hardest task requires 4 man-hours, so there are tasks left that require 1, 2 and 3 man-hours. If the first developer spends an hour on the first task and an hour on the third one, the second developer spends two hours on the second task and the third developer spends two hours on the third task, then they are done, since every developer worked exactly for two hours.
3. [3;4] — the hardest task requires 4 man-hours, so there is only one task left that requires 3 man-hours. A solution is to make each developer work for an hour.
Submitted Solution:
```
class Node:
def __init__(self, value):
self.max = value
self.sum = value
def add_leaf(left, right, node, leaf, value):
if left == right:
tree[node] = Node(value)
return
m = (left + right) // 2
if leaf <= m :
add_leaf(left, m, node * 2, leaf, value)
if tree[node] == None:
tree[node] = Node(value)
else:
tree[node].sum += value
if tree[node].max < tree[node * 2].max:
tree[node].max = tree[node * 2].max
else:
add_leaf(m + 1, right, node * 2 + 1, leaf, value)
if tree[node] == None:
tree[node] = Node(value)
else:
tree[node].sum += value
if tree[node].max < tree[node * 2 + 1].max:
tree[node].max = tree[node * 2 + 1].max
def query(left, right, qleft, qright, node):
aux = [0,0]
auxl = [0,0]
auxr = [0,0]
if qleft <= left and right <= qright:
aux = [tree[node].sum, tree[node].max]
return aux
m = (left + right) // 2
if qleft <= m and m <= qright:
auxl = query(left, m, qleft, qright, node * 2)
if qright >= m+1 and m + 1 >= qleft:
auxr = query(m + 1, right, qleft, qright, node * 2 + 1)
aux[0] = auxl[0] + auxr[0]
aux[1] = max(auxl[1],auxr[1])
return aux
level = 1
line = input().split()
n = int(line[0])
k = int(line[1])
while (1 << level) < n:
level += 1
level += 1
nodes = (1 << level) - 1
tree = [None] * (nodes + 1)
line = input().split()
for i in range(n):
add_leaf(1, n, 1, i + 1, int(line[i]))
rez = 0
for i in range(n):
for j in range(i+1,n):
aux = query(1,n, i + 1, j + 1, 1)
if (aux[0] - aux[1]) % k == 0:
rez += 1
print(rez)
``` | instruction | 0 | 18,096 | 11 | 36,192 |
No | output | 1 | 18,096 | 11 | 36,193 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yura has a team of k developers and a list of n tasks numbered from 1 to n. Yura is going to choose some tasks to be done this week. Due to strange Looksery habits the numbers of chosen tasks should be a segment of consecutive integers containing no less than 2 numbers, i. e. a sequence of form l, l + 1, ..., r for some 1 ≤ l < r ≤ n.
Every task i has an integer number ai associated with it denoting how many man-hours are required to complete the i-th task. Developers are not self-confident at all, and they are actually afraid of difficult tasks. Knowing that, Yura decided to pick up a hardest task (the one that takes the biggest number of man-hours to be completed, among several hardest tasks with same difficulty level he chooses arbitrary one) and complete it on his own. So, if tasks with numbers [l, r] are chosen then the developers are left with r - l tasks to be done by themselves.
Every developer can spend any integer amount of hours over any task, but when they are done with the whole assignment there should be exactly ai man-hours spent over the i-th task.
The last, but not the least problem with developers is that one gets angry if he works more than another developer. A set of tasks [l, r] is considered good if it is possible to find such a distribution of work that allows to complete all the tasks and to have every developer working for the same amount of time (amount of work performed by Yura doesn't matter for other workers as well as for him).
For example, let's suppose that Yura have chosen tasks with following difficulties: a = [1, 2, 3, 4], and he has three developers in his disposal. He takes the hardest fourth task to finish by himself, and the developers are left with tasks with difficulties [1, 2, 3]. If the first one spends an hour on the first task and an hour on the third one, the second developer spends two hours on the second task and the third developer spends two hours on the third task, then they are done, since every developer worked exactly for two hours and every task has been worked over for the required amount of time. As another example, if the first task required two hours instead of one to be completed then it would be impossible to assign the tasks in a way described above.
Besides work, Yura is fond of problem solving. He wonders how many pairs (l, r) (1 ≤ l < r ≤ n) exists such that a segment [l, r] is good? Yura has already solved this problem, but he has no time to write the code. Please, help Yura and implement the solution for this problem.
Input
The first line of input contains two positive integers: n and k (1 ≤ n ≤ 300 000, 1 ≤ k ≤ 1 000 000), the number of tasks in the list and the number of developers in Yura's disposal.
The second line contains n integers ai (1 ≤ ai ≤ 109).
Output
Output a single integer — the number of pairs (l, r) satisfying the conditions from the statement.
Examples
Input
4 3
1 2 3 4
Output
3
Input
4 2
4 4 7 4
Output
6
Note
In the first sample there are three good segments:
1. [1;3] — the hardest task requires 3 man-hours, so there are tasks left that require 1 and 2 man-hours. A solution is to make first developer work on the first task for an hour, while second and third developers work on the second task. Each developer works exactly one hour.
2. [1;4] — the hardest task requires 4 man-hours, so there are tasks left that require 1, 2 and 3 man-hours. If the first developer spends an hour on the first task and an hour on the third one, the second developer spends two hours on the second task and the third developer spends two hours on the third task, then they are done, since every developer worked exactly for two hours.
3. [3;4] — the hardest task requires 4 man-hours, so there is only one task left that requires 3 man-hours. A solution is to make each developer work for an hour.
Submitted Solution:
```
import sys
sys.setrecursionlimit(100000000)
s=input().split()
n=int(s[0])
k=int(s[1])
s=input().split()
t=[]
v=0
for i in range(n):
t+=[int(s[i])]
s=[0,t[0]]
for i in range(1,n):
s+=[s[i]+t[i]]
m=[(0,t[0])]
for i in range(1,n):
(a,b)=m[i-1]
if t[i]>b:
m+=[(i,t[i])]
else:
m+=[(a,b)]
def vMax(l,r):
(a,b)=m[l]
(c,d)=m[r]
if d>b:return(c,d)
else:
x=(l+r)//2
v=t[x]
j=x
for i in range(x+1,r+1):
if t[i]>v:
v=t[i]
j=i
for i in range(l+1,x):
if t[i]>v:
v=t[i]
j=i
return(j,v)
def nb(l,r):
if r-l<1:return(0)
elif r-l>20:
j,v=vMax(l,r)
x=0
z=[0 for i in range(k)]
for i in range(l,j+1):
z[s[i]%k]+=1
for i in range(j,r+1):
x+=z[(s[i+1]-v)%k]
return(x+nb(l,j-1)+nb(j+1,r)-1)
else:
j,v=vMax(l,r)
x=0
z=[]
for i in range(l,j+1):
z+=[s[i]%k]
for i in range(j,r+1):
for y in z:
if y==((s[i+1]-v)%k):x+=1
return(x+nb(l,j-1)+nb(j+1,r)-1)
print(nb(0,n-1))
``` | instruction | 0 | 18,097 | 11 | 36,194 |
No | output | 1 | 18,097 | 11 | 36,195 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N people living on a number line.
The i-th person lives at coordinate X_i.
You are going to hold a meeting that all N people have to attend.
The meeting can be held at any integer coordinate. If you choose to hold the meeting at coordinate P, the i-th person will spend (X_i - P)^2 points of stamina to attend the meeting.
Find the minimum total points of stamina the N people have to spend.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq X_i \leq 100
Input
Input is given from Standard Input in the following format:
N
X_1 X_2 ... X_N
Output
Print the minimum total stamina the N people have to spend.
Examples
Input
2
1 4
Output
5
Input
7
14 14 2 13 56 2 37
Output
2354
Submitted Solution:
```
n=int(input())
x=list(map(int,input().split()))
avr=round(sum(x)/n)
sum=0
for i in range(n):
sum+=(x[i]-avr)**2
print(sum)
``` | instruction | 0 | 18,315 | 11 | 36,630 |
Yes | output | 1 | 18,315 | 11 | 36,631 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N people living on a number line.
The i-th person lives at coordinate X_i.
You are going to hold a meeting that all N people have to attend.
The meeting can be held at any integer coordinate. If you choose to hold the meeting at coordinate P, the i-th person will spend (X_i - P)^2 points of stamina to attend the meeting.
Find the minimum total points of stamina the N people have to spend.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq X_i \leq 100
Input
Input is given from Standard Input in the following format:
N
X_1 X_2 ... X_N
Output
Print the minimum total stamina the N people have to spend.
Examples
Input
2
1 4
Output
5
Input
7
14 14 2 13 56 2 37
Output
2354
Submitted Solution:
```
n = int(input())
xi = list(map(int, input().split()))
p = round(sum(xi) / n)
hp = 0
for x in xi:
hp += (x - p) ** 2
print(hp)
``` | instruction | 0 | 18,316 | 11 | 36,632 |
Yes | output | 1 | 18,316 | 11 | 36,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N people living on a number line.
The i-th person lives at coordinate X_i.
You are going to hold a meeting that all N people have to attend.
The meeting can be held at any integer coordinate. If you choose to hold the meeting at coordinate P, the i-th person will spend (X_i - P)^2 points of stamina to attend the meeting.
Find the minimum total points of stamina the N people have to spend.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq X_i \leq 100
Input
Input is given from Standard Input in the following format:
N
X_1 X_2 ... X_N
Output
Print the minimum total stamina the N people have to spend.
Examples
Input
2
1 4
Output
5
Input
7
14 14 2 13 56 2 37
Output
2354
Submitted Solution:
```
N = int(input())
A = list(map(int, input().split()))
ave = round(sum(A)/len(A))
print(sum([(i-ave)**2 for i in A]))
``` | instruction | 0 | 18,317 | 11 | 36,634 |
Yes | output | 1 | 18,317 | 11 | 36,635 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N people living on a number line.
The i-th person lives at coordinate X_i.
You are going to hold a meeting that all N people have to attend.
The meeting can be held at any integer coordinate. If you choose to hold the meeting at coordinate P, the i-th person will spend (X_i - P)^2 points of stamina to attend the meeting.
Find the minimum total points of stamina the N people have to spend.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq X_i \leq 100
Input
Input is given from Standard Input in the following format:
N
X_1 X_2 ... X_N
Output
Print the minimum total stamina the N people have to spend.
Examples
Input
2
1 4
Output
5
Input
7
14 14 2 13 56 2 37
Output
2354
Submitted Solution:
```
n = int(input())
x = list(map(int,input().split()))
ave = round(sum(x)/n)
print(sum([(i-ave)**2 for i in x]))
``` | instruction | 0 | 18,318 | 11 | 36,636 |
Yes | output | 1 | 18,318 | 11 | 36,637 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N people living on a number line.
The i-th person lives at coordinate X_i.
You are going to hold a meeting that all N people have to attend.
The meeting can be held at any integer coordinate. If you choose to hold the meeting at coordinate P, the i-th person will spend (X_i - P)^2 points of stamina to attend the meeting.
Find the minimum total points of stamina the N people have to spend.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq X_i \leq 100
Input
Input is given from Standard Input in the following format:
N
X_1 X_2 ... X_N
Output
Print the minimum total stamina the N people have to spend.
Examples
Input
2
1 4
Output
5
Input
7
14 14 2 13 56 2 37
Output
2354
Submitted Solution:
```
def resolve():
import statistics
n = input()
x = list(map(int, input().split()))
m = round(statistics.mean(x))
sum_val = 0
for val in x:
sum_val += (val - m) ** 2
print(sum_val)
if __name__ == '__main__':
resolve()
``` | instruction | 0 | 18,319 | 11 | 36,638 |
No | output | 1 | 18,319 | 11 | 36,639 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N people living on a number line.
The i-th person lives at coordinate X_i.
You are going to hold a meeting that all N people have to attend.
The meeting can be held at any integer coordinate. If you choose to hold the meeting at coordinate P, the i-th person will spend (X_i - P)^2 points of stamina to attend the meeting.
Find the minimum total points of stamina the N people have to spend.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq X_i \leq 100
Input
Input is given from Standard Input in the following format:
N
X_1 X_2 ... X_N
Output
Print the minimum total stamina the N people have to spend.
Examples
Input
2
1 4
Output
5
Input
7
14 14 2 13 56 2 37
Output
2354
Submitted Solution:
```
import sys
input = sys.stdin.readline
N = int(input())
X = [int(x) for x in input().split()]
P = int(sum(X) / N) + 1
print(sum(X))
print(sum(X) / N)
total = 0
for x in X:
kyori = (x - P) * (x - P)
total += kyori
print(total)
``` | instruction | 0 | 18,320 | 11 | 36,640 |
No | output | 1 | 18,320 | 11 | 36,641 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N people living on a number line.
The i-th person lives at coordinate X_i.
You are going to hold a meeting that all N people have to attend.
The meeting can be held at any integer coordinate. If you choose to hold the meeting at coordinate P, the i-th person will spend (X_i - P)^2 points of stamina to attend the meeting.
Find the minimum total points of stamina the N people have to spend.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq X_i \leq 100
Input
Input is given from Standard Input in the following format:
N
X_1 X_2 ... X_N
Output
Print the minimum total stamina the N people have to spend.
Examples
Input
2
1 4
Output
5
Input
7
14 14 2 13 56 2 37
Output
2354
Submitted Solution:
```
N=int(input())
lst=list(map(int,input().split()))
P=round(lst/N)
print(sum[(i-P)**2 for i in lst])
``` | instruction | 0 | 18,321 | 11 | 36,642 |
No | output | 1 | 18,321 | 11 | 36,643 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N people living on a number line.
The i-th person lives at coordinate X_i.
You are going to hold a meeting that all N people have to attend.
The meeting can be held at any integer coordinate. If you choose to hold the meeting at coordinate P, the i-th person will spend (X_i - P)^2 points of stamina to attend the meeting.
Find the minimum total points of stamina the N people have to spend.
Constraints
* All values in input are integers.
* 1 \leq N \leq 100
* 1 \leq X_i \leq 100
Input
Input is given from Standard Input in the following format:
N
X_1 X_2 ... X_N
Output
Print the minimum total stamina the N people have to spend.
Examples
Input
2
1 4
Output
5
Input
7
14 14 2 13 56 2 37
Output
2354
Submitted Solution:
```
N = int(input())
X = list(map(int, input().split()))
ans = 10e9
for i in range(1,max(X)):
tmp = 0
for j in range(N):
tmp += (X[j]-i)**2
if tmp < ans:
ans = tmp
# print(tmp)
print(ans)
``` | instruction | 0 | 18,322 | 11 | 36,644 |
No | output | 1 | 18,322 | 11 | 36,645 |
Provide a correct Python 3 solution for this coding contest problem.
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.
Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.
* A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.
* An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.
A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these.
Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
Constraints
* 1≦N,A,B≦100000
* A+B≦N
* S is N characters long.
* S consists only of the letters `a`, `b` and `c`.
Input
Inputs are provided from Standard Input in the following form.
N A B
S
Output
Output N lines. On the i-th line, output `Yes` if the i-th participant passed the Qualification contests or `No` if that participant did not pass.
Examples
Input
10 2 3
abccabaabb
Output
Yes
Yes
No
No
Yes
Yes
Yes
No
No
No
Input
12 5 2
cabbabaacaba
Output
No
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
No
No
Input
5 2 2
ccccc
Output
No
No
No
No
No | instruction | 0 | 18,435 | 11 | 36,870 |
"Correct Solution:
```
N,A,B=map(int,input().split())
S=input()
an=0
bn=0
for i in list(S):
if i == "a" and an+bn<A+B :
print("Yes")
an+=1
elif i == "b" and an+bn<A+B and bn<B:
print("Yes")
bn+=1
else:
print("No")
``` | output | 1 | 18,435 | 11 | 36,871 |
Provide a correct Python 3 solution for this coding contest problem.
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.
Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.
* A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.
* An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.
A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these.
Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
Constraints
* 1≦N,A,B≦100000
* A+B≦N
* S is N characters long.
* S consists only of the letters `a`, `b` and `c`.
Input
Inputs are provided from Standard Input in the following form.
N A B
S
Output
Output N lines. On the i-th line, output `Yes` if the i-th participant passed the Qualification contests or `No` if that participant did not pass.
Examples
Input
10 2 3
abccabaabb
Output
Yes
Yes
No
No
Yes
Yes
Yes
No
No
No
Input
12 5 2
cabbabaacaba
Output
No
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
No
No
Input
5 2 2
ccccc
Output
No
No
No
No
No | instruction | 0 | 18,436 | 11 | 36,872 |
"Correct Solution:
```
N,A,B = map(int, input().split())
*S, = input()
count = 0
bi = 0
for s in S:
ans = 'No'
if count < A+B:
if s == 'a':
ans = 'Yes'
count += 1
if s == 'b' and bi < B:
ans = 'Yes'
count += 1
bi += 1
print(ans)
``` | output | 1 | 18,436 | 11 | 36,873 |
Provide a correct Python 3 solution for this coding contest problem.
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.
Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.
* A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.
* An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.
A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these.
Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
Constraints
* 1≦N,A,B≦100000
* A+B≦N
* S is N characters long.
* S consists only of the letters `a`, `b` and `c`.
Input
Inputs are provided from Standard Input in the following form.
N A B
S
Output
Output N lines. On the i-th line, output `Yes` if the i-th participant passed the Qualification contests or `No` if that participant did not pass.
Examples
Input
10 2 3
abccabaabb
Output
Yes
Yes
No
No
Yes
Yes
Yes
No
No
No
Input
12 5 2
cabbabaacaba
Output
No
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
No
No
Input
5 2 2
ccccc
Output
No
No
No
No
No | instruction | 0 | 18,437 | 11 | 36,874 |
"Correct Solution:
```
N,A,B = map(int,input().split())
S = input()
b = 0
q = 0
for s in S:
if (s == 'a' and q < A+B) or (s == 'b' and q < A+B and b < B):
print('Yes')
q += 1
else:
print('No')
if s == 'b':
b += 1
``` | output | 1 | 18,437 | 11 | 36,875 |
Provide a correct Python 3 solution for this coding contest problem.
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.
Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.
* A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.
* An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.
A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these.
Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
Constraints
* 1≦N,A,B≦100000
* A+B≦N
* S is N characters long.
* S consists only of the letters `a`, `b` and `c`.
Input
Inputs are provided from Standard Input in the following form.
N A B
S
Output
Output N lines. On the i-th line, output `Yes` if the i-th participant passed the Qualification contests or `No` if that participant did not pass.
Examples
Input
10 2 3
abccabaabb
Output
Yes
Yes
No
No
Yes
Yes
Yes
No
No
No
Input
12 5 2
cabbabaacaba
Output
No
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
No
No
Input
5 2 2
ccccc
Output
No
No
No
No
No | instruction | 0 | 18,438 | 11 | 36,876 |
"Correct Solution:
```
N,A,B=map(int,input().split())
S=' '+input()
AB=A+B
j=1
num=0
for i in range(1,N+1):
if S[i]=='b' and num<AB and j<=B:
print('Yes')
j+=1
num+=1
elif S[i]=='a' and num<AB:
print('Yes')
num+=1
else:
print('No')
``` | output | 1 | 18,438 | 11 | 36,877 |
Provide a correct Python 3 solution for this coding contest problem.
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.
Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.
* A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.
* An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.
A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these.
Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
Constraints
* 1≦N,A,B≦100000
* A+B≦N
* S is N characters long.
* S consists only of the letters `a`, `b` and `c`.
Input
Inputs are provided from Standard Input in the following form.
N A B
S
Output
Output N lines. On the i-th line, output `Yes` if the i-th participant passed the Qualification contests or `No` if that participant did not pass.
Examples
Input
10 2 3
abccabaabb
Output
Yes
Yes
No
No
Yes
Yes
Yes
No
No
No
Input
12 5 2
cabbabaacaba
Output
No
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
No
No
Input
5 2 2
ccccc
Output
No
No
No
No
No | instruction | 0 | 18,439 | 11 | 36,878 |
"Correct Solution:
```
N, A, B = map(int, input().split())
S = input()
a = b = 0
for s in S:
if s == 'a' and a+b < A+B:
print('Yes')
a += 1
elif s == 'b' and a+b < A+B and b < B:
print('Yes')
b += 1
else:
print('No')
``` | output | 1 | 18,439 | 11 | 36,879 |
Provide a correct Python 3 solution for this coding contest problem.
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.
Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.
* A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.
* An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.
A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these.
Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
Constraints
* 1≦N,A,B≦100000
* A+B≦N
* S is N characters long.
* S consists only of the letters `a`, `b` and `c`.
Input
Inputs are provided from Standard Input in the following form.
N A B
S
Output
Output N lines. On the i-th line, output `Yes` if the i-th participant passed the Qualification contests or `No` if that participant did not pass.
Examples
Input
10 2 3
abccabaabb
Output
Yes
Yes
No
No
Yes
Yes
Yes
No
No
No
Input
12 5 2
cabbabaacaba
Output
No
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
No
No
Input
5 2 2
ccccc
Output
No
No
No
No
No | instruction | 0 | 18,440 | 11 | 36,880 |
"Correct Solution:
```
n, a, b = map(int, input().split())
s = input()
ab = a + b
cnt = 0
fcnt = 1
for i in range(n):
if s[i] == 'a' and ab > cnt:
cnt += 1
print('Yes')
elif s[i] == 'b' and (ab > cnt and b >= fcnt):
cnt += 1
fcnt += 1
print('Yes')
else:
print('No')
``` | output | 1 | 18,440 | 11 | 36,881 |
Provide a correct Python 3 solution for this coding contest problem.
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.
Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.
* A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.
* An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.
A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these.
Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
Constraints
* 1≦N,A,B≦100000
* A+B≦N
* S is N characters long.
* S consists only of the letters `a`, `b` and `c`.
Input
Inputs are provided from Standard Input in the following form.
N A B
S
Output
Output N lines. On the i-th line, output `Yes` if the i-th participant passed the Qualification contests or `No` if that participant did not pass.
Examples
Input
10 2 3
abccabaabb
Output
Yes
Yes
No
No
Yes
Yes
Yes
No
No
No
Input
12 5 2
cabbabaacaba
Output
No
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
No
No
Input
5 2 2
ccccc
Output
No
No
No
No
No | instruction | 0 | 18,441 | 11 | 36,882 |
"Correct Solution:
```
n,a,b =map(int, input().split())
s =input()
ca = 0
cb = 0
for i in s:
if i=='a' and ca+cb<a+b:
ca+=1
print('Yes')
elif i=='b' and ca+cb<a+b and cb<b:
cb+=1
print('Yes')
else:
print('No')
``` | output | 1 | 18,441 | 11 | 36,883 |
Provide a correct Python 3 solution for this coding contest problem.
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.
Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.
* A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.
* An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.
A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these.
Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
Constraints
* 1≦N,A,B≦100000
* A+B≦N
* S is N characters long.
* S consists only of the letters `a`, `b` and `c`.
Input
Inputs are provided from Standard Input in the following form.
N A B
S
Output
Output N lines. On the i-th line, output `Yes` if the i-th participant passed the Qualification contests or `No` if that participant did not pass.
Examples
Input
10 2 3
abccabaabb
Output
Yes
Yes
No
No
Yes
Yes
Yes
No
No
No
Input
12 5 2
cabbabaacaba
Output
No
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
No
No
Input
5 2 2
ccccc
Output
No
No
No
No
No | instruction | 0 | 18,442 | 11 | 36,884 |
"Correct Solution:
```
n,A,B = map(int, input().split())
s = input()
for i in s:
if (i == "a" and A + B > 0):
print("Yes")
A -= 1
elif (i == "b" and A + B > 0 and B > 0):
print("Yes")
B -= 1
else:
print("No")
``` | output | 1 | 18,442 | 11 | 36,885 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.
Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.
* A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.
* An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.
A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these.
Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
Constraints
* 1≦N,A,B≦100000
* A+B≦N
* S is N characters long.
* S consists only of the letters `a`, `b` and `c`.
Input
Inputs are provided from Standard Input in the following form.
N A B
S
Output
Output N lines. On the i-th line, output `Yes` if the i-th participant passed the Qualification contests or `No` if that participant did not pass.
Examples
Input
10 2 3
abccabaabb
Output
Yes
Yes
No
No
Yes
Yes
Yes
No
No
No
Input
12 5 2
cabbabaacaba
Output
No
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
No
No
Input
5 2 2
ccccc
Output
No
No
No
No
No
Submitted Solution:
```
n,a,b = list(map(int,input().split()))
s = input()
y = 0
z = 0
for x in s:
if x == 'a' and y+z<(a+b):
print('Yes')
y+=1
elif x == 'b' and y+z<(a+b) and z<b:
print('Yes')
z+=1
else:
print('No')
``` | instruction | 0 | 18,443 | 11 | 36,886 |
Yes | output | 1 | 18,443 | 11 | 36,887 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.
Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.
* A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.
* An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.
A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these.
Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
Constraints
* 1≦N,A,B≦100000
* A+B≦N
* S is N characters long.
* S consists only of the letters `a`, `b` and `c`.
Input
Inputs are provided from Standard Input in the following form.
N A B
S
Output
Output N lines. On the i-th line, output `Yes` if the i-th participant passed the Qualification contests or `No` if that participant did not pass.
Examples
Input
10 2 3
abccabaabb
Output
Yes
Yes
No
No
Yes
Yes
Yes
No
No
No
Input
12 5 2
cabbabaacaba
Output
No
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
No
No
Input
5 2 2
ccccc
Output
No
No
No
No
No
Submitted Solution:
```
N,A,B=map(int,input().split())
S=input()
for i in range(N):
if S[i]=="a":
if (A+B)>0:
print("Yes")
A-=1
else:
print("No")
if S[i]=="b":
if (A+B)>0 and B>0:
print("Yes")
B-=1
else:
print("No")
if S[i]=="c":
print("No")
``` | instruction | 0 | 18,444 | 11 | 36,888 |
Yes | output | 1 | 18,444 | 11 | 36,889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.
Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.
* A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.
* An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.
A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these.
Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
Constraints
* 1≦N,A,B≦100000
* A+B≦N
* S is N characters long.
* S consists only of the letters `a`, `b` and `c`.
Input
Inputs are provided from Standard Input in the following form.
N A B
S
Output
Output N lines. On the i-th line, output `Yes` if the i-th participant passed the Qualification contests or `No` if that participant did not pass.
Examples
Input
10 2 3
abccabaabb
Output
Yes
Yes
No
No
Yes
Yes
Yes
No
No
No
Input
12 5 2
cabbabaacaba
Output
No
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
No
No
Input
5 2 2
ccccc
Output
No
No
No
No
No
Submitted Solution:
```
N, A, B = map(int, input().split())
a = 0
b = 0
S = input()
for s in S:
if s == "a":
if a + b < A + B:
a += 1
print("Yes")
else:
print("No")
elif s=="b":
if a + b < A + B and b < B:
b += 1
print("Yes")
else:
print("No")
else:
print("No")
``` | instruction | 0 | 18,445 | 11 | 36,890 |
Yes | output | 1 | 18,445 | 11 | 36,891 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.
Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.
* A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.
* An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.
A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these.
Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
Constraints
* 1≦N,A,B≦100000
* A+B≦N
* S is N characters long.
* S consists only of the letters `a`, `b` and `c`.
Input
Inputs are provided from Standard Input in the following form.
N A B
S
Output
Output N lines. On the i-th line, output `Yes` if the i-th participant passed the Qualification contests or `No` if that participant did not pass.
Examples
Input
10 2 3
abccabaabb
Output
Yes
Yes
No
No
Yes
Yes
Yes
No
No
No
Input
12 5 2
cabbabaacaba
Output
No
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
No
No
Input
5 2 2
ccccc
Output
No
No
No
No
No
Submitted Solution:
```
n,A,B=map(int,input().split())
s=list(input())
cnt=0
cnt_b=0
for i in range(n):
if s[i]=='a' and cnt<A+B:
print('Yes')
cnt+=1
elif s[i]=='b' and cnt<A+B and cnt_b<B:
print('Yes')
cnt_b+=1
cnt+=1
else:
print('No')
``` | instruction | 0 | 18,446 | 11 | 36,892 |
Yes | output | 1 | 18,446 | 11 | 36,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.
Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.
* A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.
* An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.
A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these.
Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
Constraints
* 1≦N,A,B≦100000
* A+B≦N
* S is N characters long.
* S consists only of the letters `a`, `b` and `c`.
Input
Inputs are provided from Standard Input in the following form.
N A B
S
Output
Output N lines. On the i-th line, output `Yes` if the i-th participant passed the Qualification contests or `No` if that participant did not pass.
Examples
Input
10 2 3
abccabaabb
Output
Yes
Yes
No
No
Yes
Yes
Yes
No
No
No
Input
12 5 2
cabbabaacaba
Output
No
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
No
No
Input
5 2 2
ccccc
Output
No
No
No
No
No
Submitted Solution:
```
n,a,b = tuple(map(int,input().split()))
s = list(input())
count = 0
foreign = 0
for s_i in s:
if s_i=='a':
if count<=a+b:
count+=1
print("Yes")
elif s_i=='b':
if count<=a+b and foreign <= b:
count+=1
foreign+=1
print("Yes")
else:
print("No")
``` | instruction | 0 | 18,447 | 11 | 36,894 |
No | output | 1 | 18,447 | 11 | 36,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.
Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.
* A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.
* An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.
A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these.
Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
Constraints
* 1≦N,A,B≦100000
* A+B≦N
* S is N characters long.
* S consists only of the letters `a`, `b` and `c`.
Input
Inputs are provided from Standard Input in the following form.
N A B
S
Output
Output N lines. On the i-th line, output `Yes` if the i-th participant passed the Qualification contests or `No` if that participant did not pass.
Examples
Input
10 2 3
abccabaabb
Output
Yes
Yes
No
No
Yes
Yes
Yes
No
No
No
Input
12 5 2
cabbabaacaba
Output
No
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
No
No
Input
5 2 2
ccccc
Output
No
No
No
No
No
Submitted Solution:
```
n,a,b=map(int,input().split())
s=input().strip()
njp=0
nop=0
for i in range(len(s)):
if s[i]=='a':
if njp<(a+b):
njp+=1
print('Yes')
else:
print('No')
elif s[i]=='b':
if njp<(a+b):
if nop<=b:
nop+=1
njp+=1
print('yes')
else:
print('No')
else:
print('No')
else:
print('No')
``` | instruction | 0 | 18,448 | 11 | 36,896 |
No | output | 1 | 18,448 | 11 | 36,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.
Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.
* A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.
* An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.
A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these.
Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
Constraints
* 1≦N,A,B≦100000
* A+B≦N
* S is N characters long.
* S consists only of the letters `a`, `b` and `c`.
Input
Inputs are provided from Standard Input in the following form.
N A B
S
Output
Output N lines. On the i-th line, output `Yes` if the i-th participant passed the Qualification contests or `No` if that participant did not pass.
Examples
Input
10 2 3
abccabaabb
Output
Yes
Yes
No
No
Yes
Yes
Yes
No
No
No
Input
12 5 2
cabbabaacaba
Output
No
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
No
No
Input
5 2 2
ccccc
Output
No
No
No
No
No
Submitted Solution:
```
n,a,b = map(int,input().split())
s = list(input())
A = 0 #国内の合格者
B = 0 #海外の合格者
for i in range(n):
if s[i] == "c":
print("No")
# print("1")
elif s[i] == "a" and (A + B) < a + b:
print("Yes")
A += 1
# print("2")
elif s[i] == "b" and (A + B) < a + b and B <= b:
print("Yes")
B += 1
# print("3")
else:
print("No")
# print("4")
``` | instruction | 0 | 18,449 | 11 | 36,898 |
No | output | 1 | 18,449 | 11 | 36,899 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.
Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.
* A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.
* An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.
A string S is assigned indicating attributes of all participants. If the i-th character of string S is `a`, this means the participant ranked i-th in the Qualification contests is a Japanese student; `b` means the participant ranked i-th is an overseas student; and `c` means the participant ranked i-th is neither of these.
Write a program that outputs for all the participants in descending rank either `Yes` if they passed the Qualification contests or `No` if they did not pass.
Constraints
* 1≦N,A,B≦100000
* A+B≦N
* S is N characters long.
* S consists only of the letters `a`, `b` and `c`.
Input
Inputs are provided from Standard Input in the following form.
N A B
S
Output
Output N lines. On the i-th line, output `Yes` if the i-th participant passed the Qualification contests or `No` if that participant did not pass.
Examples
Input
10 2 3
abccabaabb
Output
Yes
Yes
No
No
Yes
Yes
Yes
No
No
No
Input
12 5 2
cabbabaacaba
Output
No
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
No
No
Input
5 2 2
ccccc
Output
No
No
No
No
No
Submitted Solution:
```
n,x,y=map(int,input().split())
s= input()
c=0
d=x+y
f= False
for i in range(n):
if s[i]=="a":
if c<d:
f=True
elif s[i]=="b":
if c<d and c<=y:
f=True
if f:
print("YES")
c+=1
else:
print("NO")
f=False
``` | instruction | 0 | 18,450 | 11 | 36,900 |
No | output | 1 | 18,450 | 11 | 36,901 |
Provide a correct Python 3 solution for this coding contest problem.
Income Inequality
We often compute the average as the first step in processing statistical data. Yes, the average is a good tendency measure of data, but it is not always the best. In some cases, the average may hinder the understanding of the data.
For example, consider the national income of a country. As the term income inequality suggests, a small number of people earn a good portion of the gross national income in many countries. In such cases, the average income computes much higher than the income of the vast majority. It is not appropriate to regard the average as the income of typical people.
Let us observe the above-mentioned phenomenon in some concrete data. Incomes of n people, a1, ... , an, are given. You are asked to write a program that reports the number of people whose incomes are less than or equal to the average (a1 + ... + an) / n.
Input
The input consists of multiple datasets, each in the following format.
> n
> a1 a2 ... an
A dataset consists of two lines. In the first line, the number of people n is given. n is an integer satisfying 2 ≤ n ≤ 10 000. In the second line, incomes of n people are given. ai (1 ≤ i ≤ n) is the income of the i-th person. This value is an integer greater than or equal to 1 and less than or equal to 100 000.
The end of the input is indicated by a line containing a zero. The sum of n's of all the datasets does not exceed 50 000.
Output
For each dataset, output the number of people whose incomes are less than or equal to the average.
Sample Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output for the Sample Input
7
3
9
1
4
Example
Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output
7
3
9
1
4 | instruction | 0 | 18,508 | 11 | 37,016 |
"Correct Solution:
```
while True:
n = int(input())
if n==0:
break
a = list(map(int, input().split()))
ave = int(sum(a) // n)
cnt = 0
for i in a:
if i<=ave:
cnt += 1
print(cnt)
``` | output | 1 | 18,508 | 11 | 37,017 |
Provide a correct Python 3 solution for this coding contest problem.
Income Inequality
We often compute the average as the first step in processing statistical data. Yes, the average is a good tendency measure of data, but it is not always the best. In some cases, the average may hinder the understanding of the data.
For example, consider the national income of a country. As the term income inequality suggests, a small number of people earn a good portion of the gross national income in many countries. In such cases, the average income computes much higher than the income of the vast majority. It is not appropriate to regard the average as the income of typical people.
Let us observe the above-mentioned phenomenon in some concrete data. Incomes of n people, a1, ... , an, are given. You are asked to write a program that reports the number of people whose incomes are less than or equal to the average (a1 + ... + an) / n.
Input
The input consists of multiple datasets, each in the following format.
> n
> a1 a2 ... an
A dataset consists of two lines. In the first line, the number of people n is given. n is an integer satisfying 2 ≤ n ≤ 10 000. In the second line, incomes of n people are given. ai (1 ≤ i ≤ n) is the income of the i-th person. This value is an integer greater than or equal to 1 and less than or equal to 100 000.
The end of the input is indicated by a line containing a zero. The sum of n's of all the datasets does not exceed 50 000.
Output
For each dataset, output the number of people whose incomes are less than or equal to the average.
Sample Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output for the Sample Input
7
3
9
1
4
Example
Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output
7
3
9
1
4 | instruction | 0 | 18,509 | 11 | 37,018 |
"Correct Solution:
```
#!/usr/bin/python3
import array
from fractions import Fraction
import functools
import itertools
import math
import os
import sys
def main():
while True:
N = read_int()
if N == 0:
break
A = read_ints()
print(solve(N, A))
def solve(N, A):
avg = Fraction(sum(A), N)
return len([a for a in A if a <= avg])
###############################################################################
# AUXILIARY FUNCTIONS
DEBUG = 'DEBUG' in os.environ
def inp():
return sys.stdin.readline().rstrip()
def read_int():
return int(inp())
def read_ints():
return [int(e) for e in inp().split()]
def dprint(*value, sep=' ', end='\n'):
if DEBUG:
print(*value, sep=sep, end=end)
if __name__ == '__main__':
main()
``` | output | 1 | 18,509 | 11 | 37,019 |
Provide a correct Python 3 solution for this coding contest problem.
Income Inequality
We often compute the average as the first step in processing statistical data. Yes, the average is a good tendency measure of data, but it is not always the best. In some cases, the average may hinder the understanding of the data.
For example, consider the national income of a country. As the term income inequality suggests, a small number of people earn a good portion of the gross national income in many countries. In such cases, the average income computes much higher than the income of the vast majority. It is not appropriate to regard the average as the income of typical people.
Let us observe the above-mentioned phenomenon in some concrete data. Incomes of n people, a1, ... , an, are given. You are asked to write a program that reports the number of people whose incomes are less than or equal to the average (a1 + ... + an) / n.
Input
The input consists of multiple datasets, each in the following format.
> n
> a1 a2 ... an
A dataset consists of two lines. In the first line, the number of people n is given. n is an integer satisfying 2 ≤ n ≤ 10 000. In the second line, incomes of n people are given. ai (1 ≤ i ≤ n) is the income of the i-th person. This value is an integer greater than or equal to 1 and less than or equal to 100 000.
The end of the input is indicated by a line containing a zero. The sum of n's of all the datasets does not exceed 50 000.
Output
For each dataset, output the number of people whose incomes are less than or equal to the average.
Sample Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output for the Sample Input
7
3
9
1
4
Example
Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output
7
3
9
1
4 | instruction | 0 | 18,510 | 11 | 37,020 |
"Correct Solution:
```
while True:
n = int(input())
if n == 0:
break
a = list(map(int, input().split()))
s = int(sum(a) // n)
ans = 0
for x in a:
if x <= s:
ans += 1
print(ans)
``` | output | 1 | 18,510 | 11 | 37,021 |
Provide a correct Python 3 solution for this coding contest problem.
Income Inequality
We often compute the average as the first step in processing statistical data. Yes, the average is a good tendency measure of data, but it is not always the best. In some cases, the average may hinder the understanding of the data.
For example, consider the national income of a country. As the term income inequality suggests, a small number of people earn a good portion of the gross national income in many countries. In such cases, the average income computes much higher than the income of the vast majority. It is not appropriate to regard the average as the income of typical people.
Let us observe the above-mentioned phenomenon in some concrete data. Incomes of n people, a1, ... , an, are given. You are asked to write a program that reports the number of people whose incomes are less than or equal to the average (a1 + ... + an) / n.
Input
The input consists of multiple datasets, each in the following format.
> n
> a1 a2 ... an
A dataset consists of two lines. In the first line, the number of people n is given. n is an integer satisfying 2 ≤ n ≤ 10 000. In the second line, incomes of n people are given. ai (1 ≤ i ≤ n) is the income of the i-th person. This value is an integer greater than or equal to 1 and less than or equal to 100 000.
The end of the input is indicated by a line containing a zero. The sum of n's of all the datasets does not exceed 50 000.
Output
For each dataset, output the number of people whose incomes are less than or equal to the average.
Sample Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output for the Sample Input
7
3
9
1
4
Example
Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output
7
3
9
1
4 | instruction | 0 | 18,511 | 11 | 37,022 |
"Correct Solution:
```
n = int(input())
while n != 0:
list = input().split()
for i in range(len(list)):
list[i] = int(list[i])
avgs = sum(list) / n
i = 0
a = 0
for i in range(len(list)):
if list[i] <= avgs:
a = a + 1
print(a)
n = int(input())
``` | output | 1 | 18,511 | 11 | 37,023 |
Provide a correct Python 3 solution for this coding contest problem.
Income Inequality
We often compute the average as the first step in processing statistical data. Yes, the average is a good tendency measure of data, but it is not always the best. In some cases, the average may hinder the understanding of the data.
For example, consider the national income of a country. As the term income inequality suggests, a small number of people earn a good portion of the gross national income in many countries. In such cases, the average income computes much higher than the income of the vast majority. It is not appropriate to regard the average as the income of typical people.
Let us observe the above-mentioned phenomenon in some concrete data. Incomes of n people, a1, ... , an, are given. You are asked to write a program that reports the number of people whose incomes are less than or equal to the average (a1 + ... + an) / n.
Input
The input consists of multiple datasets, each in the following format.
> n
> a1 a2 ... an
A dataset consists of two lines. In the first line, the number of people n is given. n is an integer satisfying 2 ≤ n ≤ 10 000. In the second line, incomes of n people are given. ai (1 ≤ i ≤ n) is the income of the i-th person. This value is an integer greater than or equal to 1 and less than or equal to 100 000.
The end of the input is indicated by a line containing a zero. The sum of n's of all the datasets does not exceed 50 000.
Output
For each dataset, output the number of people whose incomes are less than or equal to the average.
Sample Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output for the Sample Input
7
3
9
1
4
Example
Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output
7
3
9
1
4 | instruction | 0 | 18,512 | 11 | 37,024 |
"Correct Solution:
```
while True:
n=int(input())
if n == 0:
break
A = list(map(int,input().split()))
# print(A)
a =sum(A)
bar = a/n
# print(bar)
count=0
for i in A:
if i<=bar:
count+=1
print(count)
``` | output | 1 | 18,512 | 11 | 37,025 |
Provide a correct Python 3 solution for this coding contest problem.
Income Inequality
We often compute the average as the first step in processing statistical data. Yes, the average is a good tendency measure of data, but it is not always the best. In some cases, the average may hinder the understanding of the data.
For example, consider the national income of a country. As the term income inequality suggests, a small number of people earn a good portion of the gross national income in many countries. In such cases, the average income computes much higher than the income of the vast majority. It is not appropriate to regard the average as the income of typical people.
Let us observe the above-mentioned phenomenon in some concrete data. Incomes of n people, a1, ... , an, are given. You are asked to write a program that reports the number of people whose incomes are less than or equal to the average (a1 + ... + an) / n.
Input
The input consists of multiple datasets, each in the following format.
> n
> a1 a2 ... an
A dataset consists of two lines. In the first line, the number of people n is given. n is an integer satisfying 2 ≤ n ≤ 10 000. In the second line, incomes of n people are given. ai (1 ≤ i ≤ n) is the income of the i-th person. This value is an integer greater than or equal to 1 and less than or equal to 100 000.
The end of the input is indicated by a line containing a zero. The sum of n's of all the datasets does not exceed 50 000.
Output
For each dataset, output the number of people whose incomes are less than or equal to the average.
Sample Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output for the Sample Input
7
3
9
1
4
Example
Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output
7
3
9
1
4 | instruction | 0 | 18,513 | 11 | 37,026 |
"Correct Solution:
```
while True:
n = int(input())
if n == 0:
break
a = list(map(int, input().split()))
avg = sum(a)/n
print(len([i for i in a if avg >= i]))
``` | output | 1 | 18,513 | 11 | 37,027 |
Provide a correct Python 3 solution for this coding contest problem.
Income Inequality
We often compute the average as the first step in processing statistical data. Yes, the average is a good tendency measure of data, but it is not always the best. In some cases, the average may hinder the understanding of the data.
For example, consider the national income of a country. As the term income inequality suggests, a small number of people earn a good portion of the gross national income in many countries. In such cases, the average income computes much higher than the income of the vast majority. It is not appropriate to regard the average as the income of typical people.
Let us observe the above-mentioned phenomenon in some concrete data. Incomes of n people, a1, ... , an, are given. You are asked to write a program that reports the number of people whose incomes are less than or equal to the average (a1 + ... + an) / n.
Input
The input consists of multiple datasets, each in the following format.
> n
> a1 a2 ... an
A dataset consists of two lines. In the first line, the number of people n is given. n is an integer satisfying 2 ≤ n ≤ 10 000. In the second line, incomes of n people are given. ai (1 ≤ i ≤ n) is the income of the i-th person. This value is an integer greater than or equal to 1 and less than or equal to 100 000.
The end of the input is indicated by a line containing a zero. The sum of n's of all the datasets does not exceed 50 000.
Output
For each dataset, output the number of people whose incomes are less than or equal to the average.
Sample Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output for the Sample Input
7
3
9
1
4
Example
Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output
7
3
9
1
4 | instruction | 0 | 18,514 | 11 | 37,028 |
"Correct Solution:
```
#!/usr/bin/env python3
while True:
n = int(input())
if n == 0: break
a = list(map(int, input().split()))
ave_num = sum(a) / n
cnt = 0
for item in a:
if item <= ave_num:
cnt += 1
print(cnt)
``` | output | 1 | 18,514 | 11 | 37,029 |
Provide a correct Python 3 solution for this coding contest problem.
Income Inequality
We often compute the average as the first step in processing statistical data. Yes, the average is a good tendency measure of data, but it is not always the best. In some cases, the average may hinder the understanding of the data.
For example, consider the national income of a country. As the term income inequality suggests, a small number of people earn a good portion of the gross national income in many countries. In such cases, the average income computes much higher than the income of the vast majority. It is not appropriate to regard the average as the income of typical people.
Let us observe the above-mentioned phenomenon in some concrete data. Incomes of n people, a1, ... , an, are given. You are asked to write a program that reports the number of people whose incomes are less than or equal to the average (a1 + ... + an) / n.
Input
The input consists of multiple datasets, each in the following format.
> n
> a1 a2 ... an
A dataset consists of two lines. In the first line, the number of people n is given. n is an integer satisfying 2 ≤ n ≤ 10 000. In the second line, incomes of n people are given. ai (1 ≤ i ≤ n) is the income of the i-th person. This value is an integer greater than or equal to 1 and less than or equal to 100 000.
The end of the input is indicated by a line containing a zero. The sum of n's of all the datasets does not exceed 50 000.
Output
For each dataset, output the number of people whose incomes are less than or equal to the average.
Sample Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output for the Sample Input
7
3
9
1
4
Example
Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output
7
3
9
1
4 | instruction | 0 | 18,515 | 11 | 37,030 |
"Correct Solution:
```
while True:
x=int(input())
if x==0:
break
alst = list(map(int,input().split()))
num = sum(alst) / x
print(sum(a <= num for a in alst))
``` | output | 1 | 18,515 | 11 | 37,031 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Income Inequality
We often compute the average as the first step in processing statistical data. Yes, the average is a good tendency measure of data, but it is not always the best. In some cases, the average may hinder the understanding of the data.
For example, consider the national income of a country. As the term income inequality suggests, a small number of people earn a good portion of the gross national income in many countries. In such cases, the average income computes much higher than the income of the vast majority. It is not appropriate to regard the average as the income of typical people.
Let us observe the above-mentioned phenomenon in some concrete data. Incomes of n people, a1, ... , an, are given. You are asked to write a program that reports the number of people whose incomes are less than or equal to the average (a1 + ... + an) / n.
Input
The input consists of multiple datasets, each in the following format.
> n
> a1 a2 ... an
A dataset consists of two lines. In the first line, the number of people n is given. n is an integer satisfying 2 ≤ n ≤ 10 000. In the second line, incomes of n people are given. ai (1 ≤ i ≤ n) is the income of the i-th person. This value is an integer greater than or equal to 1 and less than or equal to 100 000.
The end of the input is indicated by a line containing a zero. The sum of n's of all the datasets does not exceed 50 000.
Output
For each dataset, output the number of people whose incomes are less than or equal to the average.
Sample Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output for the Sample Input
7
3
9
1
4
Example
Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output
7
3
9
1
4
Submitted Solution:
```
a = int(input())
su = 0
while a != 0:
b = list(map(int,input().split()))
c = sum(b)/len(b)
for i in range(len(b)):
if b[i] <= c:
su += 1
print(su)
su = 0
a = int(input())
``` | instruction | 0 | 18,516 | 11 | 37,032 |
Yes | output | 1 | 18,516 | 11 | 37,033 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Income Inequality
We often compute the average as the first step in processing statistical data. Yes, the average is a good tendency measure of data, but it is not always the best. In some cases, the average may hinder the understanding of the data.
For example, consider the national income of a country. As the term income inequality suggests, a small number of people earn a good portion of the gross national income in many countries. In such cases, the average income computes much higher than the income of the vast majority. It is not appropriate to regard the average as the income of typical people.
Let us observe the above-mentioned phenomenon in some concrete data. Incomes of n people, a1, ... , an, are given. You are asked to write a program that reports the number of people whose incomes are less than or equal to the average (a1 + ... + an) / n.
Input
The input consists of multiple datasets, each in the following format.
> n
> a1 a2 ... an
A dataset consists of two lines. In the first line, the number of people n is given. n is an integer satisfying 2 ≤ n ≤ 10 000. In the second line, incomes of n people are given. ai (1 ≤ i ≤ n) is the income of the i-th person. This value is an integer greater than or equal to 1 and less than or equal to 100 000.
The end of the input is indicated by a line containing a zero. The sum of n's of all the datasets does not exceed 50 000.
Output
For each dataset, output the number of people whose incomes are less than or equal to the average.
Sample Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output for the Sample Input
7
3
9
1
4
Example
Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output
7
3
9
1
4
Submitted Solution:
```
while True :
n=int(input())
if n==0 : break
a=input().split()
tot=0;ans=0
for i in a : tot+=int(i)
for i in a :
if (tot/n)>=int(i) : ans+=1
print(ans)
``` | instruction | 0 | 18,517 | 11 | 37,034 |
Yes | output | 1 | 18,517 | 11 | 37,035 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Income Inequality
We often compute the average as the first step in processing statistical data. Yes, the average is a good tendency measure of data, but it is not always the best. In some cases, the average may hinder the understanding of the data.
For example, consider the national income of a country. As the term income inequality suggests, a small number of people earn a good portion of the gross national income in many countries. In such cases, the average income computes much higher than the income of the vast majority. It is not appropriate to regard the average as the income of typical people.
Let us observe the above-mentioned phenomenon in some concrete data. Incomes of n people, a1, ... , an, are given. You are asked to write a program that reports the number of people whose incomes are less than or equal to the average (a1 + ... + an) / n.
Input
The input consists of multiple datasets, each in the following format.
> n
> a1 a2 ... an
A dataset consists of two lines. In the first line, the number of people n is given. n is an integer satisfying 2 ≤ n ≤ 10 000. In the second line, incomes of n people are given. ai (1 ≤ i ≤ n) is the income of the i-th person. This value is an integer greater than or equal to 1 and less than or equal to 100 000.
The end of the input is indicated by a line containing a zero. The sum of n's of all the datasets does not exceed 50 000.
Output
For each dataset, output the number of people whose incomes are less than or equal to the average.
Sample Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output for the Sample Input
7
3
9
1
4
Example
Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output
7
3
9
1
4
Submitted Solution:
```
# coding=utf-8
###
### for python program
###
import sys
import math
# math class
class mymath:
### pi
pi = 3.14159265358979323846264338
### Prime Number
def pnum_eratosthenes(self, n):
ptable = [0 for i in range(n+1)]
plist = []
for i in range(2, n+1):
if ptable[i]==0:
plist.append(i)
for j in range(i+i, n+1, i):
ptable[j] = 1
return plist
def pnum_check(self, n):
if (n==1):
return False
elif (n==2):
return True
else:
for x in range(2,n):
if(n % x==0):
return False
return True
### GCD
def gcd(self, a, b):
if b == 0:
return a
return self.gcd(b, a%b)
### LCM
def lcm(self, a, b):
return (a*b)//self.gcd(a,b)
### Mat Multiplication
def mul(self, A, B):
ans = []
for a in A:
c = 0
for j, row in enumerate(a):
c += row*B[j]
ans.append(c)
return ans
### intチェック
def is_integer(self, n):
try:
float(n)
except ValueError:
return False
else:
return float(n).is_integer()
### 幾何学問題用
def dist(self, A, B):
d = 0
for i in range(len(A)):
d += (A[i]-B[i])**2
d = d**(1/2)
return d
### 絶対値
def abs(self, n):
if n >= 0:
return n
else:
return -n
mymath = mymath()
### output class
class output:
### list
def list(self, l):
l = list(l)
#print(" ", end="")
for i, num in enumerate(l):
print(num, end="")
if i != len(l)-1:
print(" ", end="")
print()
output = output()
### input sample
#i = input()
#N = int(input())
#A, B, C = [x for x in input().split()]
#N, K = [int(x) for x in input().split()]
#inlist = [int(w) for w in input().split()]
#R = float(input())
#A.append(list(map(int,input().split())))
#for line in sys.stdin.readlines():
# x, y = [int(temp) for temp in line.split()]
#abc list
#abc = [chr(ord('a') + i) for i in range(26)]
### output sample
# print("{0} {1} {2:.5f}".format(A//B, A%B, A/B))
# print("{0:.6f} {1:.6f}".format(R*R*math.pi,R*2*math.pi))
# print(" {}".format(i), end="")
def printA(A):
N = len(A)
for i, n in enumerate(A):
print(n, end='')
if i != N-1:
print(' ', end='')
print()
# リスト内包表記 ifあり
# [x-k if x != 0 else x for x in C]
# ソート
# N = N.sort()
# 10000個の素数リスト
# P = mymath.pnum_eratosthenes(105000)
def get_input():
N = []
while True:
try:
N.append(input())
#N.append(int(input()))
#N.append(float(input()))
except EOFError:
break
return N
while True:
N = int(input())
if N==0:
break
A = [int(x) for x in input().split()]
ave = sum(A)/N
count = 0
for i in A:
if i <= ave:
count += 1
print(count)
``` | instruction | 0 | 18,518 | 11 | 37,036 |
Yes | output | 1 | 18,518 | 11 | 37,037 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Income Inequality
We often compute the average as the first step in processing statistical data. Yes, the average is a good tendency measure of data, but it is not always the best. In some cases, the average may hinder the understanding of the data.
For example, consider the national income of a country. As the term income inequality suggests, a small number of people earn a good portion of the gross national income in many countries. In such cases, the average income computes much higher than the income of the vast majority. It is not appropriate to regard the average as the income of typical people.
Let us observe the above-mentioned phenomenon in some concrete data. Incomes of n people, a1, ... , an, are given. You are asked to write a program that reports the number of people whose incomes are less than or equal to the average (a1 + ... + an) / n.
Input
The input consists of multiple datasets, each in the following format.
> n
> a1 a2 ... an
A dataset consists of two lines. In the first line, the number of people n is given. n is an integer satisfying 2 ≤ n ≤ 10 000. In the second line, incomes of n people are given. ai (1 ≤ i ≤ n) is the income of the i-th person. This value is an integer greater than or equal to 1 and less than or equal to 100 000.
The end of the input is indicated by a line containing a zero. The sum of n's of all the datasets does not exceed 50 000.
Output
For each dataset, output the number of people whose incomes are less than or equal to the average.
Sample Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output for the Sample Input
7
3
9
1
4
Example
Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output
7
3
9
1
4
Submitted Solution:
```
while True:
n=int(input())
c=0
if n==0:
break
ns=list(map(int, input().split()))
for i in range(n):
if ns[i]<=sum(ns)/n:
c+=1
print(c)
``` | instruction | 0 | 18,519 | 11 | 37,038 |
Yes | output | 1 | 18,519 | 11 | 37,039 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Income Inequality
We often compute the average as the first step in processing statistical data. Yes, the average is a good tendency measure of data, but it is not always the best. In some cases, the average may hinder the understanding of the data.
For example, consider the national income of a country. As the term income inequality suggests, a small number of people earn a good portion of the gross national income in many countries. In such cases, the average income computes much higher than the income of the vast majority. It is not appropriate to regard the average as the income of typical people.
Let us observe the above-mentioned phenomenon in some concrete data. Incomes of n people, a1, ... , an, are given. You are asked to write a program that reports the number of people whose incomes are less than or equal to the average (a1 + ... + an) / n.
Input
The input consists of multiple datasets, each in the following format.
> n
> a1 a2 ... an
A dataset consists of two lines. In the first line, the number of people n is given. n is an integer satisfying 2 ≤ n ≤ 10 000. In the second line, incomes of n people are given. ai (1 ≤ i ≤ n) is the income of the i-th person. This value is an integer greater than or equal to 1 and less than or equal to 100 000.
The end of the input is indicated by a line containing a zero. The sum of n's of all the datasets does not exceed 50 000.
Output
For each dataset, output the number of people whose incomes are less than or equal to the average.
Sample Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output for the Sample Input
7
3
9
1
4
Example
Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output
7
3
9
1
4
Submitted Solution:
```
from statistics import mean
while True:
n=int(input())
c=0
if n==0:
break
ns=list(map(int, input().split()))
for i in range(n):
if ns[i]<=mean(ns):
c+=1
print(c)
``` | instruction | 0 | 18,520 | 11 | 37,040 |
No | output | 1 | 18,520 | 11 | 37,041 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Income Inequality
We often compute the average as the first step in processing statistical data. Yes, the average is a good tendency measure of data, but it is not always the best. In some cases, the average may hinder the understanding of the data.
For example, consider the national income of a country. As the term income inequality suggests, a small number of people earn a good portion of the gross national income in many countries. In such cases, the average income computes much higher than the income of the vast majority. It is not appropriate to regard the average as the income of typical people.
Let us observe the above-mentioned phenomenon in some concrete data. Incomes of n people, a1, ... , an, are given. You are asked to write a program that reports the number of people whose incomes are less than or equal to the average (a1 + ... + an) / n.
Input
The input consists of multiple datasets, each in the following format.
> n
> a1 a2 ... an
A dataset consists of two lines. In the first line, the number of people n is given. n is an integer satisfying 2 ≤ n ≤ 10 000. In the second line, incomes of n people are given. ai (1 ≤ i ≤ n) is the income of the i-th person. This value is an integer greater than or equal to 1 and less than or equal to 100 000.
The end of the input is indicated by a line containing a zero. The sum of n's of all the datasets does not exceed 50 000.
Output
For each dataset, output the number of people whose incomes are less than or equal to the average.
Sample Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output for the Sample Input
7
3
9
1
4
Example
Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output
7
3
9
1
4
Submitted Solution:
```
from statistics import mean
while True:
n=int(input())
c=0
if n==0:
break
ns=list(map(int, input().split()))
for i in range(n):
if ns[i]<=mean(ns):
c+=1
else:
pass
print(c)
``` | instruction | 0 | 18,521 | 11 | 37,042 |
No | output | 1 | 18,521 | 11 | 37,043 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Income Inequality
We often compute the average as the first step in processing statistical data. Yes, the average is a good tendency measure of data, but it is not always the best. In some cases, the average may hinder the understanding of the data.
For example, consider the national income of a country. As the term income inequality suggests, a small number of people earn a good portion of the gross national income in many countries. In such cases, the average income computes much higher than the income of the vast majority. It is not appropriate to regard the average as the income of typical people.
Let us observe the above-mentioned phenomenon in some concrete data. Incomes of n people, a1, ... , an, are given. You are asked to write a program that reports the number of people whose incomes are less than or equal to the average (a1 + ... + an) / n.
Input
The input consists of multiple datasets, each in the following format.
> n
> a1 a2 ... an
A dataset consists of two lines. In the first line, the number of people n is given. n is an integer satisfying 2 ≤ n ≤ 10 000. In the second line, incomes of n people are given. ai (1 ≤ i ≤ n) is the income of the i-th person. This value is an integer greater than or equal to 1 and less than or equal to 100 000.
The end of the input is indicated by a line containing a zero. The sum of n's of all the datasets does not exceed 50 000.
Output
For each dataset, output the number of people whose incomes are less than or equal to the average.
Sample Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output for the Sample Input
7
3
9
1
4
Example
Input
7
15 15 15 15 15 15 15
4
10 20 30 60
10
1 1 1 1 1 1 1 1 1 100
7
90 90 90 90 90 90 10
7
2 7 1 8 2 8 4
0
Output
7
3
9
1
4
Submitted Solution:
```
n=int(input())
a=[int(i) for i in input().split]
total=0
for i in a:
total+=i
hei=total/n
count=0
for i in a:
if i <= hei:
count+=1
print(count)
``` | instruction | 0 | 18,522 | 11 | 37,044 |
No | output | 1 | 18,522 | 11 | 37,045 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's denote a m-free matrix as a binary (that is, consisting of only 1's and 0's) matrix such that every square submatrix of size m × m of this matrix contains at least one zero.
Consider the following problem:
You are given two integers n and m. You have to construct an m-free square matrix of size n × n such that the number of 1's in this matrix is maximum possible. Print the maximum possible number of 1's in such matrix.
You don't have to solve this problem. Instead, you have to construct a few tests for it.
You will be given t numbers x1, x2, ..., xt. For every <image>, find two integers ni and mi (ni ≥ mi) such that the answer for the aforementioned problem is exactly xi if we set n = ni and m = mi.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of tests you have to construct.
Then t lines follow, i-th line containing one integer xi (0 ≤ xi ≤ 109).
Note that in hacks you have to set t = 1.
Output
For each test you have to construct, output two positive numbers ni and mi (1 ≤ mi ≤ ni ≤ 109) such that the maximum number of 1's in a mi-free ni × ni matrix is exactly xi. If there are multiple solutions, you may output any of them; and if this is impossible to construct a test, output a single integer - 1.
Example
Input
3
21
0
1
Output
5 2
1 1
-1
Submitted Solution:
```
n = int(input())
for x in range(n):
x = int(input())
if x == 0:
print(1, 1)
else:
flag = True
for d in range(1, int(x**0.5)+2):
if x % d == 0:
s = x // d
n = (s + d)//2
if int(n) != n:
continue
nm = s - n
if nm == 0:
continue
m = n//nm
if (m>0) and (n**2 - (n//m)**2 == x):
print(n, m)
flag = False
break
if flag:
print(-1)
``` | instruction | 0 | 19,101 | 11 | 38,202 |
Yes | output | 1 | 19,101 | 11 | 38,203 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board.
The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns.
How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?
Constraints
* 1 \leq H, W \leq N \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
H
W
Output
Print the answer.
Examples
Input
3
2
3
Output
2
Input
100
1
1
Output
10000
Input
5
4
2
Output
8
Submitted Solution:
```
n = int(input())
h = int(input())
w = int(input())
ans = (n-w+1)*(n-h+1)
print(ans)
``` | instruction | 0 | 19,194 | 11 | 38,388 |
Yes | output | 1 | 19,194 | 11 | 38,389 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board.
The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns.
How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?
Constraints
* 1 \leq H, W \leq N \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
H
W
Output
Print the answer.
Examples
Input
3
2
3
Output
2
Input
100
1
1
Output
10000
Input
5
4
2
Output
8
Submitted Solution:
```
n=int(input())
h=int(input())
w=int(input())
a=h-1
b=w-1
f=(n-a)*(n-b)
print(f)
``` | instruction | 0 | 19,195 | 11 | 38,390 |
Yes | output | 1 | 19,195 | 11 | 38,391 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board.
The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns.
How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?
Constraints
* 1 \leq H, W \leq N \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
H
W
Output
Print the answer.
Examples
Input
3
2
3
Output
2
Input
100
1
1
Output
10000
Input
5
4
2
Output
8
Submitted Solution:
```
n = int(input())+1
a = int(input())
b = int(input())
print((n-a)*(n-b))
``` | instruction | 0 | 19,196 | 11 | 38,392 |
Yes | output | 1 | 19,196 | 11 | 38,393 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board.
The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns.
How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?
Constraints
* 1 \leq H, W \leq N \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
H
W
Output
Print the answer.
Examples
Input
3
2
3
Output
2
Input
100
1
1
Output
10000
Input
5
4
2
Output
8
Submitted Solution:
```
N,H,W = int(input()),int(input()),int(input())
print(max(0,N-W+1)*max(0,N-H+1))
``` | instruction | 0 | 19,197 | 11 | 38,394 |
Yes | output | 1 | 19,197 | 11 | 38,395 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board.
The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns.
How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?
Constraints
* 1 \leq H, W \leq N \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
H
W
Output
Print the answer.
Examples
Input
3
2
3
Output
2
Input
100
1
1
Output
10000
Input
5
4
2
Output
8
Submitted Solution:
```
from math import *
log = lambda *x: print(*x)
n, = cin(int)
h, = cin(int)
w, = cin(int)
tw, th = 0, ceil(n/h)
while w <= n:
tw += 1
w += 1
log(th * tw)
``` | instruction | 0 | 19,198 | 11 | 38,396 |
No | output | 1 | 19,198 | 11 | 38,397 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board.
The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns.
How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?
Constraints
* 1 \leq H, W \leq N \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
H
W
Output
Print the answer.
Examples
Input
3
2
3
Output
2
Input
100
1
1
Output
10000
Input
5
4
2
Output
8
Submitted Solution:
```
n = int(input())
h = int(input())
w = int(input())
count_h = 0
count_w = 0
while True:
nn=n
if nn>=h:
nn -=h
count_h +=1
elif nn<h or nn==0:
break
while True:
nn=n
if nn>=w:
nn -=w
count_w +=1
elif nn<w or nn==0:
break
print(count_h * count_w)
``` | instruction | 0 | 19,199 | 11 | 38,398 |
No | output | 1 | 19,199 | 11 | 38,399 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board.
The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns.
How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?
Constraints
* 1 \leq H, W \leq N \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
H
W
Output
Print the answer.
Examples
Input
3
2
3
Output
2
Input
100
1
1
Output
10000
Input
5
4
2
Output
8
Submitted Solution:
```
h = int(input())
w = int(input())
n = int(input())
print((h-n+1)*(w-n+1))
``` | instruction | 0 | 19,200 | 11 | 38,400 |
No | output | 1 | 19,200 | 11 | 38,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It has been decided that a programming contest sponsored by company A will be held, so we will post the notice on a bulletin board.
The bulletin board is in the form of a grid with N rows and N columns, and the notice will occupy a rectangular region with H rows and W columns.
How many ways are there to choose where to put the notice so that it completely covers exactly HW squares?
Constraints
* 1 \leq H, W \leq N \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
H
W
Output
Print the answer.
Examples
Input
3
2
3
Output
2
Input
100
1
1
Output
10000
Input
5
4
2
Output
8
Submitted Solution:
```
N = int(input())
H = int(input())
W = int(input())
NH = 0
NW = 0
if(N - H != 0):
NH = N - H + 1
if(N - W != 0):
NW = N - W + 1
print(NH * NW)
``` | instruction | 0 | 19,201 | 11 | 38,402 |
No | output | 1 | 19,201 | 11 | 38,403 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.