message stringlengths 2 43.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 853 107k | cluster float64 24 24 | __index_level_0__ int64 1.71k 214k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp's workday lasts exactly n minutes. He loves chocolate bars and can eat one bar in one minute. Today Polycarp has k bars at the beginning of the workday.
In some minutes of the workday Polycarp has important things to do and in such minutes he is not able to eat a chocolate bar. In other minutes he can either eat or not eat one chocolate bar. It is guaranteed, that in the first and in the last minutes of the workday Polycarp has no important things to do and he will always eat bars in this minutes to gladden himself at the begining and at the end of the workday. Also it is guaranteed, that k is strictly greater than 1.
Your task is to determine such an order of eating chocolate bars that the maximum break time between eating bars is as minimum as possible.
Consider that Polycarp eats a bar in the minute x and the next bar in the minute y (x < y). Then the break time is equal to y - x - 1 minutes. It is not necessary for Polycarp to eat all bars he has.
Input
The first line contains two integers n and k (2 β€ n β€ 200 000, 2 β€ k β€ n) β the length of the workday in minutes and the number of chocolate bars, which Polycarp has in the beginning of the workday.
The second line contains the string with length n consisting of zeros and ones. If the i-th symbol in the string equals to zero, Polycarp has no important things to do in the minute i and he can eat a chocolate bar. In the other case, Polycarp is busy in the minute i and can not eat a chocolate bar. It is guaranteed, that the first and the last characters of the string are equal to zero, and Polycarp always eats chocolate bars in these minutes.
Output
Print the minimum possible break in minutes between eating chocolate bars.
Examples
Input
3 3
010
Output
1
Input
8 3
01010110
Output
3
Note
In the first example Polycarp can not eat the chocolate bar in the second minute, so the time of the break equals to one minute.
In the second example Polycarp will eat bars in the minutes 1 and 8 anyway, also he needs to eat the chocolate bar in the minute 5, so that the time of the maximum break will be equal to 3 minutes.
Submitted Solution:
```
a,b = map(int,input().split())
c = list(map(int,input()))
d = []
e = []
f = 1
if c.count(0) > b:f += c.count(0) - b
for x in range(len(c)):
if c[x] == 0:d.append(x)
for x in d[1: -1]:
if d.index(x) - f < 0:k = (x - d[0]) - 1
else:k = (x - d[d.index(x) - f]) - 1
if d.index(x) + f > len(d) - 1:l = (d[-1] - x) - 1
else: l = (d[d.index(x) + f] - x) - 1
if k > l:e.append(k)
else:e.append(l)
else:e.append(0)
e.sort()
print(e[0])
``` | instruction | 0 | 65,337 | 24 | 130,674 |
No | output | 1 | 65,337 | 24 | 130,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp's workday lasts exactly n minutes. He loves chocolate bars and can eat one bar in one minute. Today Polycarp has k bars at the beginning of the workday.
In some minutes of the workday Polycarp has important things to do and in such minutes he is not able to eat a chocolate bar. In other minutes he can either eat or not eat one chocolate bar. It is guaranteed, that in the first and in the last minutes of the workday Polycarp has no important things to do and he will always eat bars in this minutes to gladden himself at the begining and at the end of the workday. Also it is guaranteed, that k is strictly greater than 1.
Your task is to determine such an order of eating chocolate bars that the maximum break time between eating bars is as minimum as possible.
Consider that Polycarp eats a bar in the minute x and the next bar in the minute y (x < y). Then the break time is equal to y - x - 1 minutes. It is not necessary for Polycarp to eat all bars he has.
Input
The first line contains two integers n and k (2 β€ n β€ 200 000, 2 β€ k β€ n) β the length of the workday in minutes and the number of chocolate bars, which Polycarp has in the beginning of the workday.
The second line contains the string with length n consisting of zeros and ones. If the i-th symbol in the string equals to zero, Polycarp has no important things to do in the minute i and he can eat a chocolate bar. In the other case, Polycarp is busy in the minute i and can not eat a chocolate bar. It is guaranteed, that the first and the last characters of the string are equal to zero, and Polycarp always eats chocolate bars in these minutes.
Output
Print the minimum possible break in minutes between eating chocolate bars.
Examples
Input
3 3
010
Output
1
Input
8 3
01010110
Output
3
Note
In the first example Polycarp can not eat the chocolate bar in the second minute, so the time of the break equals to one minute.
In the second example Polycarp will eat bars in the minutes 1 and 8 anyway, also he needs to eat the chocolate bar in the minute 5, so that the time of the maximum break will be equal to 3 minutes.
Submitted Solution:
```
n, k = map(int, input().split())
s = input()
l, r = int(-1), int(n)
while r - l > 1:
m = (l+r)//2
c, p = 1, 0
cond = True
while p < n and c < m:
i = p + m + 1
while i >= p and (i >= n or s[i] == '1'):
i = i - 1;
if (i == p):
break
c = c + 1
p = i
cond = cond and (c <= m and n-1 - p - 1 <= m)
if cond:
r = m
else:
l = m
print(int(r))
``` | instruction | 0 | 65,338 | 24 | 130,676 |
No | output | 1 | 65,338 | 24 | 130,677 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp's workday lasts exactly n minutes. He loves chocolate bars and can eat one bar in one minute. Today Polycarp has k bars at the beginning of the workday.
In some minutes of the workday Polycarp has important things to do and in such minutes he is not able to eat a chocolate bar. In other minutes he can either eat or not eat one chocolate bar. It is guaranteed, that in the first and in the last minutes of the workday Polycarp has no important things to do and he will always eat bars in this minutes to gladden himself at the begining and at the end of the workday. Also it is guaranteed, that k is strictly greater than 1.
Your task is to determine such an order of eating chocolate bars that the maximum break time between eating bars is as minimum as possible.
Consider that Polycarp eats a bar in the minute x and the next bar in the minute y (x < y). Then the break time is equal to y - x - 1 minutes. It is not necessary for Polycarp to eat all bars he has.
Input
The first line contains two integers n and k (2 β€ n β€ 200 000, 2 β€ k β€ n) β the length of the workday in minutes and the number of chocolate bars, which Polycarp has in the beginning of the workday.
The second line contains the string with length n consisting of zeros and ones. If the i-th symbol in the string equals to zero, Polycarp has no important things to do in the minute i and he can eat a chocolate bar. In the other case, Polycarp is busy in the minute i and can not eat a chocolate bar. It is guaranteed, that the first and the last characters of the string are equal to zero, and Polycarp always eats chocolate bars in these minutes.
Output
Print the minimum possible break in minutes between eating chocolate bars.
Examples
Input
3 3
010
Output
1
Input
8 3
01010110
Output
3
Note
In the first example Polycarp can not eat the chocolate bar in the second minute, so the time of the break equals to one minute.
In the second example Polycarp will eat bars in the minutes 1 and 8 anyway, also he needs to eat the chocolate bar in the minute 5, so that the time of the maximum break will be equal to 3 minutes.
Submitted Solution:
```
def ok(x, k, s, n):
i = 0
while i < len(s):
k -= 1
if(i + x + 1 >= len(s)):
break
j = n[i + x + 1]
if(j == i):
return 0
i = j
return k >= 0
n, k = map(int, input().split())
s = input()
n = [0]
for i in range(1, len(s)):
if s[i] == '0':
n.append(i)
else:
n.append(n[-1])
l = -1
r = len(s)
while r - l > 1:
m = (l + r)//2
if ok(m, k, s, n):
r = m
else:
l = m
print(r)
``` | instruction | 0 | 65,339 | 24 | 130,678 |
No | output | 1 | 65,339 | 24 | 130,679 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp's workday lasts exactly n minutes. He loves chocolate bars and can eat one bar in one minute. Today Polycarp has k bars at the beginning of the workday.
In some minutes of the workday Polycarp has important things to do and in such minutes he is not able to eat a chocolate bar. In other minutes he can either eat or not eat one chocolate bar. It is guaranteed, that in the first and in the last minutes of the workday Polycarp has no important things to do and he will always eat bars in this minutes to gladden himself at the begining and at the end of the workday. Also it is guaranteed, that k is strictly greater than 1.
Your task is to determine such an order of eating chocolate bars that the maximum break time between eating bars is as minimum as possible.
Consider that Polycarp eats a bar in the minute x and the next bar in the minute y (x < y). Then the break time is equal to y - x - 1 minutes. It is not necessary for Polycarp to eat all bars he has.
Input
The first line contains two integers n and k (2 β€ n β€ 200 000, 2 β€ k β€ n) β the length of the workday in minutes and the number of chocolate bars, which Polycarp has in the beginning of the workday.
The second line contains the string with length n consisting of zeros and ones. If the i-th symbol in the string equals to zero, Polycarp has no important things to do in the minute i and he can eat a chocolate bar. In the other case, Polycarp is busy in the minute i and can not eat a chocolate bar. It is guaranteed, that the first and the last characters of the string are equal to zero, and Polycarp always eats chocolate bars in these minutes.
Output
Print the minimum possible break in minutes between eating chocolate bars.
Examples
Input
3 3
010
Output
1
Input
8 3
01010110
Output
3
Note
In the first example Polycarp can not eat the chocolate bar in the second minute, so the time of the break equals to one minute.
In the second example Polycarp will eat bars in the minutes 1 and 8 anyway, also he needs to eat the chocolate bar in the minute 5, so that the time of the maximum break will be equal to 3 minutes.
Submitted Solution:
```
n, k = map(int, input().split())
s = input()
l, r = int(-1), int(n)
while r - l > 1:
m = (l+r)//2
c, p = 1, 0
cond = True
while p < n - 1 and c < m:
i = p + m + 1
while i >= p and (i >= n or s[i] == '1'):
i = i - 1;
if (i == p):
cond = False
break
c = c + 1
p = i
cond = cond and (c <= m and n-1 - p - 1 <= m)
if cond:
r = m
else:
l = m
print(int(r))
``` | instruction | 0 | 65,340 | 24 | 130,680 |
No | output | 1 | 65,340 | 24 | 130,681 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has spent the entire day preparing problems for you. Now he has to sleep for at least a minutes to feel refreshed.
Polycarp can only wake up by hearing the sound of his alarm. So he has just fallen asleep and his first alarm goes off in b minutes.
Every time Polycarp wakes up, he decides if he wants to sleep for some more time or not. If he's slept for less than a minutes in total, then he sets his alarm to go off in c minutes after it is reset and spends d minutes to fall asleep again. Otherwise, he gets out of his bed and proceeds with the day.
If the alarm goes off while Polycarp is falling asleep, then he resets his alarm to go off in another c minutes and tries to fall asleep for d minutes again.
You just want to find out when will Polycarp get out of his bed or report that it will never happen.
Please check out the notes for some explanations of the example.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of testcases.
The only line of each testcase contains four integers a, b, c, d (1 β€ a, b, c, d β€ 10^9) β the time Polycarp has to sleep for to feel refreshed, the time before the first alarm goes off, the time before every succeeding alarm goes off and the time Polycarp spends to fall asleep.
Output
For each test case print one integer. If Polycarp never gets out of his bed then print -1. Otherwise, print the time it takes for Polycarp to get out of his bed.
Example
Input
7
10 3 6 4
11 3 6 4
5 9 4 10
6 5 2 3
1 1 1 1
3947465 47342 338129 123123
234123843 13 361451236 361451000
Output
27
27
9
-1
1
6471793
358578060125049
Note
In the first testcase Polycarp wakes up after 3 minutes. He only rested for 3 minutes out of 10 minutes he needed. So after that he sets his alarm to go off in 6 minutes and spends 4 minutes falling asleep. Thus, he rests for 2 more minutes, totaling in 3+2=5 minutes of sleep. Then he repeats the procedure three more times and ends up with 11 minutes of sleep. Finally, he gets out of his bed. He spent 3 minutes before the first alarm and then reset his alarm four times. The answer is 3+4 β
6 = 27.
The second example is almost like the first one but Polycarp needs 11 minutes of sleep instead of 10. However, that changes nothing because he gets 11 minutes with these alarm parameters anyway.
In the third testcase Polycarp wakes up rested enough after the first alarm. Thus, the answer is b=9.
In the fourth testcase Polycarp wakes up after 5 minutes. Unfortunately, he keeps resetting his alarm infinitely being unable to rest for even a single minute :( | instruction | 0 | 65,738 | 24 | 131,476 |
Tags: math
Correct Solution:
```
for _ in range(int(input())):
import math
a, b, c, d = map(int, input().split())
if b >= a:
print(b)
else:
if (d >= c):
print("-1")
else:
x = a - b
y = c - d
t = math.ceil(x/y)
print(b + t*c)
``` | output | 1 | 65,738 | 24 | 131,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has spent the entire day preparing problems for you. Now he has to sleep for at least a minutes to feel refreshed.
Polycarp can only wake up by hearing the sound of his alarm. So he has just fallen asleep and his first alarm goes off in b minutes.
Every time Polycarp wakes up, he decides if he wants to sleep for some more time or not. If he's slept for less than a minutes in total, then he sets his alarm to go off in c minutes after it is reset and spends d minutes to fall asleep again. Otherwise, he gets out of his bed and proceeds with the day.
If the alarm goes off while Polycarp is falling asleep, then he resets his alarm to go off in another c minutes and tries to fall asleep for d minutes again.
You just want to find out when will Polycarp get out of his bed or report that it will never happen.
Please check out the notes for some explanations of the example.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of testcases.
The only line of each testcase contains four integers a, b, c, d (1 β€ a, b, c, d β€ 10^9) β the time Polycarp has to sleep for to feel refreshed, the time before the first alarm goes off, the time before every succeeding alarm goes off and the time Polycarp spends to fall asleep.
Output
For each test case print one integer. If Polycarp never gets out of his bed then print -1. Otherwise, print the time it takes for Polycarp to get out of his bed.
Example
Input
7
10 3 6 4
11 3 6 4
5 9 4 10
6 5 2 3
1 1 1 1
3947465 47342 338129 123123
234123843 13 361451236 361451000
Output
27
27
9
-1
1
6471793
358578060125049
Note
In the first testcase Polycarp wakes up after 3 minutes. He only rested for 3 minutes out of 10 minutes he needed. So after that he sets his alarm to go off in 6 minutes and spends 4 minutes falling asleep. Thus, he rests for 2 more minutes, totaling in 3+2=5 minutes of sleep. Then he repeats the procedure three more times and ends up with 11 minutes of sleep. Finally, he gets out of his bed. He spent 3 minutes before the first alarm and then reset his alarm four times. The answer is 3+4 β
6 = 27.
The second example is almost like the first one but Polycarp needs 11 minutes of sleep instead of 10. However, that changes nothing because he gets 11 minutes with these alarm parameters anyway.
In the third testcase Polycarp wakes up rested enough after the first alarm. Thus, the answer is b=9.
In the fourth testcase Polycarp wakes up after 5 minutes. Unfortunately, he keeps resetting his alarm infinitely being unable to rest for even a single minute :( | instruction | 0 | 65,739 | 24 | 131,478 |
Tags: math
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Fri Jul 31 22:43:03 2020
@author: Admin
"""
for _ in range(int(input())):
arr = list(map(int, input().rstrip().split()))[:4]
a,b,c,d,count= arr[0],arr[1],arr[2],arr[3],0
if a<=b:
print(b)
else:
if c<=d:
print(-1)
else:
y,x=a-b,c-d
if y%x!=0:
an = b+((int(y/x)+1)*c)
else:
an = b+(int(y/x)*c)
print(an)
``` | output | 1 | 65,739 | 24 | 131,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has spent the entire day preparing problems for you. Now he has to sleep for at least a minutes to feel refreshed.
Polycarp can only wake up by hearing the sound of his alarm. So he has just fallen asleep and his first alarm goes off in b minutes.
Every time Polycarp wakes up, he decides if he wants to sleep for some more time or not. If he's slept for less than a minutes in total, then he sets his alarm to go off in c minutes after it is reset and spends d minutes to fall asleep again. Otherwise, he gets out of his bed and proceeds with the day.
If the alarm goes off while Polycarp is falling asleep, then he resets his alarm to go off in another c minutes and tries to fall asleep for d minutes again.
You just want to find out when will Polycarp get out of his bed or report that it will never happen.
Please check out the notes for some explanations of the example.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of testcases.
The only line of each testcase contains four integers a, b, c, d (1 β€ a, b, c, d β€ 10^9) β the time Polycarp has to sleep for to feel refreshed, the time before the first alarm goes off, the time before every succeeding alarm goes off and the time Polycarp spends to fall asleep.
Output
For each test case print one integer. If Polycarp never gets out of his bed then print -1. Otherwise, print the time it takes for Polycarp to get out of his bed.
Example
Input
7
10 3 6 4
11 3 6 4
5 9 4 10
6 5 2 3
1 1 1 1
3947465 47342 338129 123123
234123843 13 361451236 361451000
Output
27
27
9
-1
1
6471793
358578060125049
Note
In the first testcase Polycarp wakes up after 3 minutes. He only rested for 3 minutes out of 10 minutes he needed. So after that he sets his alarm to go off in 6 minutes and spends 4 minutes falling asleep. Thus, he rests for 2 more minutes, totaling in 3+2=5 minutes of sleep. Then he repeats the procedure three more times and ends up with 11 minutes of sleep. Finally, he gets out of his bed. He spent 3 minutes before the first alarm and then reset his alarm four times. The answer is 3+4 β
6 = 27.
The second example is almost like the first one but Polycarp needs 11 minutes of sleep instead of 10. However, that changes nothing because he gets 11 minutes with these alarm parameters anyway.
In the third testcase Polycarp wakes up rested enough after the first alarm. Thus, the answer is b=9.
In the fourth testcase Polycarp wakes up after 5 minutes. Unfortunately, he keeps resetting his alarm infinitely being unable to rest for even a single minute :( | instruction | 0 | 65,740 | 24 | 131,480 |
Tags: math
Correct Solution:
```
for _ in range(int(input())):
l = list(map(int,input().split()))
a = l[0]
b = l[1]
c = l[2]
d = l[3]
if d>=c and a>b:
print(-1)
elif a<=b:
print(b)
else:
x = a-b
y = c-d
if x%y == 0:
count = int(x//y)
else:
count = int((x//y)+1)
r = b + (count*c)
print(r)
``` | output | 1 | 65,740 | 24 | 131,481 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has spent the entire day preparing problems for you. Now he has to sleep for at least a minutes to feel refreshed.
Polycarp can only wake up by hearing the sound of his alarm. So he has just fallen asleep and his first alarm goes off in b minutes.
Every time Polycarp wakes up, he decides if he wants to sleep for some more time or not. If he's slept for less than a minutes in total, then he sets his alarm to go off in c minutes after it is reset and spends d minutes to fall asleep again. Otherwise, he gets out of his bed and proceeds with the day.
If the alarm goes off while Polycarp is falling asleep, then he resets his alarm to go off in another c minutes and tries to fall asleep for d minutes again.
You just want to find out when will Polycarp get out of his bed or report that it will never happen.
Please check out the notes for some explanations of the example.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of testcases.
The only line of each testcase contains four integers a, b, c, d (1 β€ a, b, c, d β€ 10^9) β the time Polycarp has to sleep for to feel refreshed, the time before the first alarm goes off, the time before every succeeding alarm goes off and the time Polycarp spends to fall asleep.
Output
For each test case print one integer. If Polycarp never gets out of his bed then print -1. Otherwise, print the time it takes for Polycarp to get out of his bed.
Example
Input
7
10 3 6 4
11 3 6 4
5 9 4 10
6 5 2 3
1 1 1 1
3947465 47342 338129 123123
234123843 13 361451236 361451000
Output
27
27
9
-1
1
6471793
358578060125049
Note
In the first testcase Polycarp wakes up after 3 minutes. He only rested for 3 minutes out of 10 minutes he needed. So after that he sets his alarm to go off in 6 minutes and spends 4 minutes falling asleep. Thus, he rests for 2 more minutes, totaling in 3+2=5 minutes of sleep. Then he repeats the procedure three more times and ends up with 11 minutes of sleep. Finally, he gets out of his bed. He spent 3 minutes before the first alarm and then reset his alarm four times. The answer is 3+4 β
6 = 27.
The second example is almost like the first one but Polycarp needs 11 minutes of sleep instead of 10. However, that changes nothing because he gets 11 minutes with these alarm parameters anyway.
In the third testcase Polycarp wakes up rested enough after the first alarm. Thus, the answer is b=9.
In the fourth testcase Polycarp wakes up after 5 minutes. Unfortunately, he keeps resetting his alarm infinitely being unable to rest for even a single minute :( | instruction | 0 | 65,741 | 24 | 131,482 |
Tags: math
Correct Solution:
```
def run():
import sys
sys.stdin = open('/home/punit/Documents/Cpp Programs/input.txt', 'r')
sys.stdout = open('/home/punit/Documents/Cpp Programs/output.txt', 'w')
# run()
from math import ceil,sqrt,floor
for _ in range(int(input())):
a,b,c,d = map(int,input().split())
if(a<=b):
print(b)
elif(c<=d):
print(-1)
else:
Sum = b
x = int(ceil((a-b)/(c-d)))
x = x*c
Sum+=x
print(Sum)
``` | output | 1 | 65,741 | 24 | 131,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has spent the entire day preparing problems for you. Now he has to sleep for at least a minutes to feel refreshed.
Polycarp can only wake up by hearing the sound of his alarm. So he has just fallen asleep and his first alarm goes off in b minutes.
Every time Polycarp wakes up, he decides if he wants to sleep for some more time or not. If he's slept for less than a minutes in total, then he sets his alarm to go off in c minutes after it is reset and spends d minutes to fall asleep again. Otherwise, he gets out of his bed and proceeds with the day.
If the alarm goes off while Polycarp is falling asleep, then he resets his alarm to go off in another c minutes and tries to fall asleep for d minutes again.
You just want to find out when will Polycarp get out of his bed or report that it will never happen.
Please check out the notes for some explanations of the example.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of testcases.
The only line of each testcase contains four integers a, b, c, d (1 β€ a, b, c, d β€ 10^9) β the time Polycarp has to sleep for to feel refreshed, the time before the first alarm goes off, the time before every succeeding alarm goes off and the time Polycarp spends to fall asleep.
Output
For each test case print one integer. If Polycarp never gets out of his bed then print -1. Otherwise, print the time it takes for Polycarp to get out of his bed.
Example
Input
7
10 3 6 4
11 3 6 4
5 9 4 10
6 5 2 3
1 1 1 1
3947465 47342 338129 123123
234123843 13 361451236 361451000
Output
27
27
9
-1
1
6471793
358578060125049
Note
In the first testcase Polycarp wakes up after 3 minutes. He only rested for 3 minutes out of 10 minutes he needed. So after that he sets his alarm to go off in 6 minutes and spends 4 minutes falling asleep. Thus, he rests for 2 more minutes, totaling in 3+2=5 minutes of sleep. Then he repeats the procedure three more times and ends up with 11 minutes of sleep. Finally, he gets out of his bed. He spent 3 minutes before the first alarm and then reset his alarm four times. The answer is 3+4 β
6 = 27.
The second example is almost like the first one but Polycarp needs 11 minutes of sleep instead of 10. However, that changes nothing because he gets 11 minutes with these alarm parameters anyway.
In the third testcase Polycarp wakes up rested enough after the first alarm. Thus, the answer is b=9.
In the fourth testcase Polycarp wakes up after 5 minutes. Unfortunately, he keeps resetting his alarm infinitely being unable to rest for even a single minute :( | instruction | 0 | 65,742 | 24 | 131,484 |
Tags: math
Correct Solution:
```
t=int(input())
for _ in range(t):
a,b,c,d = map(int,input().split())
if a<=b:print(b)
else:
if c<=d:print('-1')
else:
r = (a - b)%(c - d)
p = (a - b)//(c - d)
if r==0:tot = b + p*c
else:tot = b + (p+1)*c
print(tot)
``` | output | 1 | 65,742 | 24 | 131,485 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has spent the entire day preparing problems for you. Now he has to sleep for at least a minutes to feel refreshed.
Polycarp can only wake up by hearing the sound of his alarm. So he has just fallen asleep and his first alarm goes off in b minutes.
Every time Polycarp wakes up, he decides if he wants to sleep for some more time or not. If he's slept for less than a minutes in total, then he sets his alarm to go off in c minutes after it is reset and spends d minutes to fall asleep again. Otherwise, he gets out of his bed and proceeds with the day.
If the alarm goes off while Polycarp is falling asleep, then he resets his alarm to go off in another c minutes and tries to fall asleep for d minutes again.
You just want to find out when will Polycarp get out of his bed or report that it will never happen.
Please check out the notes for some explanations of the example.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of testcases.
The only line of each testcase contains four integers a, b, c, d (1 β€ a, b, c, d β€ 10^9) β the time Polycarp has to sleep for to feel refreshed, the time before the first alarm goes off, the time before every succeeding alarm goes off and the time Polycarp spends to fall asleep.
Output
For each test case print one integer. If Polycarp never gets out of his bed then print -1. Otherwise, print the time it takes for Polycarp to get out of his bed.
Example
Input
7
10 3 6 4
11 3 6 4
5 9 4 10
6 5 2 3
1 1 1 1
3947465 47342 338129 123123
234123843 13 361451236 361451000
Output
27
27
9
-1
1
6471793
358578060125049
Note
In the first testcase Polycarp wakes up after 3 minutes. He only rested for 3 minutes out of 10 minutes he needed. So after that he sets his alarm to go off in 6 minutes and spends 4 minutes falling asleep. Thus, he rests for 2 more minutes, totaling in 3+2=5 minutes of sleep. Then he repeats the procedure three more times and ends up with 11 minutes of sleep. Finally, he gets out of his bed. He spent 3 minutes before the first alarm and then reset his alarm four times. The answer is 3+4 β
6 = 27.
The second example is almost like the first one but Polycarp needs 11 minutes of sleep instead of 10. However, that changes nothing because he gets 11 minutes with these alarm parameters anyway.
In the third testcase Polycarp wakes up rested enough after the first alarm. Thus, the answer is b=9.
In the fourth testcase Polycarp wakes up after 5 minutes. Unfortunately, he keeps resetting his alarm infinitely being unable to rest for even a single minute :( | instruction | 0 | 65,743 | 24 | 131,486 |
Tags: math
Correct Solution:
```
from math import ceil
for _ in range(int(input())):
a,b,c,d = map(int,input().split())
if a<=b:
print(b)
elif c<=d:
print('-1')
else:
left = a-b
slept = c-d
multi = ceil(left/slept)
ans = b + multi*c
print(ans)
``` | output | 1 | 65,743 | 24 | 131,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has spent the entire day preparing problems for you. Now he has to sleep for at least a minutes to feel refreshed.
Polycarp can only wake up by hearing the sound of his alarm. So he has just fallen asleep and his first alarm goes off in b minutes.
Every time Polycarp wakes up, he decides if he wants to sleep for some more time or not. If he's slept for less than a minutes in total, then he sets his alarm to go off in c minutes after it is reset and spends d minutes to fall asleep again. Otherwise, he gets out of his bed and proceeds with the day.
If the alarm goes off while Polycarp is falling asleep, then he resets his alarm to go off in another c minutes and tries to fall asleep for d minutes again.
You just want to find out when will Polycarp get out of his bed or report that it will never happen.
Please check out the notes for some explanations of the example.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of testcases.
The only line of each testcase contains four integers a, b, c, d (1 β€ a, b, c, d β€ 10^9) β the time Polycarp has to sleep for to feel refreshed, the time before the first alarm goes off, the time before every succeeding alarm goes off and the time Polycarp spends to fall asleep.
Output
For each test case print one integer. If Polycarp never gets out of his bed then print -1. Otherwise, print the time it takes for Polycarp to get out of his bed.
Example
Input
7
10 3 6 4
11 3 6 4
5 9 4 10
6 5 2 3
1 1 1 1
3947465 47342 338129 123123
234123843 13 361451236 361451000
Output
27
27
9
-1
1
6471793
358578060125049
Note
In the first testcase Polycarp wakes up after 3 minutes. He only rested for 3 minutes out of 10 minutes he needed. So after that he sets his alarm to go off in 6 minutes and spends 4 minutes falling asleep. Thus, he rests for 2 more minutes, totaling in 3+2=5 minutes of sleep. Then he repeats the procedure three more times and ends up with 11 minutes of sleep. Finally, he gets out of his bed. He spent 3 minutes before the first alarm and then reset his alarm four times. The answer is 3+4 β
6 = 27.
The second example is almost like the first one but Polycarp needs 11 minutes of sleep instead of 10. However, that changes nothing because he gets 11 minutes with these alarm parameters anyway.
In the third testcase Polycarp wakes up rested enough after the first alarm. Thus, the answer is b=9.
In the fourth testcase Polycarp wakes up after 5 minutes. Unfortunately, he keeps resetting his alarm infinitely being unable to rest for even a single minute :( | instruction | 0 | 65,744 | 24 | 131,488 |
Tags: math
Correct Solution:
```
t = int(input())
for i in range(t):
a, b, c, d = map(int, input().split())
if b < a:
if c <= d:
print(-1)
else:
print(b + (a-b+c-d-1)//(c-d)*c)
else:
print(b)
``` | output | 1 | 65,744 | 24 | 131,489 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has spent the entire day preparing problems for you. Now he has to sleep for at least a minutes to feel refreshed.
Polycarp can only wake up by hearing the sound of his alarm. So he has just fallen asleep and his first alarm goes off in b minutes.
Every time Polycarp wakes up, he decides if he wants to sleep for some more time or not. If he's slept for less than a minutes in total, then he sets his alarm to go off in c minutes after it is reset and spends d minutes to fall asleep again. Otherwise, he gets out of his bed and proceeds with the day.
If the alarm goes off while Polycarp is falling asleep, then he resets his alarm to go off in another c minutes and tries to fall asleep for d minutes again.
You just want to find out when will Polycarp get out of his bed or report that it will never happen.
Please check out the notes for some explanations of the example.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of testcases.
The only line of each testcase contains four integers a, b, c, d (1 β€ a, b, c, d β€ 10^9) β the time Polycarp has to sleep for to feel refreshed, the time before the first alarm goes off, the time before every succeeding alarm goes off and the time Polycarp spends to fall asleep.
Output
For each test case print one integer. If Polycarp never gets out of his bed then print -1. Otherwise, print the time it takes for Polycarp to get out of his bed.
Example
Input
7
10 3 6 4
11 3 6 4
5 9 4 10
6 5 2 3
1 1 1 1
3947465 47342 338129 123123
234123843 13 361451236 361451000
Output
27
27
9
-1
1
6471793
358578060125049
Note
In the first testcase Polycarp wakes up after 3 minutes. He only rested for 3 minutes out of 10 minutes he needed. So after that he sets his alarm to go off in 6 minutes and spends 4 minutes falling asleep. Thus, he rests for 2 more minutes, totaling in 3+2=5 minutes of sleep. Then he repeats the procedure three more times and ends up with 11 minutes of sleep. Finally, he gets out of his bed. He spent 3 minutes before the first alarm and then reset his alarm four times. The answer is 3+4 β
6 = 27.
The second example is almost like the first one but Polycarp needs 11 minutes of sleep instead of 10. However, that changes nothing because he gets 11 minutes with these alarm parameters anyway.
In the third testcase Polycarp wakes up rested enough after the first alarm. Thus, the answer is b=9.
In the fourth testcase Polycarp wakes up after 5 minutes. Unfortunately, he keeps resetting his alarm infinitely being unable to rest for even a single minute :( | instruction | 0 | 65,745 | 24 | 131,490 |
Tags: math
Correct Solution:
```
import math
t=int(input())
for _ in range(t):
a,b,c,d=map(int,input().split())
if a-b>0 and d>=c:
print(-1)
elif a-b<=0:
print(b)
else:
x=c-d
y=math.ceil((a-b)/x)
ans=b+y*c
print(ans)
``` | output | 1 | 65,745 | 24 | 131,491 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has spent the entire day preparing problems for you. Now he has to sleep for at least a minutes to feel refreshed.
Polycarp can only wake up by hearing the sound of his alarm. So he has just fallen asleep and his first alarm goes off in b minutes.
Every time Polycarp wakes up, he decides if he wants to sleep for some more time or not. If he's slept for less than a minutes in total, then he sets his alarm to go off in c minutes after it is reset and spends d minutes to fall asleep again. Otherwise, he gets out of his bed and proceeds with the day.
If the alarm goes off while Polycarp is falling asleep, then he resets his alarm to go off in another c minutes and tries to fall asleep for d minutes again.
You just want to find out when will Polycarp get out of his bed or report that it will never happen.
Please check out the notes for some explanations of the example.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of testcases.
The only line of each testcase contains four integers a, b, c, d (1 β€ a, b, c, d β€ 10^9) β the time Polycarp has to sleep for to feel refreshed, the time before the first alarm goes off, the time before every succeeding alarm goes off and the time Polycarp spends to fall asleep.
Output
For each test case print one integer. If Polycarp never gets out of his bed then print -1. Otherwise, print the time it takes for Polycarp to get out of his bed.
Example
Input
7
10 3 6 4
11 3 6 4
5 9 4 10
6 5 2 3
1 1 1 1
3947465 47342 338129 123123
234123843 13 361451236 361451000
Output
27
27
9
-1
1
6471793
358578060125049
Note
In the first testcase Polycarp wakes up after 3 minutes. He only rested for 3 minutes out of 10 minutes he needed. So after that he sets his alarm to go off in 6 minutes and spends 4 minutes falling asleep. Thus, he rests for 2 more minutes, totaling in 3+2=5 minutes of sleep. Then he repeats the procedure three more times and ends up with 11 minutes of sleep. Finally, he gets out of his bed. He spent 3 minutes before the first alarm and then reset his alarm four times. The answer is 3+4 β
6 = 27.
The second example is almost like the first one but Polycarp needs 11 minutes of sleep instead of 10. However, that changes nothing because he gets 11 minutes with these alarm parameters anyway.
In the third testcase Polycarp wakes up rested enough after the first alarm. Thus, the answer is b=9.
In the fourth testcase Polycarp wakes up after 5 minutes. Unfortunately, he keeps resetting his alarm infinitely being unable to rest for even a single minute :(
Submitted Solution:
```
from sys import stdin, stdout
from sys import maxsize
#input = stdin.readline().strip
from math import ceil
def solve():
pass
test = 1
test = int(input())
for t in range(0, test):
# brr = [list(map(int,input().split())) for i in range(rows)] # 2D array row-wise input
# n = int(input())
# s = list(input()) # String Input, converted to mutable list.
a,b,c,d = list(map(int, input().split()))
# arr = [int(x) for x in input().split()]
if(b>=a):
print(b)
else:
rest=a-b
try:
n=ceil(rest/(c-d))
except:
n=-1
if(n<=0):
print(-1)
else:
print(b+n*c)
ans = solve()
'''
rows, cols = (5, 5)
arr = [[0]*cols for j in range(rows)] # 2D array initialization
b=input().split() # list created by spliting about spaces
brr = [[int(b[cols*i+j]) for j in range(cols)] for i in range(rows)] # 2D array Linear Input
rows,cols=len(brr),len(brr[0]) # no of rows/cols for 2D array
arr.sort(key = lambda x : x[1]) # sort list of tuples by 2nd element, Default priority - 1st Element then 2nd Element
s=set() # empty set
a=maxsize # initializing infinity
b=-maxsize # initializing -infinity
mapped=list(map(function,input)) # to apply function to list element-wise
try: # Error handling
#code 1
except: # ex. to stop at EOF
#code 2 , if error occurs
'''
``` | instruction | 0 | 65,746 | 24 | 131,492 |
Yes | output | 1 | 65,746 | 24 | 131,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has spent the entire day preparing problems for you. Now he has to sleep for at least a minutes to feel refreshed.
Polycarp can only wake up by hearing the sound of his alarm. So he has just fallen asleep and his first alarm goes off in b minutes.
Every time Polycarp wakes up, he decides if he wants to sleep for some more time or not. If he's slept for less than a minutes in total, then he sets his alarm to go off in c minutes after it is reset and spends d minutes to fall asleep again. Otherwise, he gets out of his bed and proceeds with the day.
If the alarm goes off while Polycarp is falling asleep, then he resets his alarm to go off in another c minutes and tries to fall asleep for d minutes again.
You just want to find out when will Polycarp get out of his bed or report that it will never happen.
Please check out the notes for some explanations of the example.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of testcases.
The only line of each testcase contains four integers a, b, c, d (1 β€ a, b, c, d β€ 10^9) β the time Polycarp has to sleep for to feel refreshed, the time before the first alarm goes off, the time before every succeeding alarm goes off and the time Polycarp spends to fall asleep.
Output
For each test case print one integer. If Polycarp never gets out of his bed then print -1. Otherwise, print the time it takes for Polycarp to get out of his bed.
Example
Input
7
10 3 6 4
11 3 6 4
5 9 4 10
6 5 2 3
1 1 1 1
3947465 47342 338129 123123
234123843 13 361451236 361451000
Output
27
27
9
-1
1
6471793
358578060125049
Note
In the first testcase Polycarp wakes up after 3 minutes. He only rested for 3 minutes out of 10 minutes he needed. So after that he sets his alarm to go off in 6 minutes and spends 4 minutes falling asleep. Thus, he rests for 2 more minutes, totaling in 3+2=5 minutes of sleep. Then he repeats the procedure three more times and ends up with 11 minutes of sleep. Finally, he gets out of his bed. He spent 3 minutes before the first alarm and then reset his alarm four times. The answer is 3+4 β
6 = 27.
The second example is almost like the first one but Polycarp needs 11 minutes of sleep instead of 10. However, that changes nothing because he gets 11 minutes with these alarm parameters anyway.
In the third testcase Polycarp wakes up rested enough after the first alarm. Thus, the answer is b=9.
In the fourth testcase Polycarp wakes up after 5 minutes. Unfortunately, he keeps resetting his alarm infinitely being unable to rest for even a single minute :(
Submitted Solution:
```
t=int(input())
for _ in range(t):
a,b,c,d=map(int,input().split())
if b>=a:
print(b)
continue
else:
if d>=c:
print(-1)
continue
else:
to_sleep=a-b
per_cycle=c-d
cycles=to_sleep//per_cycle #here we may be 1 cycle behind due to //
to_sleep=to_sleep-cycles*(c-d)
if to_sleep!=0:
final=b+(cycles+1)*c#*(c)+d+to_sleep
else:
final=b+cycles*c
print(final)
``` | instruction | 0 | 65,747 | 24 | 131,494 |
Yes | output | 1 | 65,747 | 24 | 131,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has spent the entire day preparing problems for you. Now he has to sleep for at least a minutes to feel refreshed.
Polycarp can only wake up by hearing the sound of his alarm. So he has just fallen asleep and his first alarm goes off in b minutes.
Every time Polycarp wakes up, he decides if he wants to sleep for some more time or not. If he's slept for less than a minutes in total, then he sets his alarm to go off in c minutes after it is reset and spends d minutes to fall asleep again. Otherwise, he gets out of his bed and proceeds with the day.
If the alarm goes off while Polycarp is falling asleep, then he resets his alarm to go off in another c minutes and tries to fall asleep for d minutes again.
You just want to find out when will Polycarp get out of his bed or report that it will never happen.
Please check out the notes for some explanations of the example.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of testcases.
The only line of each testcase contains four integers a, b, c, d (1 β€ a, b, c, d β€ 10^9) β the time Polycarp has to sleep for to feel refreshed, the time before the first alarm goes off, the time before every succeeding alarm goes off and the time Polycarp spends to fall asleep.
Output
For each test case print one integer. If Polycarp never gets out of his bed then print -1. Otherwise, print the time it takes for Polycarp to get out of his bed.
Example
Input
7
10 3 6 4
11 3 6 4
5 9 4 10
6 5 2 3
1 1 1 1
3947465 47342 338129 123123
234123843 13 361451236 361451000
Output
27
27
9
-1
1
6471793
358578060125049
Note
In the first testcase Polycarp wakes up after 3 minutes. He only rested for 3 minutes out of 10 minutes he needed. So after that he sets his alarm to go off in 6 minutes and spends 4 minutes falling asleep. Thus, he rests for 2 more minutes, totaling in 3+2=5 minutes of sleep. Then he repeats the procedure three more times and ends up with 11 minutes of sleep. Finally, he gets out of his bed. He spent 3 minutes before the first alarm and then reset his alarm four times. The answer is 3+4 β
6 = 27.
The second example is almost like the first one but Polycarp needs 11 minutes of sleep instead of 10. However, that changes nothing because he gets 11 minutes with these alarm parameters anyway.
In the third testcase Polycarp wakes up rested enough after the first alarm. Thus, the answer is b=9.
In the fourth testcase Polycarp wakes up after 5 minutes. Unfortunately, he keeps resetting his alarm infinitely being unable to rest for even a single minute :(
Submitted Solution:
```
for i in range(int(input())):
a,b,c,d = map(int,input().split())
if a<=b:
print(b)
else:
if d>=c:
print(-1)
else:
q = (a-b)//(c-d)
if (a-b)%(c-d)!=0:
q+=1
print(b+q*c)
``` | instruction | 0 | 65,748 | 24 | 131,496 |
Yes | output | 1 | 65,748 | 24 | 131,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has spent the entire day preparing problems for you. Now he has to sleep for at least a minutes to feel refreshed.
Polycarp can only wake up by hearing the sound of his alarm. So he has just fallen asleep and his first alarm goes off in b minutes.
Every time Polycarp wakes up, he decides if he wants to sleep for some more time or not. If he's slept for less than a minutes in total, then he sets his alarm to go off in c minutes after it is reset and spends d minutes to fall asleep again. Otherwise, he gets out of his bed and proceeds with the day.
If the alarm goes off while Polycarp is falling asleep, then he resets his alarm to go off in another c minutes and tries to fall asleep for d minutes again.
You just want to find out when will Polycarp get out of his bed or report that it will never happen.
Please check out the notes for some explanations of the example.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of testcases.
The only line of each testcase contains four integers a, b, c, d (1 β€ a, b, c, d β€ 10^9) β the time Polycarp has to sleep for to feel refreshed, the time before the first alarm goes off, the time before every succeeding alarm goes off and the time Polycarp spends to fall asleep.
Output
For each test case print one integer. If Polycarp never gets out of his bed then print -1. Otherwise, print the time it takes for Polycarp to get out of his bed.
Example
Input
7
10 3 6 4
11 3 6 4
5 9 4 10
6 5 2 3
1 1 1 1
3947465 47342 338129 123123
234123843 13 361451236 361451000
Output
27
27
9
-1
1
6471793
358578060125049
Note
In the first testcase Polycarp wakes up after 3 minutes. He only rested for 3 minutes out of 10 minutes he needed. So after that he sets his alarm to go off in 6 minutes and spends 4 minutes falling asleep. Thus, he rests for 2 more minutes, totaling in 3+2=5 minutes of sleep. Then he repeats the procedure three more times and ends up with 11 minutes of sleep. Finally, he gets out of his bed. He spent 3 minutes before the first alarm and then reset his alarm four times. The answer is 3+4 β
6 = 27.
The second example is almost like the first one but Polycarp needs 11 minutes of sleep instead of 10. However, that changes nothing because he gets 11 minutes with these alarm parameters anyway.
In the third testcase Polycarp wakes up rested enough after the first alarm. Thus, the answer is b=9.
In the fourth testcase Polycarp wakes up after 5 minutes. Unfortunately, he keeps resetting his alarm infinitely being unable to rest for even a single minute :(
Submitted Solution:
```
from sys import stdin, stdout
import heapq
import cProfile
from collections import Counter, defaultdict, deque
from functools import reduce
import math
def get_int():
return int(stdin.readline().strip())
def get_tuple():
return map(int, stdin.readline().split())
def get_list():
return list(map(int, stdin.readline().split()))
def solve():
a,b,c,d = get_tuple()
if a<=b:
print(b)
elif a>b and c<=d:
print(-1)
else:
times = math.ceil((a-b)/(c-d))
print(b+c*times)
def main():
solve()
TestCases = True
if TestCases:
for i in range(get_int()):
main()
else:
main()
``` | instruction | 0 | 65,749 | 24 | 131,498 |
Yes | output | 1 | 65,749 | 24 | 131,499 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has spent the entire day preparing problems for you. Now he has to sleep for at least a minutes to feel refreshed.
Polycarp can only wake up by hearing the sound of his alarm. So he has just fallen asleep and his first alarm goes off in b minutes.
Every time Polycarp wakes up, he decides if he wants to sleep for some more time or not. If he's slept for less than a minutes in total, then he sets his alarm to go off in c minutes after it is reset and spends d minutes to fall asleep again. Otherwise, he gets out of his bed and proceeds with the day.
If the alarm goes off while Polycarp is falling asleep, then he resets his alarm to go off in another c minutes and tries to fall asleep for d minutes again.
You just want to find out when will Polycarp get out of his bed or report that it will never happen.
Please check out the notes for some explanations of the example.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of testcases.
The only line of each testcase contains four integers a, b, c, d (1 β€ a, b, c, d β€ 10^9) β the time Polycarp has to sleep for to feel refreshed, the time before the first alarm goes off, the time before every succeeding alarm goes off and the time Polycarp spends to fall asleep.
Output
For each test case print one integer. If Polycarp never gets out of his bed then print -1. Otherwise, print the time it takes for Polycarp to get out of his bed.
Example
Input
7
10 3 6 4
11 3 6 4
5 9 4 10
6 5 2 3
1 1 1 1
3947465 47342 338129 123123
234123843 13 361451236 361451000
Output
27
27
9
-1
1
6471793
358578060125049
Note
In the first testcase Polycarp wakes up after 3 minutes. He only rested for 3 minutes out of 10 minutes he needed. So after that he sets his alarm to go off in 6 minutes and spends 4 minutes falling asleep. Thus, he rests for 2 more minutes, totaling in 3+2=5 minutes of sleep. Then he repeats the procedure three more times and ends up with 11 minutes of sleep. Finally, he gets out of his bed. He spent 3 minutes before the first alarm and then reset his alarm four times. The answer is 3+4 β
6 = 27.
The second example is almost like the first one but Polycarp needs 11 minutes of sleep instead of 10. However, that changes nothing because he gets 11 minutes with these alarm parameters anyway.
In the third testcase Polycarp wakes up rested enough after the first alarm. Thus, the answer is b=9.
In the fourth testcase Polycarp wakes up after 5 minutes. Unfortunately, he keeps resetting his alarm infinitely being unable to rest for even a single minute :(
Submitted Solution:
```
for i in range(int(input())):
a,b,c,d = map(int,input().split())
if a>b:
if c>d:
e = (a-b)//(c-d) + (a-b)%(c-d)
print(b + e*d)
else:
print(-1)
else:
print(a)
``` | instruction | 0 | 65,750 | 24 | 131,500 |
No | output | 1 | 65,750 | 24 | 131,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has spent the entire day preparing problems for you. Now he has to sleep for at least a minutes to feel refreshed.
Polycarp can only wake up by hearing the sound of his alarm. So he has just fallen asleep and his first alarm goes off in b minutes.
Every time Polycarp wakes up, he decides if he wants to sleep for some more time or not. If he's slept for less than a minutes in total, then he sets his alarm to go off in c minutes after it is reset and spends d minutes to fall asleep again. Otherwise, he gets out of his bed and proceeds with the day.
If the alarm goes off while Polycarp is falling asleep, then he resets his alarm to go off in another c minutes and tries to fall asleep for d minutes again.
You just want to find out when will Polycarp get out of his bed or report that it will never happen.
Please check out the notes for some explanations of the example.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of testcases.
The only line of each testcase contains four integers a, b, c, d (1 β€ a, b, c, d β€ 10^9) β the time Polycarp has to sleep for to feel refreshed, the time before the first alarm goes off, the time before every succeeding alarm goes off and the time Polycarp spends to fall asleep.
Output
For each test case print one integer. If Polycarp never gets out of his bed then print -1. Otherwise, print the time it takes for Polycarp to get out of his bed.
Example
Input
7
10 3 6 4
11 3 6 4
5 9 4 10
6 5 2 3
1 1 1 1
3947465 47342 338129 123123
234123843 13 361451236 361451000
Output
27
27
9
-1
1
6471793
358578060125049
Note
In the first testcase Polycarp wakes up after 3 minutes. He only rested for 3 minutes out of 10 minutes he needed. So after that he sets his alarm to go off in 6 minutes and spends 4 minutes falling asleep. Thus, he rests for 2 more minutes, totaling in 3+2=5 minutes of sleep. Then he repeats the procedure three more times and ends up with 11 minutes of sleep. Finally, he gets out of his bed. He spent 3 minutes before the first alarm and then reset his alarm four times. The answer is 3+4 β
6 = 27.
The second example is almost like the first one but Polycarp needs 11 minutes of sleep instead of 10. However, that changes nothing because he gets 11 minutes with these alarm parameters anyway.
In the third testcase Polycarp wakes up rested enough after the first alarm. Thus, the answer is b=9.
In the fourth testcase Polycarp wakes up after 5 minutes. Unfortunately, he keeps resetting his alarm infinitely being unable to rest for even a single minute :(
Submitted Solution:
```
n=int(input())
for i in range(n):
l=list(map(int, input().split()))
if l[0]<=l[1]:
print(l[1])
else:
hrs=l[1]
rem=l[0]-l[1]
sleep=l[2]-l[3]
if sleep<=0:
print(-1)
else:
x=rem/sleep
y=0
if x.is_integer():
y=x
else:
y=int(x)+1
print(hrs+(l[2]*y))
``` | instruction | 0 | 65,751 | 24 | 131,502 |
No | output | 1 | 65,751 | 24 | 131,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has spent the entire day preparing problems for you. Now he has to sleep for at least a minutes to feel refreshed.
Polycarp can only wake up by hearing the sound of his alarm. So he has just fallen asleep and his first alarm goes off in b minutes.
Every time Polycarp wakes up, he decides if he wants to sleep for some more time or not. If he's slept for less than a minutes in total, then he sets his alarm to go off in c minutes after it is reset and spends d minutes to fall asleep again. Otherwise, he gets out of his bed and proceeds with the day.
If the alarm goes off while Polycarp is falling asleep, then he resets his alarm to go off in another c minutes and tries to fall asleep for d minutes again.
You just want to find out when will Polycarp get out of his bed or report that it will never happen.
Please check out the notes for some explanations of the example.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of testcases.
The only line of each testcase contains four integers a, b, c, d (1 β€ a, b, c, d β€ 10^9) β the time Polycarp has to sleep for to feel refreshed, the time before the first alarm goes off, the time before every succeeding alarm goes off and the time Polycarp spends to fall asleep.
Output
For each test case print one integer. If Polycarp never gets out of his bed then print -1. Otherwise, print the time it takes for Polycarp to get out of his bed.
Example
Input
7
10 3 6 4
11 3 6 4
5 9 4 10
6 5 2 3
1 1 1 1
3947465 47342 338129 123123
234123843 13 361451236 361451000
Output
27
27
9
-1
1
6471793
358578060125049
Note
In the first testcase Polycarp wakes up after 3 minutes. He only rested for 3 minutes out of 10 minutes he needed. So after that he sets his alarm to go off in 6 minutes and spends 4 minutes falling asleep. Thus, he rests for 2 more minutes, totaling in 3+2=5 minutes of sleep. Then he repeats the procedure three more times and ends up with 11 minutes of sleep. Finally, he gets out of his bed. He spent 3 minutes before the first alarm and then reset his alarm four times. The answer is 3+4 β
6 = 27.
The second example is almost like the first one but Polycarp needs 11 minutes of sleep instead of 10. However, that changes nothing because he gets 11 minutes with these alarm parameters anyway.
In the third testcase Polycarp wakes up rested enough after the first alarm. Thus, the answer is b=9.
In the fourth testcase Polycarp wakes up after 5 minutes. Unfortunately, he keeps resetting his alarm infinitely being unable to rest for even a single minute :(
Submitted Solution:
```
t = int(input())
c = 0
while c<=t:
a,b,c,d = map(int, input().split())
count = 0
slept = b
if a<=b:
print(b)
elif b<a and c>=d:
while slept < a:
slept += (c-d)
count += 1
total = b + count*c
print(total)
else:
print(-1)
c+=1
``` | instruction | 0 | 65,752 | 24 | 131,504 |
No | output | 1 | 65,752 | 24 | 131,505 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has spent the entire day preparing problems for you. Now he has to sleep for at least a minutes to feel refreshed.
Polycarp can only wake up by hearing the sound of his alarm. So he has just fallen asleep and his first alarm goes off in b minutes.
Every time Polycarp wakes up, he decides if he wants to sleep for some more time or not. If he's slept for less than a minutes in total, then he sets his alarm to go off in c minutes after it is reset and spends d minutes to fall asleep again. Otherwise, he gets out of his bed and proceeds with the day.
If the alarm goes off while Polycarp is falling asleep, then he resets his alarm to go off in another c minutes and tries to fall asleep for d minutes again.
You just want to find out when will Polycarp get out of his bed or report that it will never happen.
Please check out the notes for some explanations of the example.
Input
The first line contains one integer t (1 β€ t β€ 1000) β the number of testcases.
The only line of each testcase contains four integers a, b, c, d (1 β€ a, b, c, d β€ 10^9) β the time Polycarp has to sleep for to feel refreshed, the time before the first alarm goes off, the time before every succeeding alarm goes off and the time Polycarp spends to fall asleep.
Output
For each test case print one integer. If Polycarp never gets out of his bed then print -1. Otherwise, print the time it takes for Polycarp to get out of his bed.
Example
Input
7
10 3 6 4
11 3 6 4
5 9 4 10
6 5 2 3
1 1 1 1
3947465 47342 338129 123123
234123843 13 361451236 361451000
Output
27
27
9
-1
1
6471793
358578060125049
Note
In the first testcase Polycarp wakes up after 3 minutes. He only rested for 3 minutes out of 10 minutes he needed. So after that he sets his alarm to go off in 6 minutes and spends 4 minutes falling asleep. Thus, he rests for 2 more minutes, totaling in 3+2=5 minutes of sleep. Then he repeats the procedure three more times and ends up with 11 minutes of sleep. Finally, he gets out of his bed. He spent 3 minutes before the first alarm and then reset his alarm four times. The answer is 3+4 β
6 = 27.
The second example is almost like the first one but Polycarp needs 11 minutes of sleep instead of 10. However, that changes nothing because he gets 11 minutes with these alarm parameters anyway.
In the third testcase Polycarp wakes up rested enough after the first alarm. Thus, the answer is b=9.
In the fourth testcase Polycarp wakes up after 5 minutes. Unfortunately, he keeps resetting his alarm infinitely being unable to rest for even a single minute :(
Submitted Solution:
```
from sys import stdin
import math
def readline():
return stdin.readline()
tests = int(readline())
def solve(a, b, c, d):
if b >= a:
return b
if c == d:
m = 1
else:
m = math.ceil((a - b) / (c - d))
answer = b + c * m
if answer < a:
return -1
return answer
for t in range(0, tests):
#n = int(readline())
[a, b, c, d] = list(map(int, readline().split(' ')))
print(solve(a, b, c, d))
``` | instruction | 0 | 65,753 | 24 | 131,506 |
No | output | 1 | 65,753 | 24 | 131,507 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp is playing a game called "Running Over The Bridges". In this game he has to run over n bridges from the left to the right. Bridges are arranged one after the other, so the i-th bridge begins where the (i - 1)-th bridge ends.
You have the following data about bridges: li and ti β the length of the i-th bridge and the maximum allowed time which Polycarp can spend running over the i-th bridge. Thus, if Polycarp is in the beginning of the bridge i at the time T then he has to leave it at the time T + ti or earlier. It is allowed to reach the right end of a bridge exactly at the time T + ti.
Polycarp can run from the left side to the right one with speed 0.5, so he will run over a bridge with length s in time 2Β·s. Besides, he has several magical drinks. If he uses one drink, his speed increases twice (i.e. to value 1) for r seconds. All magical drinks are identical. Please note that Polycarp can use a drink only at integer moments of time, and he drinks it instantly and completely. Additionally, if Polycarp uses a drink at the moment T he can use the next drink not earlier than at the moment T + r.
What is the minimal number of drinks Polycarp has to use to run over all n bridges? If this number is not greater than 105, then you have to find out the moments of time when Polycarp has to use each magical drink.
Input
The first line contains two integers n and r (1 β€ n β€ 2Β·105, 1 β€ r β€ 1012) β the number of bridges and the duration of the effect of a magical drink.
The second line contains a sequence of integers l1, l2, ..., ln (1 β€ li β€ 5Β·106), where li is equal to the length of the i-th bridge.
The third line contains a sequence of integers t1, t2, ..., tn (1 β€ ti β€ 107), where ti is equal to the maximum allowed time which Polycarp can spend running over the i-th bridge.
Output
The first line of the output should contain k β the minimal number of drinks which Polycarp has to use, or -1 if there is no solution.
If the solution exists and the value of k is not greater than 105 then output k integers on the next line β moments of time from beginning of the game when Polycarp has to use drinks. Print the moments of time in chronological order. If there are several solutions, you can output any of them.
Examples
Input
1 3
7
10
Output
2
0 3
Input
3 3
3 3 3
3 3 2
Output
-1
Input
3 100000
5 5 5
5 7 8
Output
1
0
Input
4 1000
1 2 3 4
10 9 10 9
Output
0
Note
In the first case, there is only one bridge and it is clear that Polycarp cannot run over it without magical drinks. So, if he will use one magical drink on start (moment of time 0), and the second one β three seconds later (moment of time 3), he will be able to reach the end of the bridge in time. Please note, in this case there are several possible answers to the problem. For example, Polycarp can use the first drink at the moment of time 4 and the second one β at the moment of time 7.
In the second case, Polycarp cannot run over all bridges even if he will use magical drinks. So, answer in this case is -1.
In the fourth case, Polycarp can run over all bridges without magical drinks.
Submitted Solution:
```
n,r = map(int, input().split())
l = list(map(int, input().split()))
t = list(map(int, input().split()))
cas=0
ans=[]
ansc=0
potion=0
for i in range(n):
if 2*(max(l[i]-potion,0))+potion >t[i]:
if l[i]>t[i]:
print(-1)
exit()
else:
b=cas+potion
pc=0
while(l[i]-max(l[i]-(b-cas),0)+2*max(l[i]-(b-cas),0)>t[i]):
#print(b,cas)
ans.append(b)
b+=r
pc+=1
potion=max(0,b-t[i])
cas=min( t[i] if b>t[i] else cas+2*(l[i]-(b-cas))+(b-cas),b)
else:
cas=cas+2*(max(l[i]-potion,0))
potion=max(potion-2*l[i],0)
print(len(ans))
print(*ans)
``` | instruction | 0 | 66,081 | 24 | 132,162 |
No | output | 1 | 66,081 | 24 | 132,163 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp is playing a game called "Running Over The Bridges". In this game he has to run over n bridges from the left to the right. Bridges are arranged one after the other, so the i-th bridge begins where the (i - 1)-th bridge ends.
You have the following data about bridges: li and ti β the length of the i-th bridge and the maximum allowed time which Polycarp can spend running over the i-th bridge. Thus, if Polycarp is in the beginning of the bridge i at the time T then he has to leave it at the time T + ti or earlier. It is allowed to reach the right end of a bridge exactly at the time T + ti.
Polycarp can run from the left side to the right one with speed 0.5, so he will run over a bridge with length s in time 2Β·s. Besides, he has several magical drinks. If he uses one drink, his speed increases twice (i.e. to value 1) for r seconds. All magical drinks are identical. Please note that Polycarp can use a drink only at integer moments of time, and he drinks it instantly and completely. Additionally, if Polycarp uses a drink at the moment T he can use the next drink not earlier than at the moment T + r.
What is the minimal number of drinks Polycarp has to use to run over all n bridges? If this number is not greater than 105, then you have to find out the moments of time when Polycarp has to use each magical drink.
Input
The first line contains two integers n and r (1 β€ n β€ 2Β·105, 1 β€ r β€ 1012) β the number of bridges and the duration of the effect of a magical drink.
The second line contains a sequence of integers l1, l2, ..., ln (1 β€ li β€ 5Β·106), where li is equal to the length of the i-th bridge.
The third line contains a sequence of integers t1, t2, ..., tn (1 β€ ti β€ 107), where ti is equal to the maximum allowed time which Polycarp can spend running over the i-th bridge.
Output
The first line of the output should contain k β the minimal number of drinks which Polycarp has to use, or -1 if there is no solution.
If the solution exists and the value of k is not greater than 105 then output k integers on the next line β moments of time from beginning of the game when Polycarp has to use drinks. Print the moments of time in chronological order. If there are several solutions, you can output any of them.
Examples
Input
1 3
7
10
Output
2
0 3
Input
3 3
3 3 3
3 3 2
Output
-1
Input
3 100000
5 5 5
5 7 8
Output
1
0
Input
4 1000
1 2 3 4
10 9 10 9
Output
0
Note
In the first case, there is only one bridge and it is clear that Polycarp cannot run over it without magical drinks. So, if he will use one magical drink on start (moment of time 0), and the second one β three seconds later (moment of time 3), he will be able to reach the end of the bridge in time. Please note, in this case there are several possible answers to the problem. For example, Polycarp can use the first drink at the moment of time 4 and the second one β at the moment of time 7.
In the second case, Polycarp cannot run over all bridges even if he will use magical drinks. So, answer in this case is -1.
In the fourth case, Polycarp can run over all bridges without magical drinks.
Submitted Solution:
```
n,r = map(int, input().split())
l = list(map(int, input().split()))
t = list(map(int, input().split()))
cas=0
ans=[]
ansc=0
potion=0
for i in range(n):
if 2*(max(l[i]-potion,0))+potion >t[i]:
if l[i]>t[i]:
print(-1)
exit()
else:
b=cas+potion
pc=0
while(l[i]-max(l[i]-(b-cas),0)+2*max(l[i]-(b-cas),0)>t[i]):
#print(l[i]-max(l[i]-(b-cas),0)+2*max(l[i]-(b-cas),0),t[i])
#print(b,cas)
ans.append(b)
b+=r
pc+=1
potion=max(0,b-t[i])
cas=l[i]-max(l[i]-(b-cas),0)+2*max(l[i]-(b-cas),0)
else:
cas=cas+l[i]-max(l[i]-potion,0)+2*max(l[i]-potion,0)
potion=max(potion-l[i],0)
#print(i,cas)
print(len(ans))
print(*ans)
``` | instruction | 0 | 66,082 | 24 | 132,164 |
No | output | 1 | 66,082 | 24 | 132,165 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the educational reform Polycarp studies only two subjects at school, Safety Studies and PE (Physical Education). During the long months of the fourth term, he received n marks in them. When teachers wrote a mark in the journal, they didn't write in what subject the mark was for, they just wrote the mark.
Now it's time to show the journal to his strict parents. Polycarp knows that recently at the Parent Meeting the parents were told that he received a Safety Studies marks and b PE marks (a + b = n). Now Polycarp wants to write a subject's name in front of each mark so that:
* there are exactly a Safety Studies marks,
* there are exactly b PE marks,
* the total average score in both subjects is maximum.
An average subject grade is the sum of all marks in it, divided by the number of them. Of course, the division is performed in real numbers without rounding up or down. Polycarp aims to maximize the x1 + x2, where x1 is the average score in the first subject (Safety Studies), and x2 is the average score in the second one (Physical Education).
Input
The first line contains an integer n (2 β€ n β€ 105), n is the number of marks in Polycarp's Journal. The second line contains two positive integers a, b (1 β€ a, b β€ n - 1, a + b = n). The third line contains a sequence of integers t1, t2, ..., tn (1 β€ ti β€ 5), they are Polycarp's marks.
Output
Print the sequence of integers f1, f2, ..., fn, where fi (1 β€ fi β€ 2) is the number of a subject to which the i-th mark should be attributed. If there are several possible solutions, then print such that the sequence f1, f2, ..., fn is the smallest lexicographically.
The sequence p1, p2, ..., pn is lexicographically less than q1, q2, ..., qn if there exists such j (1 β€ j β€ n) that pi = qi for all 1 β€ i < j, Π°nd pj < qj.
Examples
Input
5
3 2
4 4 5 4 4
Output
1 1 2 1 2
Input
4
2 2
3 5 4 5
Output
1 1 2 2
Input
6
1 5
4 4 4 5 4 4
Output
2 2 2 1 2 2
Note
In the first sample the average score in the first subject is equal to 4, and in the second one β to 4.5. The total average score is 8.5. | instruction | 0 | 66,114 | 24 | 132,228 |
Tags: greedy, math, sortings
Correct Solution:
```
n=int(input())
a,b = map(int,input().split())
l = list(map(int,input().split()))
arr =[(l[i],i) for i in range(n) ]
if a<b:
arr.sort(key= lambda x:(-x[0] , x[1]))
else:
arr.sort(key= lambda x:(-x[0], -x[1]))
ans=[0]*n
if a<b:
for i in range(a):
ans[arr[i][1]]=1
for i in range(a,a+b):
ans[arr[i][1]]=2
elif a==b:
for i in range(a):
ans[i]=1
for i in range(a,a+a):
ans[i]=2
else:
for i in range(b):
ans[arr[i][1]]=2
for i in range(b,b+a):
ans[arr[i][1]]=1
print(*ans)
``` | output | 1 | 66,114 | 24 | 132,229 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the educational reform Polycarp studies only two subjects at school, Safety Studies and PE (Physical Education). During the long months of the fourth term, he received n marks in them. When teachers wrote a mark in the journal, they didn't write in what subject the mark was for, they just wrote the mark.
Now it's time to show the journal to his strict parents. Polycarp knows that recently at the Parent Meeting the parents were told that he received a Safety Studies marks and b PE marks (a + b = n). Now Polycarp wants to write a subject's name in front of each mark so that:
* there are exactly a Safety Studies marks,
* there are exactly b PE marks,
* the total average score in both subjects is maximum.
An average subject grade is the sum of all marks in it, divided by the number of them. Of course, the division is performed in real numbers without rounding up or down. Polycarp aims to maximize the x1 + x2, where x1 is the average score in the first subject (Safety Studies), and x2 is the average score in the second one (Physical Education).
Input
The first line contains an integer n (2 β€ n β€ 105), n is the number of marks in Polycarp's Journal. The second line contains two positive integers a, b (1 β€ a, b β€ n - 1, a + b = n). The third line contains a sequence of integers t1, t2, ..., tn (1 β€ ti β€ 5), they are Polycarp's marks.
Output
Print the sequence of integers f1, f2, ..., fn, where fi (1 β€ fi β€ 2) is the number of a subject to which the i-th mark should be attributed. If there are several possible solutions, then print such that the sequence f1, f2, ..., fn is the smallest lexicographically.
The sequence p1, p2, ..., pn is lexicographically less than q1, q2, ..., qn if there exists such j (1 β€ j β€ n) that pi = qi for all 1 β€ i < j, Π°nd pj < qj.
Examples
Input
5
3 2
4 4 5 4 4
Output
1 1 2 1 2
Input
4
2 2
3 5 4 5
Output
1 1 2 2
Input
6
1 5
4 4 4 5 4 4
Output
2 2 2 1 2 2
Note
In the first sample the average score in the first subject is equal to 4, and in the second one β to 4.5. The total average score is 8.5. | instruction | 0 | 66,115 | 24 | 132,230 |
Tags: greedy, math, sortings
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
import threading
threading.stack_size(10**8)
sys.setrecursionlimit(300000)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
#-------------------------------------------------------------------------
mod=10**9+7
n=int(input())
a,b=map(int,input().split())
l=list(map(int,input().split()))
arr = [(l[i], i) for i in range(n)]
if a <b:
arr.sort(key = lambda x : (-x[0],x[1]))
else:
arr.sort(key = lambda x : (-x[0],-x[1]))
# if a > b:
# a,b = b,a
ans = [0]*n
# print(arr)
if a <b:
for i in range(a):
ans[arr[i][1]] = 1
for i in range(a, a+b):
ans[arr[i][1]] = 2
elif a == b:
for i in range(a):
ans[i] = 1
for i in range(a, a+a):
ans[i] = 2
else:
for i in range(b):
ans[arr[i][1]] = 2
for i in range(b, a+b):
ans[arr[i][1]] = 1
print(*ans)
``` | output | 1 | 66,115 | 24 | 132,231 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the educational reform Polycarp studies only two subjects at school, Safety Studies and PE (Physical Education). During the long months of the fourth term, he received n marks in them. When teachers wrote a mark in the journal, they didn't write in what subject the mark was for, they just wrote the mark.
Now it's time to show the journal to his strict parents. Polycarp knows that recently at the Parent Meeting the parents were told that he received a Safety Studies marks and b PE marks (a + b = n). Now Polycarp wants to write a subject's name in front of each mark so that:
* there are exactly a Safety Studies marks,
* there are exactly b PE marks,
* the total average score in both subjects is maximum.
An average subject grade is the sum of all marks in it, divided by the number of them. Of course, the division is performed in real numbers without rounding up or down. Polycarp aims to maximize the x1 + x2, where x1 is the average score in the first subject (Safety Studies), and x2 is the average score in the second one (Physical Education).
Input
The first line contains an integer n (2 β€ n β€ 105), n is the number of marks in Polycarp's Journal. The second line contains two positive integers a, b (1 β€ a, b β€ n - 1, a + b = n). The third line contains a sequence of integers t1, t2, ..., tn (1 β€ ti β€ 5), they are Polycarp's marks.
Output
Print the sequence of integers f1, f2, ..., fn, where fi (1 β€ fi β€ 2) is the number of a subject to which the i-th mark should be attributed. If there are several possible solutions, then print such that the sequence f1, f2, ..., fn is the smallest lexicographically.
The sequence p1, p2, ..., pn is lexicographically less than q1, q2, ..., qn if there exists such j (1 β€ j β€ n) that pi = qi for all 1 β€ i < j, Π°nd pj < qj.
Examples
Input
5
3 2
4 4 5 4 4
Output
1 1 2 1 2
Input
4
2 2
3 5 4 5
Output
1 1 2 2
Input
6
1 5
4 4 4 5 4 4
Output
2 2 2 1 2 2
Note
In the first sample the average score in the first subject is equal to 4, and in the second one β to 4.5. The total average score is 8.5. | instruction | 0 | 66,116 | 24 | 132,232 |
Tags: greedy, math, sortings
Correct Solution:
```
n = int(input())
a, b = map(int, input().split())
if a == b: print('1 ' * a + '2 ' * b)
else:
t = [[] for i in range(6)]
for i, j in enumerate(map(int, input().split())): t[j].append(i)
if b < a:
t = t[1] + t[2] + t[3] + t[4] + t[5]
t.reverse()
p = ['1'] * n
for i in range(b): p[t[i]] = '2'
print(' '.join(p))
else:
t = t[5] + t[4] + t[3] + t[2] + t[1]
p = ['2'] * n
for i in range(a): p[t[i]] = '1'
print(' '.join(p))
``` | output | 1 | 66,116 | 24 | 132,233 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the educational reform Polycarp studies only two subjects at school, Safety Studies and PE (Physical Education). During the long months of the fourth term, he received n marks in them. When teachers wrote a mark in the journal, they didn't write in what subject the mark was for, they just wrote the mark.
Now it's time to show the journal to his strict parents. Polycarp knows that recently at the Parent Meeting the parents were told that he received a Safety Studies marks and b PE marks (a + b = n). Now Polycarp wants to write a subject's name in front of each mark so that:
* there are exactly a Safety Studies marks,
* there are exactly b PE marks,
* the total average score in both subjects is maximum.
An average subject grade is the sum of all marks in it, divided by the number of them. Of course, the division is performed in real numbers without rounding up or down. Polycarp aims to maximize the x1 + x2, where x1 is the average score in the first subject (Safety Studies), and x2 is the average score in the second one (Physical Education).
Input
The first line contains an integer n (2 β€ n β€ 105), n is the number of marks in Polycarp's Journal. The second line contains two positive integers a, b (1 β€ a, b β€ n - 1, a + b = n). The third line contains a sequence of integers t1, t2, ..., tn (1 β€ ti β€ 5), they are Polycarp's marks.
Output
Print the sequence of integers f1, f2, ..., fn, where fi (1 β€ fi β€ 2) is the number of a subject to which the i-th mark should be attributed. If there are several possible solutions, then print such that the sequence f1, f2, ..., fn is the smallest lexicographically.
The sequence p1, p2, ..., pn is lexicographically less than q1, q2, ..., qn if there exists such j (1 β€ j β€ n) that pi = qi for all 1 β€ i < j, Π°nd pj < qj.
Examples
Input
5
3 2
4 4 5 4 4
Output
1 1 2 1 2
Input
4
2 2
3 5 4 5
Output
1 1 2 2
Input
6
1 5
4 4 4 5 4 4
Output
2 2 2 1 2 2
Note
In the first sample the average score in the first subject is equal to 4, and in the second one β to 4.5. The total average score is 8.5. | instruction | 0 | 66,117 | 24 | 132,234 |
Tags: greedy, math, sortings
Correct Solution:
```
import sys
from itertools import *
from math import *
def solve():
n = int(input())
a, b = map(int, input().split())
if a == b:
for i in range(a): print(1, end = ' ')
for i in range(a): print(2, end = ' ')
return
first = 2
if a > b:
a, b = b, a
first = 1
l = list(map(int, input().split()))
lobjs = [(i, val) for i, val in enumerate(l)]
lobjs.sort(key = lambda j: (j[1], j[0] if first == 1 else -j[0]), reverse = True)
res = [0] * n
for index, obj in enumerate(lobjs):
if index < a: res[obj[0]] = 3 - first
else: res[obj[0]] = first
print(' '.join(map(str, res)))
if sys.hexversion == 50594544 : sys.stdin = open("test.txt")
solve()
``` | output | 1 | 66,117 | 24 | 132,235 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the educational reform Polycarp studies only two subjects at school, Safety Studies and PE (Physical Education). During the long months of the fourth term, he received n marks in them. When teachers wrote a mark in the journal, they didn't write in what subject the mark was for, they just wrote the mark.
Now it's time to show the journal to his strict parents. Polycarp knows that recently at the Parent Meeting the parents were told that he received a Safety Studies marks and b PE marks (a + b = n). Now Polycarp wants to write a subject's name in front of each mark so that:
* there are exactly a Safety Studies marks,
* there are exactly b PE marks,
* the total average score in both subjects is maximum.
An average subject grade is the sum of all marks in it, divided by the number of them. Of course, the division is performed in real numbers without rounding up or down. Polycarp aims to maximize the x1 + x2, where x1 is the average score in the first subject (Safety Studies), and x2 is the average score in the second one (Physical Education).
Input
The first line contains an integer n (2 β€ n β€ 105), n is the number of marks in Polycarp's Journal. The second line contains two positive integers a, b (1 β€ a, b β€ n - 1, a + b = n). The third line contains a sequence of integers t1, t2, ..., tn (1 β€ ti β€ 5), they are Polycarp's marks.
Output
Print the sequence of integers f1, f2, ..., fn, where fi (1 β€ fi β€ 2) is the number of a subject to which the i-th mark should be attributed. If there are several possible solutions, then print such that the sequence f1, f2, ..., fn is the smallest lexicographically.
The sequence p1, p2, ..., pn is lexicographically less than q1, q2, ..., qn if there exists such j (1 β€ j β€ n) that pi = qi for all 1 β€ i < j, Π°nd pj < qj.
Examples
Input
5
3 2
4 4 5 4 4
Output
1 1 2 1 2
Input
4
2 2
3 5 4 5
Output
1 1 2 2
Input
6
1 5
4 4 4 5 4 4
Output
2 2 2 1 2 2
Note
In the first sample the average score in the first subject is equal to 4, and in the second one β to 4.5. The total average score is 8.5. | instruction | 0 | 66,118 | 24 | 132,236 |
Tags: greedy, math, sortings
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 10**9+7
Ri = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
n = int(ri())
a,b = Ri()
arr = Ri()
arr = [(arr[i], i) for i in range(n)]
if a <b:
arr.sort(key = lambda x : (-x[0],x[1]))
else:
arr.sort(key = lambda x : (-x[0],-x[1]))
# if a > b:
# a,b = b,a
ans = [0]*n
# print(arr)
if a <b:
for i in range(a):
ans[arr[i][1]] = 1
for i in range(a, a+b):
ans[arr[i][1]] = 2
elif a == b:
for i in range(a):
ans[i] = 1
for i in range(a, a+a):
ans[i] = 2
else:
for i in range(b):
ans[arr[i][1]] = 2
for i in range(b, a+b):
ans[arr[i][1]] = 1
print(*ans)
``` | output | 1 | 66,118 | 24 | 132,237 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the educational reform Polycarp studies only two subjects at school, Safety Studies and PE (Physical Education). During the long months of the fourth term, he received n marks in them. When teachers wrote a mark in the journal, they didn't write in what subject the mark was for, they just wrote the mark.
Now it's time to show the journal to his strict parents. Polycarp knows that recently at the Parent Meeting the parents were told that he received a Safety Studies marks and b PE marks (a + b = n). Now Polycarp wants to write a subject's name in front of each mark so that:
* there are exactly a Safety Studies marks,
* there are exactly b PE marks,
* the total average score in both subjects is maximum.
An average subject grade is the sum of all marks in it, divided by the number of them. Of course, the division is performed in real numbers without rounding up or down. Polycarp aims to maximize the x1 + x2, where x1 is the average score in the first subject (Safety Studies), and x2 is the average score in the second one (Physical Education).
Input
The first line contains an integer n (2 β€ n β€ 105), n is the number of marks in Polycarp's Journal. The second line contains two positive integers a, b (1 β€ a, b β€ n - 1, a + b = n). The third line contains a sequence of integers t1, t2, ..., tn (1 β€ ti β€ 5), they are Polycarp's marks.
Output
Print the sequence of integers f1, f2, ..., fn, where fi (1 β€ fi β€ 2) is the number of a subject to which the i-th mark should be attributed. If there are several possible solutions, then print such that the sequence f1, f2, ..., fn is the smallest lexicographically.
The sequence p1, p2, ..., pn is lexicographically less than q1, q2, ..., qn if there exists such j (1 β€ j β€ n) that pi = qi for all 1 β€ i < j, Π°nd pj < qj.
Examples
Input
5
3 2
4 4 5 4 4
Output
1 1 2 1 2
Input
4
2 2
3 5 4 5
Output
1 1 2 2
Input
6
1 5
4 4 4 5 4 4
Output
2 2 2 1 2 2
Note
In the first sample the average score in the first subject is equal to 4, and in the second one β to 4.5. The total average score is 8.5. | instruction | 0 | 66,119 | 24 | 132,238 |
Tags: greedy, math, sortings
Correct Solution:
```
n=int(input())
a,b=list(map(int,input().split()))
arr=list(map(int,input().split()))
if a==b:
print("1 "*a +"2 "*(a-1)+ "2")
else:
arr=[(j,i) for i,j in enumerate(arr)]
if a<b:
ans=[1]*(n)
arr.sort(key=lambda x :(x[0],-x[1]))
for i in range(b):
ans[arr[i][1]]=2
else:
ans=[2]*n
arr.sort(key=lambda x :(x[0]))
for i in range(a):
ans[arr[i][1]]=1
print((" ").join(map(str,ans)))
``` | output | 1 | 66,119 | 24 | 132,239 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the educational reform Polycarp studies only two subjects at school, Safety Studies and PE (Physical Education). During the long months of the fourth term, he received n marks in them. When teachers wrote a mark in the journal, they didn't write in what subject the mark was for, they just wrote the mark.
Now it's time to show the journal to his strict parents. Polycarp knows that recently at the Parent Meeting the parents were told that he received a Safety Studies marks and b PE marks (a + b = n). Now Polycarp wants to write a subject's name in front of each mark so that:
* there are exactly a Safety Studies marks,
* there are exactly b PE marks,
* the total average score in both subjects is maximum.
An average subject grade is the sum of all marks in it, divided by the number of them. Of course, the division is performed in real numbers without rounding up or down. Polycarp aims to maximize the x1 + x2, where x1 is the average score in the first subject (Safety Studies), and x2 is the average score in the second one (Physical Education).
Input
The first line contains an integer n (2 β€ n β€ 105), n is the number of marks in Polycarp's Journal. The second line contains two positive integers a, b (1 β€ a, b β€ n - 1, a + b = n). The third line contains a sequence of integers t1, t2, ..., tn (1 β€ ti β€ 5), they are Polycarp's marks.
Output
Print the sequence of integers f1, f2, ..., fn, where fi (1 β€ fi β€ 2) is the number of a subject to which the i-th mark should be attributed. If there are several possible solutions, then print such that the sequence f1, f2, ..., fn is the smallest lexicographically.
The sequence p1, p2, ..., pn is lexicographically less than q1, q2, ..., qn if there exists such j (1 β€ j β€ n) that pi = qi for all 1 β€ i < j, Π°nd pj < qj.
Examples
Input
5
3 2
4 4 5 4 4
Output
1 1 2 1 2
Input
4
2 2
3 5 4 5
Output
1 1 2 2
Input
6
1 5
4 4 4 5 4 4
Output
2 2 2 1 2 2
Note
In the first sample the average score in the first subject is equal to 4, and in the second one β to 4.5. The total average score is 8.5. | instruction | 0 | 66,120 | 24 | 132,240 |
Tags: greedy, math, sortings
Correct Solution:
```
import random, math
from copy import deepcopy as dc
# To Genrate Random Number for Test-Cases
def randomNumber(s, e):
return random.randint(s, e)
# To Generate Random Array for Test-Cases
def randomArray(s, e, s_size, e_size):
size = random.randint(s_size, e_size)
arr = [randomNumber(s, e) for i in range(size)]
return arr
# To Generate Question Specific Test-Cases
def testcase():
pass
# Brute Force Approach to check Solution
def brute():
pass
# Efficient Approach for problem
def effe():
pass
# Function to call the actual solution
def solution(li, a, b):
li1 = [[li[i], i] for i in range(len(li))]
n = len(li)
li3 = sorted(li1, key = lambda x: (x[0], x[1]))
li1 = sorted(li1, key = lambda x: (x[0], n-x[1]))
li2 = [0 for i in range(len(li))]
# print(li1)
if a < b:
for i in range(len(li) - a):
li2[li1[i][1]] = 2
for i in range(len(li)-a, len(li)):
li2[li1[i][1]] = 1
elif a == b:
flag = True
for i in range(len(li)-a):
li2[i] = 1
for i in range(len(li)-a, len(li)):
li2[i] = 2
else:
for i in range(len(li) - b):
li2[li3[i][1]] = 1
for i in range(len(li)-b, len(li)):
li2[li3[i][1]] = 2
return li2
# Function to take input
def input_test():
n = int(input())
a, b = map(int, input().strip().split(" "))
li = list(map(int, input().strip().split(" ")))
out = solution(li, a, b)
print(' '.join(list(map(str, out))))
# Function to check test my code
def test():
pass
input_test()
# test()
``` | output | 1 | 66,120 | 24 | 132,241 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After the educational reform Polycarp studies only two subjects at school, Safety Studies and PE (Physical Education). During the long months of the fourth term, he received n marks in them. When teachers wrote a mark in the journal, they didn't write in what subject the mark was for, they just wrote the mark.
Now it's time to show the journal to his strict parents. Polycarp knows that recently at the Parent Meeting the parents were told that he received a Safety Studies marks and b PE marks (a + b = n). Now Polycarp wants to write a subject's name in front of each mark so that:
* there are exactly a Safety Studies marks,
* there are exactly b PE marks,
* the total average score in both subjects is maximum.
An average subject grade is the sum of all marks in it, divided by the number of them. Of course, the division is performed in real numbers without rounding up or down. Polycarp aims to maximize the x1 + x2, where x1 is the average score in the first subject (Safety Studies), and x2 is the average score in the second one (Physical Education).
Input
The first line contains an integer n (2 β€ n β€ 105), n is the number of marks in Polycarp's Journal. The second line contains two positive integers a, b (1 β€ a, b β€ n - 1, a + b = n). The third line contains a sequence of integers t1, t2, ..., tn (1 β€ ti β€ 5), they are Polycarp's marks.
Output
Print the sequence of integers f1, f2, ..., fn, where fi (1 β€ fi β€ 2) is the number of a subject to which the i-th mark should be attributed. If there are several possible solutions, then print such that the sequence f1, f2, ..., fn is the smallest lexicographically.
The sequence p1, p2, ..., pn is lexicographically less than q1, q2, ..., qn if there exists such j (1 β€ j β€ n) that pi = qi for all 1 β€ i < j, Π°nd pj < qj.
Examples
Input
5
3 2
4 4 5 4 4
Output
1 1 2 1 2
Input
4
2 2
3 5 4 5
Output
1 1 2 2
Input
6
1 5
4 4 4 5 4 4
Output
2 2 2 1 2 2
Note
In the first sample the average score in the first subject is equal to 4, and in the second one β to 4.5. The total average score is 8.5. | instruction | 0 | 66,121 | 24 | 132,242 |
Tags: greedy, math, sortings
Correct Solution:
```
n = int(input())
a, b = map(int, input().split())
c = [int(i) for i in input().split()]
p = [i for i in range(n)]
Z = [x for _,x in sorted(zip(c, p))]
ans = [0] * n
if a == b:
for i in range(a):
print(1, end=' ')
for i in range(b):
print(2, end=' ')
exit()
if a > b:
Z = [x for _, x in sorted(zip(c, p))]
for i in range(a):
ans[Z[i]] = 1
for i in range(n):
if ans[i] == 0:
ans[i] = 2
print(*ans)
if a < b:
for i in range(n):
c[i] *= -1
S = [x for _, x in sorted(zip(c, p))]
for i in range(a):
ans[S[i]] = 1
for i in range(n):
if ans[i] == 0:
ans[i] = 2
print(*ans)
``` | output | 1 | 66,121 | 24 | 132,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the educational reform Polycarp studies only two subjects at school, Safety Studies and PE (Physical Education). During the long months of the fourth term, he received n marks in them. When teachers wrote a mark in the journal, they didn't write in what subject the mark was for, they just wrote the mark.
Now it's time to show the journal to his strict parents. Polycarp knows that recently at the Parent Meeting the parents were told that he received a Safety Studies marks and b PE marks (a + b = n). Now Polycarp wants to write a subject's name in front of each mark so that:
* there are exactly a Safety Studies marks,
* there are exactly b PE marks,
* the total average score in both subjects is maximum.
An average subject grade is the sum of all marks in it, divided by the number of them. Of course, the division is performed in real numbers without rounding up or down. Polycarp aims to maximize the x1 + x2, where x1 is the average score in the first subject (Safety Studies), and x2 is the average score in the second one (Physical Education).
Input
The first line contains an integer n (2 β€ n β€ 105), n is the number of marks in Polycarp's Journal. The second line contains two positive integers a, b (1 β€ a, b β€ n - 1, a + b = n). The third line contains a sequence of integers t1, t2, ..., tn (1 β€ ti β€ 5), they are Polycarp's marks.
Output
Print the sequence of integers f1, f2, ..., fn, where fi (1 β€ fi β€ 2) is the number of a subject to which the i-th mark should be attributed. If there are several possible solutions, then print such that the sequence f1, f2, ..., fn is the smallest lexicographically.
The sequence p1, p2, ..., pn is lexicographically less than q1, q2, ..., qn if there exists such j (1 β€ j β€ n) that pi = qi for all 1 β€ i < j, Π°nd pj < qj.
Examples
Input
5
3 2
4 4 5 4 4
Output
1 1 2 1 2
Input
4
2 2
3 5 4 5
Output
1 1 2 2
Input
6
1 5
4 4 4 5 4 4
Output
2 2 2 1 2 2
Note
In the first sample the average score in the first subject is equal to 4, and in the second one β to 4.5. The total average score is 8.5.
Submitted Solution:
```
import sys
from math import *
def minp():
return sys.stdin.readline().strip()
def mint():
return int(minp())
def mints():
return map(int, minp().split())
n = mint()
a, b = mints()
t = list(mints())
c = [0]*n
for i in range(n):
c[i] = (t[i] if a != b else 0, -i if a<b else i)
c.sort(reverse=a<b)
f = [0]*n
for i in range(a):
f[abs(c[i][1])] = 1
for i in range(a,a+b):
f[abs(c[i][1])] = 2
print(*f)
``` | instruction | 0 | 66,122 | 24 | 132,244 |
Yes | output | 1 | 66,122 | 24 | 132,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the educational reform Polycarp studies only two subjects at school, Safety Studies and PE (Physical Education). During the long months of the fourth term, he received n marks in them. When teachers wrote a mark in the journal, they didn't write in what subject the mark was for, they just wrote the mark.
Now it's time to show the journal to his strict parents. Polycarp knows that recently at the Parent Meeting the parents were told that he received a Safety Studies marks and b PE marks (a + b = n). Now Polycarp wants to write a subject's name in front of each mark so that:
* there are exactly a Safety Studies marks,
* there are exactly b PE marks,
* the total average score in both subjects is maximum.
An average subject grade is the sum of all marks in it, divided by the number of them. Of course, the division is performed in real numbers without rounding up or down. Polycarp aims to maximize the x1 + x2, where x1 is the average score in the first subject (Safety Studies), and x2 is the average score in the second one (Physical Education).
Input
The first line contains an integer n (2 β€ n β€ 105), n is the number of marks in Polycarp's Journal. The second line contains two positive integers a, b (1 β€ a, b β€ n - 1, a + b = n). The third line contains a sequence of integers t1, t2, ..., tn (1 β€ ti β€ 5), they are Polycarp's marks.
Output
Print the sequence of integers f1, f2, ..., fn, where fi (1 β€ fi β€ 2) is the number of a subject to which the i-th mark should be attributed. If there are several possible solutions, then print such that the sequence f1, f2, ..., fn is the smallest lexicographically.
The sequence p1, p2, ..., pn is lexicographically less than q1, q2, ..., qn if there exists such j (1 β€ j β€ n) that pi = qi for all 1 β€ i < j, Π°nd pj < qj.
Examples
Input
5
3 2
4 4 5 4 4
Output
1 1 2 1 2
Input
4
2 2
3 5 4 5
Output
1 1 2 2
Input
6
1 5
4 4 4 5 4 4
Output
2 2 2 1 2 2
Note
In the first sample the average score in the first subject is equal to 4, and in the second one β to 4.5. The total average score is 8.5.
Submitted Solution:
```
#input
n = int(input())
a,b = [int(d) for d in input().split()]
t = [int(d) for d in input().split()]
arr = [s for s in range(n)]
if a!= b: arr = sorted(arr,reverse=(a<b), key=(lambda i:t[i]))
for i in arr[0:a]: t[i] = 1
for i in arr[a:n]: t[i] = 2
for d in t: print(d, end=' ')
``` | instruction | 0 | 66,123 | 24 | 132,246 |
Yes | output | 1 | 66,123 | 24 | 132,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the educational reform Polycarp studies only two subjects at school, Safety Studies and PE (Physical Education). During the long months of the fourth term, he received n marks in them. When teachers wrote a mark in the journal, they didn't write in what subject the mark was for, they just wrote the mark.
Now it's time to show the journal to his strict parents. Polycarp knows that recently at the Parent Meeting the parents were told that he received a Safety Studies marks and b PE marks (a + b = n). Now Polycarp wants to write a subject's name in front of each mark so that:
* there are exactly a Safety Studies marks,
* there are exactly b PE marks,
* the total average score in both subjects is maximum.
An average subject grade is the sum of all marks in it, divided by the number of them. Of course, the division is performed in real numbers without rounding up or down. Polycarp aims to maximize the x1 + x2, where x1 is the average score in the first subject (Safety Studies), and x2 is the average score in the second one (Physical Education).
Input
The first line contains an integer n (2 β€ n β€ 105), n is the number of marks in Polycarp's Journal. The second line contains two positive integers a, b (1 β€ a, b β€ n - 1, a + b = n). The third line contains a sequence of integers t1, t2, ..., tn (1 β€ ti β€ 5), they are Polycarp's marks.
Output
Print the sequence of integers f1, f2, ..., fn, where fi (1 β€ fi β€ 2) is the number of a subject to which the i-th mark should be attributed. If there are several possible solutions, then print such that the sequence f1, f2, ..., fn is the smallest lexicographically.
The sequence p1, p2, ..., pn is lexicographically less than q1, q2, ..., qn if there exists such j (1 β€ j β€ n) that pi = qi for all 1 β€ i < j, Π°nd pj < qj.
Examples
Input
5
3 2
4 4 5 4 4
Output
1 1 2 1 2
Input
4
2 2
3 5 4 5
Output
1 1 2 2
Input
6
1 5
4 4 4 5 4 4
Output
2 2 2 1 2 2
Note
In the first sample the average score in the first subject is equal to 4, and in the second one β to 4.5. The total average score is 8.5.
Submitted Solution:
```
n=int(input())
a,b=map(int,input().split())
l=list(map(int,input().split()))
if a==b:
print(*[1]*a+[2]*b)
else:
h=[[l[i],i] for i in range(n)]
h.sort(key=lambda x:x[0])
ans=[0 for i in range(n)]
if b<a:
i=n-b
j=n-b-1
if h[i][0]==h[j][0]:
while i<n and h[i][0]==h[n-b][0]:
i+=1
while j>-1 and h[j][0]==h[n-b-1][0]:
j-=1
i=i-1
j=j+1
na=n-b-j;nb=i-n+b+1
k=h[j:i+1]
k.sort(key=lambda x:x[1])
for c in range(na):
k[c][0]=1
for c in range(na,na+nb):
k[c][0]=2
for c in range(na+nb):
ans[k[c][1]]=k[c][0]
else:
ans[h[j][1]]=1
ans[h[i][1]]=2
for c in range(j):
ans[h[c][1]]=1
for c in range(i+1,n):
ans[h[c][1]]=2
print(*ans)
else:
i=n-a
j=n-a-1
if h[i][0]==h[j][0]:
while i<n and h[i][0]==h[n-a][0]:
i+=1
while j>-1 and h[j][0]==h[n-a-1][0]:
j-=1
i=i-1
j=j+1
nb=n-a-j;na=i-n+a+1
k=h[j:i+1]
k.sort(key=lambda x:x[1])
for c in range(na):
k[c][0]=1
for c in range(na,na+nb):
k[c][0]=2
for c in range(na+nb):
ans[k[c][1]]=k[c][0]
else:
ans[h[j][1]]=2
ans[h[i][1]]=1
for c in range(j):
ans[h[c][1]]=2
for c in range(i+1,n):
ans[h[c][1]]=1
print(*ans)
``` | instruction | 0 | 66,124 | 24 | 132,248 |
Yes | output | 1 | 66,124 | 24 | 132,249 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the educational reform Polycarp studies only two subjects at school, Safety Studies and PE (Physical Education). During the long months of the fourth term, he received n marks in them. When teachers wrote a mark in the journal, they didn't write in what subject the mark was for, they just wrote the mark.
Now it's time to show the journal to his strict parents. Polycarp knows that recently at the Parent Meeting the parents were told that he received a Safety Studies marks and b PE marks (a + b = n). Now Polycarp wants to write a subject's name in front of each mark so that:
* there are exactly a Safety Studies marks,
* there are exactly b PE marks,
* the total average score in both subjects is maximum.
An average subject grade is the sum of all marks in it, divided by the number of them. Of course, the division is performed in real numbers without rounding up or down. Polycarp aims to maximize the x1 + x2, where x1 is the average score in the first subject (Safety Studies), and x2 is the average score in the second one (Physical Education).
Input
The first line contains an integer n (2 β€ n β€ 105), n is the number of marks in Polycarp's Journal. The second line contains two positive integers a, b (1 β€ a, b β€ n - 1, a + b = n). The third line contains a sequence of integers t1, t2, ..., tn (1 β€ ti β€ 5), they are Polycarp's marks.
Output
Print the sequence of integers f1, f2, ..., fn, where fi (1 β€ fi β€ 2) is the number of a subject to which the i-th mark should be attributed. If there are several possible solutions, then print such that the sequence f1, f2, ..., fn is the smallest lexicographically.
The sequence p1, p2, ..., pn is lexicographically less than q1, q2, ..., qn if there exists such j (1 β€ j β€ n) that pi = qi for all 1 β€ i < j, Π°nd pj < qj.
Examples
Input
5
3 2
4 4 5 4 4
Output
1 1 2 1 2
Input
4
2 2
3 5 4 5
Output
1 1 2 2
Input
6
1 5
4 4 4 5 4 4
Output
2 2 2 1 2 2
Note
In the first sample the average score in the first subject is equal to 4, and in the second one β to 4.5. The total average score is 8.5.
Submitted Solution:
```
import sys
from itertools import *
from math import *
def solve():
n = int(input())
a, b = map(int, input().split())
if a == b:
for i in range(a): print(1, end = ' ')
for i in range(a): print(2, end = ' ')
return
first = 2
if a > b:
a, b = b, a
first = 1
l = list(map(int, input().split()))
lobjs = [(i, val) for i, val in enumerate(l)]
lobjs.sort(key = lambda j: (j[1], j[0] if first == 1 else -j[0]), reverse = True)
res = [0] * n
for index, obj in enumerate(lobjs):
if index < a: res[obj[0]] = 3 - first
else: res[obj[0]] = first
print(' '.join(map(str, res)))
if sys.hexversion == 50594544 : sys.stdin = open("test.txt")
solve()
# Made By Mostafa_Khaled
``` | instruction | 0 | 66,125 | 24 | 132,250 |
Yes | output | 1 | 66,125 | 24 | 132,251 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the educational reform Polycarp studies only two subjects at school, Safety Studies and PE (Physical Education). During the long months of the fourth term, he received n marks in them. When teachers wrote a mark in the journal, they didn't write in what subject the mark was for, they just wrote the mark.
Now it's time to show the journal to his strict parents. Polycarp knows that recently at the Parent Meeting the parents were told that he received a Safety Studies marks and b PE marks (a + b = n). Now Polycarp wants to write a subject's name in front of each mark so that:
* there are exactly a Safety Studies marks,
* there are exactly b PE marks,
* the total average score in both subjects is maximum.
An average subject grade is the sum of all marks in it, divided by the number of them. Of course, the division is performed in real numbers without rounding up or down. Polycarp aims to maximize the x1 + x2, where x1 is the average score in the first subject (Safety Studies), and x2 is the average score in the second one (Physical Education).
Input
The first line contains an integer n (2 β€ n β€ 105), n is the number of marks in Polycarp's Journal. The second line contains two positive integers a, b (1 β€ a, b β€ n - 1, a + b = n). The third line contains a sequence of integers t1, t2, ..., tn (1 β€ ti β€ 5), they are Polycarp's marks.
Output
Print the sequence of integers f1, f2, ..., fn, where fi (1 β€ fi β€ 2) is the number of a subject to which the i-th mark should be attributed. If there are several possible solutions, then print such that the sequence f1, f2, ..., fn is the smallest lexicographically.
The sequence p1, p2, ..., pn is lexicographically less than q1, q2, ..., qn if there exists such j (1 β€ j β€ n) that pi = qi for all 1 β€ i < j, Π°nd pj < qj.
Examples
Input
5
3 2
4 4 5 4 4
Output
1 1 2 1 2
Input
4
2 2
3 5 4 5
Output
1 1 2 2
Input
6
1 5
4 4 4 5 4 4
Output
2 2 2 1 2 2
Note
In the first sample the average score in the first subject is equal to 4, and in the second one β to 4.5. The total average score is 8.5.
Submitted Solution:
```
n = int(input())
a, b = map(int, input().split())
c = [int(i) for i in input().split()]
p = [i for i in range(n)]
Z = [x for _,x in sorted(zip(c, p))]
c.sort()
ans = [0] * n
if a == b:
for i in range(a):
print(1, end=' ')
for i in range(b):
print(2, end=' ')
exit()
if a > b:
for i in range(a):
ans[Z[i]] = 1
for i in range(n):
if ans[i] == 0:
ans[i] = 2
print(*ans)
if a < b:
for i in range(b):
ans[Z[i]] = 2
for i in range(n):
if ans[i] == 0:
ans[i] = 1
print(*ans)
``` | instruction | 0 | 66,126 | 24 | 132,252 |
No | output | 1 | 66,126 | 24 | 132,253 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the educational reform Polycarp studies only two subjects at school, Safety Studies and PE (Physical Education). During the long months of the fourth term, he received n marks in them. When teachers wrote a mark in the journal, they didn't write in what subject the mark was for, they just wrote the mark.
Now it's time to show the journal to his strict parents. Polycarp knows that recently at the Parent Meeting the parents were told that he received a Safety Studies marks and b PE marks (a + b = n). Now Polycarp wants to write a subject's name in front of each mark so that:
* there are exactly a Safety Studies marks,
* there are exactly b PE marks,
* the total average score in both subjects is maximum.
An average subject grade is the sum of all marks in it, divided by the number of them. Of course, the division is performed in real numbers without rounding up or down. Polycarp aims to maximize the x1 + x2, where x1 is the average score in the first subject (Safety Studies), and x2 is the average score in the second one (Physical Education).
Input
The first line contains an integer n (2 β€ n β€ 105), n is the number of marks in Polycarp's Journal. The second line contains two positive integers a, b (1 β€ a, b β€ n - 1, a + b = n). The third line contains a sequence of integers t1, t2, ..., tn (1 β€ ti β€ 5), they are Polycarp's marks.
Output
Print the sequence of integers f1, f2, ..., fn, where fi (1 β€ fi β€ 2) is the number of a subject to which the i-th mark should be attributed. If there are several possible solutions, then print such that the sequence f1, f2, ..., fn is the smallest lexicographically.
The sequence p1, p2, ..., pn is lexicographically less than q1, q2, ..., qn if there exists such j (1 β€ j β€ n) that pi = qi for all 1 β€ i < j, Π°nd pj < qj.
Examples
Input
5
3 2
4 4 5 4 4
Output
1 1 2 1 2
Input
4
2 2
3 5 4 5
Output
1 1 2 2
Input
6
1 5
4 4 4 5 4 4
Output
2 2 2 1 2 2
Note
In the first sample the average score in the first subject is equal to 4, and in the second one β to 4.5. The total average score is 8.5.
Submitted Solution:
```
from operator import itemgetter
from collections import defaultdict
n=int(input())
a,b= map(int,input().split())
arr= list(map(int,input().split()))
arr = list(enumerate(arr,0))
arr=sorted(arr,key=itemgetter(1),reverse=True)
def find_min(num1,num2):
if num1<num2:
return num1
else:
return num2
result=[0]*n
if a==b:
for i in range(a):
result[i]=1
for j in range(a,n):
if result[j]==0:
result[j]=2
elif a>b:
dicta=defaultdict(list)
for i in range(n):
dicta[arr[i][1]].append(arr[i][0])
for j in range(b,n):
print(dicta[arr[j][1]])
result[dicta[arr[j][1]][0]]=1
dicta[arr[j][1]].pop(0)
for k in range(b):
result[dicta[arr[k][1]][0]]=2
dicta[arr[k][1]].pop(0)
else:
min_value= find_min(a,b)
if min_value==a:
second_range=b
val=1
else:
second_range=a
val=2
for i in range(min_value):
result[arr[i][0]] = val
if val==1:
new_val=2
else:
new_val=1
for j in range(min_value,n):
result[arr[j][0]]=new_val
for l in result:
print(l,end=" ")
# max_arr = arr[n-min_value:]
# min_arr= arr[:n-min_value]
# if min_value==a:
# for i in max_arr:
# result[i[0]]=1
# for j in min_arr:
# result[j[0]]=2
# for k in result:
# print(k,end=" ")
# else:
# for i in min_arr:
# result[i[0]]=1
# for j in max_arr:
# result[j[0]]=2
# for k in result:
# print(k,end=" ")
``` | instruction | 0 | 66,127 | 24 | 132,254 |
No | output | 1 | 66,127 | 24 | 132,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the educational reform Polycarp studies only two subjects at school, Safety Studies and PE (Physical Education). During the long months of the fourth term, he received n marks in them. When teachers wrote a mark in the journal, they didn't write in what subject the mark was for, they just wrote the mark.
Now it's time to show the journal to his strict parents. Polycarp knows that recently at the Parent Meeting the parents were told that he received a Safety Studies marks and b PE marks (a + b = n). Now Polycarp wants to write a subject's name in front of each mark so that:
* there are exactly a Safety Studies marks,
* there are exactly b PE marks,
* the total average score in both subjects is maximum.
An average subject grade is the sum of all marks in it, divided by the number of them. Of course, the division is performed in real numbers without rounding up or down. Polycarp aims to maximize the x1 + x2, where x1 is the average score in the first subject (Safety Studies), and x2 is the average score in the second one (Physical Education).
Input
The first line contains an integer n (2 β€ n β€ 105), n is the number of marks in Polycarp's Journal. The second line contains two positive integers a, b (1 β€ a, b β€ n - 1, a + b = n). The third line contains a sequence of integers t1, t2, ..., tn (1 β€ ti β€ 5), they are Polycarp's marks.
Output
Print the sequence of integers f1, f2, ..., fn, where fi (1 β€ fi β€ 2) is the number of a subject to which the i-th mark should be attributed. If there are several possible solutions, then print such that the sequence f1, f2, ..., fn is the smallest lexicographically.
The sequence p1, p2, ..., pn is lexicographically less than q1, q2, ..., qn if there exists such j (1 β€ j β€ n) that pi = qi for all 1 β€ i < j, Π°nd pj < qj.
Examples
Input
5
3 2
4 4 5 4 4
Output
1 1 2 1 2
Input
4
2 2
3 5 4 5
Output
1 1 2 2
Input
6
1 5
4 4 4 5 4 4
Output
2 2 2 1 2 2
Note
In the first sample the average score in the first subject is equal to 4, and in the second one β to 4.5. The total average score is 8.5.
Submitted Solution:
```
v = input()
while " " in v: v = v.replace(" ", " ")
i = -1
ln = len(v)
acc = ""
def next():
global i
i += 1
if i >= ln:
return None
return v[i]
while True:
c = next()
if c == None: break
if c in "0123456789":
if len(acc) > 0 and acc[-1] == ",":
acc += " "
acc += c
if c == " ":
if len(acc) > 0 and acc[-1] != " ":
acc += c
if c == ".":
if len(acc) > 0 and acc[-1] != " ":
acc += " "
next(); next();
acc += "..."
if c == ",":
if len(acc) > 1 and acc[-1] == " " and acc[-2] != ",":
acc = acc[:-1] + ","
else:
if len(acc) > 0 and acc[-1] == ",":
acc += " "
acc += ","
print(acc.strip())
``` | instruction | 0 | 66,128 | 24 | 132,256 |
No | output | 1 | 66,128 | 24 | 132,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After the educational reform Polycarp studies only two subjects at school, Safety Studies and PE (Physical Education). During the long months of the fourth term, he received n marks in them. When teachers wrote a mark in the journal, they didn't write in what subject the mark was for, they just wrote the mark.
Now it's time to show the journal to his strict parents. Polycarp knows that recently at the Parent Meeting the parents were told that he received a Safety Studies marks and b PE marks (a + b = n). Now Polycarp wants to write a subject's name in front of each mark so that:
* there are exactly a Safety Studies marks,
* there are exactly b PE marks,
* the total average score in both subjects is maximum.
An average subject grade is the sum of all marks in it, divided by the number of them. Of course, the division is performed in real numbers without rounding up or down. Polycarp aims to maximize the x1 + x2, where x1 is the average score in the first subject (Safety Studies), and x2 is the average score in the second one (Physical Education).
Input
The first line contains an integer n (2 β€ n β€ 105), n is the number of marks in Polycarp's Journal. The second line contains two positive integers a, b (1 β€ a, b β€ n - 1, a + b = n). The third line contains a sequence of integers t1, t2, ..., tn (1 β€ ti β€ 5), they are Polycarp's marks.
Output
Print the sequence of integers f1, f2, ..., fn, where fi (1 β€ fi β€ 2) is the number of a subject to which the i-th mark should be attributed. If there are several possible solutions, then print such that the sequence f1, f2, ..., fn is the smallest lexicographically.
The sequence p1, p2, ..., pn is lexicographically less than q1, q2, ..., qn if there exists such j (1 β€ j β€ n) that pi = qi for all 1 β€ i < j, Π°nd pj < qj.
Examples
Input
5
3 2
4 4 5 4 4
Output
1 1 2 1 2
Input
4
2 2
3 5 4 5
Output
1 1 2 2
Input
6
1 5
4 4 4 5 4 4
Output
2 2 2 1 2 2
Note
In the first sample the average score in the first subject is equal to 4, and in the second one β to 4.5. The total average score is 8.5.
Submitted Solution:
```
import sys
from itertools import *
from math import *
def solve():
n = int(input())
a, b = map(int, input().split())
if a == b:
for i in range(a): print(1, end = ' ')
for i in range(a): print(2, end = ' ')
return
first = 2
if a > b:
a, b = b, a
first = 1
l = list(map(int, input().split()))
lobjs = [(i, val) for i, val in enumerate(l)]
lobjs.sort(key = lambda j: (j[1], j[0]), reverse = True)
res = [0] * n
for index, obj in enumerate(lobjs):
if index < a: res[obj[0]] = 3 - first
else: res[obj[0]] = first
print(' '.join(map(str, res)))
if sys.hexversion == 50594544 : sys.stdin = open("test.txt")
solve()
``` | instruction | 0 | 66,129 | 24 | 132,258 |
No | output | 1 | 66,129 | 24 | 132,259 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain number of points (may be negative, i. e. points were subtracted). Initially the participant had some score, and each the marks were one by one added to his score. It is known that the i-th jury member gave ai points.
Polycarp does not remember how many points the participant had before this k marks were given, but he remembers that among the scores announced after each of the k judges rated the participant there were n (n β€ k) values b1, b2, ..., bn (it is guaranteed that all values bj are distinct). It is possible that Polycarp remembers not all of the scores announced, i. e. n < k. Note that the initial score wasn't announced.
Your task is to determine the number of options for the score the participant could have before the judges rated the participant.
Input
The first line contains two integers k and n (1 β€ n β€ k β€ 2 000) β the number of jury members and the number of scores Polycarp remembers.
The second line contains k integers a1, a2, ..., ak ( - 2 000 β€ ai β€ 2 000) β jury's marks in chronological order.
The third line contains n distinct integers b1, b2, ..., bn ( - 4 000 000 β€ bj β€ 4 000 000) β the values of points Polycarp remembers. Note that these values are not necessarily given in chronological order.
Output
Print the number of options for the score the participant could have before the judges rated the participant. If Polycarp messes something up and there is no options, print "0" (without quotes).
Examples
Input
4 1
-5 5 0 20
10
Output
3
Input
2 2
-2000 -2000
3998000 4000000
Output
1
Note
The answer for the first example is 3 because initially the participant could have - 10, 10 or 15 points.
In the second example there is only one correct initial score equaling to 4 002 000. | instruction | 0 | 67,813 | 24 | 135,626 |
Tags: brute force, constructive algorithms
Correct Solution:
```
def read():
return [int(x) for x in input().split()]
k,n = read()
a = read()
b = read()
sum = set()
tmp =[0]
for e in a:
tmp.append(tmp[-1]+e)
sum.add(tmp[-1])
ans = set()
for e in tmp[1:]:
init = b[0]-e
for bb in b[1:]:
if bb - init not in sum:
break
else:
ans.add(init)
print(len(ans))
``` | output | 1 | 67,813 | 24 | 135,627 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain number of points (may be negative, i. e. points were subtracted). Initially the participant had some score, and each the marks were one by one added to his score. It is known that the i-th jury member gave ai points.
Polycarp does not remember how many points the participant had before this k marks were given, but he remembers that among the scores announced after each of the k judges rated the participant there were n (n β€ k) values b1, b2, ..., bn (it is guaranteed that all values bj are distinct). It is possible that Polycarp remembers not all of the scores announced, i. e. n < k. Note that the initial score wasn't announced.
Your task is to determine the number of options for the score the participant could have before the judges rated the participant.
Input
The first line contains two integers k and n (1 β€ n β€ k β€ 2 000) β the number of jury members and the number of scores Polycarp remembers.
The second line contains k integers a1, a2, ..., ak ( - 2 000 β€ ai β€ 2 000) β jury's marks in chronological order.
The third line contains n distinct integers b1, b2, ..., bn ( - 4 000 000 β€ bj β€ 4 000 000) β the values of points Polycarp remembers. Note that these values are not necessarily given in chronological order.
Output
Print the number of options for the score the participant could have before the judges rated the participant. If Polycarp messes something up and there is no options, print "0" (without quotes).
Examples
Input
4 1
-5 5 0 20
10
Output
3
Input
2 2
-2000 -2000
3998000 4000000
Output
1
Note
The answer for the first example is 3 because initially the participant could have - 10, 10 or 15 points.
In the second example there is only one correct initial score equaling to 4 002 000. | instruction | 0 | 67,814 | 24 | 135,628 |
Tags: brute force, constructive algorithms
Correct Solution:
```
k,n = map(int,input().split())
A = list(map(int,input().split()))
B = list(map(int, input().split()))
S = [0] * k
for i in range(k):
if i == 0:
S[0] = A[0]
else:
S[i] = S[i - 1] + A[i]
S.sort()
B.sort()
ans = set()
for i,s in enumerate(S):
if i > 0 and S[i - 1] == s:
continue
t = B[0] - s
if n > 1:
j = i + 1
flag = False
for m,u in enumerate(B[1:]):
while j < len(S):
if u - S[j] == t:
if m == n - 2:
flag = True
j += 1
break
else:
j += 1
if flag == True:
ans.add(t)
else:
ans.add(t)
print(len(ans))
``` | output | 1 | 67,814 | 24 | 135,629 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain number of points (may be negative, i. e. points were subtracted). Initially the participant had some score, and each the marks were one by one added to his score. It is known that the i-th jury member gave ai points.
Polycarp does not remember how many points the participant had before this k marks were given, but he remembers that among the scores announced after each of the k judges rated the participant there were n (n β€ k) values b1, b2, ..., bn (it is guaranteed that all values bj are distinct). It is possible that Polycarp remembers not all of the scores announced, i. e. n < k. Note that the initial score wasn't announced.
Your task is to determine the number of options for the score the participant could have before the judges rated the participant.
Input
The first line contains two integers k and n (1 β€ n β€ k β€ 2 000) β the number of jury members and the number of scores Polycarp remembers.
The second line contains k integers a1, a2, ..., ak ( - 2 000 β€ ai β€ 2 000) β jury's marks in chronological order.
The third line contains n distinct integers b1, b2, ..., bn ( - 4 000 000 β€ bj β€ 4 000 000) β the values of points Polycarp remembers. Note that these values are not necessarily given in chronological order.
Output
Print the number of options for the score the participant could have before the judges rated the participant. If Polycarp messes something up and there is no options, print "0" (without quotes).
Examples
Input
4 1
-5 5 0 20
10
Output
3
Input
2 2
-2000 -2000
3998000 4000000
Output
1
Note
The answer for the first example is 3 because initially the participant could have - 10, 10 or 15 points.
In the second example there is only one correct initial score equaling to 4 002 000. | instruction | 0 | 67,815 | 24 | 135,630 |
Tags: brute force, constructive algorithms
Correct Solution:
```
s = (input()).split(" ")
k = int(s[0])
n = int(s[1])
arr = []
sum_arr = []
s = (input()).split(" ")
for i in range(k):
arr.append(int(s[i]))
sum_arr.append(arr[-1])
if i >= 1:
sum_arr[i]+=sum_arr[i-1]
arr_2 = []
s = (input()).split(" ")
for i in range(n):
arr_2.append(int(s[i]))
possible_intial = set()
for i in sum_arr:
possible_intial.add(arr_2[0]-i)
arr_2 = set(arr_2)
ctr = 0
for i in possible_intial:
after_values = set()
for j in sum_arr:
after_values.add(i+j)
flag = 0
for j in arr_2:
if j not in after_values:
flag = 1
break
if flag==0:
ctr += 1
print(ctr)
``` | output | 1 | 67,815 | 24 | 135,631 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain number of points (may be negative, i. e. points were subtracted). Initially the participant had some score, and each the marks were one by one added to his score. It is known that the i-th jury member gave ai points.
Polycarp does not remember how many points the participant had before this k marks were given, but he remembers that among the scores announced after each of the k judges rated the participant there were n (n β€ k) values b1, b2, ..., bn (it is guaranteed that all values bj are distinct). It is possible that Polycarp remembers not all of the scores announced, i. e. n < k. Note that the initial score wasn't announced.
Your task is to determine the number of options for the score the participant could have before the judges rated the participant.
Input
The first line contains two integers k and n (1 β€ n β€ k β€ 2 000) β the number of jury members and the number of scores Polycarp remembers.
The second line contains k integers a1, a2, ..., ak ( - 2 000 β€ ai β€ 2 000) β jury's marks in chronological order.
The third line contains n distinct integers b1, b2, ..., bn ( - 4 000 000 β€ bj β€ 4 000 000) β the values of points Polycarp remembers. Note that these values are not necessarily given in chronological order.
Output
Print the number of options for the score the participant could have before the judges rated the participant. If Polycarp messes something up and there is no options, print "0" (without quotes).
Examples
Input
4 1
-5 5 0 20
10
Output
3
Input
2 2
-2000 -2000
3998000 4000000
Output
1
Note
The answer for the first example is 3 because initially the participant could have - 10, 10 or 15 points.
In the second example there is only one correct initial score equaling to 4 002 000. | instruction | 0 | 67,816 | 24 | 135,632 |
Tags: brute force, constructive algorithms
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
from collections import defaultdict
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
k, n = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
d = defaultdict(int)
for i in range(n):
d[b[i]] = i
s = set()
diff = 0
for ai in a:
diff += ai
s.add(b[0]-diff)
ans = 0
for si in s:
flag = [False]*n
x = si
for ai in a:
x += ai
if x in d:
flag[d[x]] = True
if flag==[True]*n:
ans += 1
print(ans)
``` | output | 1 | 67,816 | 24 | 135,633 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain number of points (may be negative, i. e. points were subtracted). Initially the participant had some score, and each the marks were one by one added to his score. It is known that the i-th jury member gave ai points.
Polycarp does not remember how many points the participant had before this k marks were given, but he remembers that among the scores announced after each of the k judges rated the participant there were n (n β€ k) values b1, b2, ..., bn (it is guaranteed that all values bj are distinct). It is possible that Polycarp remembers not all of the scores announced, i. e. n < k. Note that the initial score wasn't announced.
Your task is to determine the number of options for the score the participant could have before the judges rated the participant.
Input
The first line contains two integers k and n (1 β€ n β€ k β€ 2 000) β the number of jury members and the number of scores Polycarp remembers.
The second line contains k integers a1, a2, ..., ak ( - 2 000 β€ ai β€ 2 000) β jury's marks in chronological order.
The third line contains n distinct integers b1, b2, ..., bn ( - 4 000 000 β€ bj β€ 4 000 000) β the values of points Polycarp remembers. Note that these values are not necessarily given in chronological order.
Output
Print the number of options for the score the participant could have before the judges rated the participant. If Polycarp messes something up and there is no options, print "0" (without quotes).
Examples
Input
4 1
-5 5 0 20
10
Output
3
Input
2 2
-2000 -2000
3998000 4000000
Output
1
Note
The answer for the first example is 3 because initially the participant could have - 10, 10 or 15 points.
In the second example there is only one correct initial score equaling to 4 002 000. | instruction | 0 | 67,817 | 24 | 135,634 |
Tags: brute force, constructive algorithms
Correct Solution:
```
'''K,N=map(int,input().split())
summ=[0]*(K+1)
a=list(map(int,input().split()))
for i in range(1,K+1):
summ[i]=a[i-1]
summ[i]+=summ[i-1]
summ.pop(0)
b=list(map(int,input().split()))
sett=[]
for i in range(N):
for j in range(K):
sett.append(b[i]-summ[j])
sett=list(set(sett))
ans=0
for i in sett:
points=[]
for j in summ:
points.append(j+i)
temp=True
for j in set(list(b)):
if j not in set(list(points)):
temp=False
if temp:
ans+=1
print(ans)'''
k, n = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
st = set()
zn = b[0]
sums = [a[0]]
for j in range(1, len(a)):
sums.append(sums[-1] + a[j])
for j in sums:
st.add(b[0] - j)
for i in b:
tmp = set()
for j in sums:
tmp.add(i - j)
st = st.intersection(tmp)
#print(st)
if len(st) == 0:
break
print(len(st))
``` | output | 1 | 67,817 | 24 | 135,635 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain number of points (may be negative, i. e. points were subtracted). Initially the participant had some score, and each the marks were one by one added to his score. It is known that the i-th jury member gave ai points.
Polycarp does not remember how many points the participant had before this k marks were given, but he remembers that among the scores announced after each of the k judges rated the participant there were n (n β€ k) values b1, b2, ..., bn (it is guaranteed that all values bj are distinct). It is possible that Polycarp remembers not all of the scores announced, i. e. n < k. Note that the initial score wasn't announced.
Your task is to determine the number of options for the score the participant could have before the judges rated the participant.
Input
The first line contains two integers k and n (1 β€ n β€ k β€ 2 000) β the number of jury members and the number of scores Polycarp remembers.
The second line contains k integers a1, a2, ..., ak ( - 2 000 β€ ai β€ 2 000) β jury's marks in chronological order.
The third line contains n distinct integers b1, b2, ..., bn ( - 4 000 000 β€ bj β€ 4 000 000) β the values of points Polycarp remembers. Note that these values are not necessarily given in chronological order.
Output
Print the number of options for the score the participant could have before the judges rated the participant. If Polycarp messes something up and there is no options, print "0" (without quotes).
Examples
Input
4 1
-5 5 0 20
10
Output
3
Input
2 2
-2000 -2000
3998000 4000000
Output
1
Note
The answer for the first example is 3 because initially the participant could have - 10, 10 or 15 points.
In the second example there is only one correct initial score equaling to 4 002 000. | instruction | 0 | 67,818 | 24 | 135,636 |
Tags: brute force, constructive algorithms
Correct Solution:
```
k, n = map(int,input().split())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
all_maybe_all_b = []
for mast_b in b:
all_maybe_for_b_i=set()
for slog in a:
mast_b -= slog
all_maybe_for_b_i.add(mast_b)
all_maybe_all_b += [all_maybe_for_b_i]
pack_ans_intersection=set.intersection(*all_maybe_all_b)
#print(all_maybe_all_b)
#print(pack_ans_intersection)
print(len(pack_ans_intersection))
``` | output | 1 | 67,818 | 24 | 135,637 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain number of points (may be negative, i. e. points were subtracted). Initially the participant had some score, and each the marks were one by one added to his score. It is known that the i-th jury member gave ai points.
Polycarp does not remember how many points the participant had before this k marks were given, but he remembers that among the scores announced after each of the k judges rated the participant there were n (n β€ k) values b1, b2, ..., bn (it is guaranteed that all values bj are distinct). It is possible that Polycarp remembers not all of the scores announced, i. e. n < k. Note that the initial score wasn't announced.
Your task is to determine the number of options for the score the participant could have before the judges rated the participant.
Input
The first line contains two integers k and n (1 β€ n β€ k β€ 2 000) β the number of jury members and the number of scores Polycarp remembers.
The second line contains k integers a1, a2, ..., ak ( - 2 000 β€ ai β€ 2 000) β jury's marks in chronological order.
The third line contains n distinct integers b1, b2, ..., bn ( - 4 000 000 β€ bj β€ 4 000 000) β the values of points Polycarp remembers. Note that these values are not necessarily given in chronological order.
Output
Print the number of options for the score the participant could have before the judges rated the participant. If Polycarp messes something up and there is no options, print "0" (without quotes).
Examples
Input
4 1
-5 5 0 20
10
Output
3
Input
2 2
-2000 -2000
3998000 4000000
Output
1
Note
The answer for the first example is 3 because initially the participant could have - 10, 10 or 15 points.
In the second example there is only one correct initial score equaling to 4 002 000. | instruction | 0 | 67,819 | 24 | 135,638 |
Tags: brute force, constructive algorithms
Correct Solution:
```
import sys
#comment these out later
#sys.stdin = open("in.in", "r")
#sys.stdout = open("out.out", "w")
inp = [int(x) for x in sys.stdin.read().split()]; ii = 0
k = inp[ii]; ii += 1
n = inp[ii]; ii += 1
marks = inp[ii:ii+k]; ii += k
scores = inp[ii:ii+n]
pre = [0]
for x in marks:
pre.append(pre[-1] + x)
ans = set()
for b in pre[1:]:
ans.add(scores[0]-b)
for a in scores[1:]:
scset = set()
for b in pre[1:]:
scset.add(a-b)
ans = ans&scset
print(len(ans))
``` | output | 1 | 67,819 | 24 | 135,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain number of points (may be negative, i. e. points were subtracted). Initially the participant had some score, and each the marks were one by one added to his score. It is known that the i-th jury member gave ai points.
Polycarp does not remember how many points the participant had before this k marks were given, but he remembers that among the scores announced after each of the k judges rated the participant there were n (n β€ k) values b1, b2, ..., bn (it is guaranteed that all values bj are distinct). It is possible that Polycarp remembers not all of the scores announced, i. e. n < k. Note that the initial score wasn't announced.
Your task is to determine the number of options for the score the participant could have before the judges rated the participant.
Input
The first line contains two integers k and n (1 β€ n β€ k β€ 2 000) β the number of jury members and the number of scores Polycarp remembers.
The second line contains k integers a1, a2, ..., ak ( - 2 000 β€ ai β€ 2 000) β jury's marks in chronological order.
The third line contains n distinct integers b1, b2, ..., bn ( - 4 000 000 β€ bj β€ 4 000 000) β the values of points Polycarp remembers. Note that these values are not necessarily given in chronological order.
Output
Print the number of options for the score the participant could have before the judges rated the participant. If Polycarp messes something up and there is no options, print "0" (without quotes).
Examples
Input
4 1
-5 5 0 20
10
Output
3
Input
2 2
-2000 -2000
3998000 4000000
Output
1
Note
The answer for the first example is 3 because initially the participant could have - 10, 10 or 15 points.
In the second example there is only one correct initial score equaling to 4 002 000. | instruction | 0 | 67,820 | 24 | 135,640 |
Tags: brute force, constructive algorithms
Correct Solution:
```
n,m = map(int,input().split())
arr = [int(x) for x in input().split()]
dist = [int(x) for x in input().split()]
for i in range(1,n):
arr[i] = arr[i]+arr[i-1]
lis = []
for i in range(m):
val = dist[i]
s = set()
for j in range(n):
s.add(val - arr[j])
lis.append(s)
print(len(set.intersection(*lis)))
``` | output | 1 | 67,820 | 24 | 135,641 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain number of points (may be negative, i. e. points were subtracted). Initially the participant had some score, and each the marks were one by one added to his score. It is known that the i-th jury member gave ai points.
Polycarp does not remember how many points the participant had before this k marks were given, but he remembers that among the scores announced after each of the k judges rated the participant there were n (n β€ k) values b1, b2, ..., bn (it is guaranteed that all values bj are distinct). It is possible that Polycarp remembers not all of the scores announced, i. e. n < k. Note that the initial score wasn't announced.
Your task is to determine the number of options for the score the participant could have before the judges rated the participant.
Input
The first line contains two integers k and n (1 β€ n β€ k β€ 2 000) β the number of jury members and the number of scores Polycarp remembers.
The second line contains k integers a1, a2, ..., ak ( - 2 000 β€ ai β€ 2 000) β jury's marks in chronological order.
The third line contains n distinct integers b1, b2, ..., bn ( - 4 000 000 β€ bj β€ 4 000 000) β the values of points Polycarp remembers. Note that these values are not necessarily given in chronological order.
Output
Print the number of options for the score the participant could have before the judges rated the participant. If Polycarp messes something up and there is no options, print "0" (without quotes).
Examples
Input
4 1
-5 5 0 20
10
Output
3
Input
2 2
-2000 -2000
3998000 4000000
Output
1
Note
The answer for the first example is 3 because initially the participant could have - 10, 10 or 15 points.
In the second example there is only one correct initial score equaling to 4 002 000.
Submitted Solution:
```
from sys import stdin
input = lambda :stdin.readline().strip()
k, n = map(int, input().split())
a = [*map(int, input().split())]
b = sorted([*map(int, input().split())])
c = [0 for i in range(k)]
c[0] = a[0]
for i in range(1, k):
c[i] = c[i - 1] + a[i]
c.sort()
if n == 1:
print(len(set(c)))
exit()
start_vals = set()
min_diff = b[-1] - b[0]
max_base = c[-1]
d = [b[i + 1] - b[i] for i in range(n - 1)]
discarded = set()
for i in range(k):
bval = b[0] - c[i]
if max_base - c[i] < min_diff or bval in start_vals or i + n > k or bval in discarded:
break
start = c[i]
di = 0
curr = c[i] + d[di]
for j in range(i + 1, k):
if curr == c[j]:
di += 1
if di == n - 1:
start_vals.add(b[0] - start)
break
curr += d[di]
print(len(start_vals))
``` | instruction | 0 | 67,821 | 24 | 135,642 |
Yes | output | 1 | 67,821 | 24 | 135,643 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain number of points (may be negative, i. e. points were subtracted). Initially the participant had some score, and each the marks were one by one added to his score. It is known that the i-th jury member gave ai points.
Polycarp does not remember how many points the participant had before this k marks were given, but he remembers that among the scores announced after each of the k judges rated the participant there were n (n β€ k) values b1, b2, ..., bn (it is guaranteed that all values bj are distinct). It is possible that Polycarp remembers not all of the scores announced, i. e. n < k. Note that the initial score wasn't announced.
Your task is to determine the number of options for the score the participant could have before the judges rated the participant.
Input
The first line contains two integers k and n (1 β€ n β€ k β€ 2 000) β the number of jury members and the number of scores Polycarp remembers.
The second line contains k integers a1, a2, ..., ak ( - 2 000 β€ ai β€ 2 000) β jury's marks in chronological order.
The third line contains n distinct integers b1, b2, ..., bn ( - 4 000 000 β€ bj β€ 4 000 000) β the values of points Polycarp remembers. Note that these values are not necessarily given in chronological order.
Output
Print the number of options for the score the participant could have before the judges rated the participant. If Polycarp messes something up and there is no options, print "0" (without quotes).
Examples
Input
4 1
-5 5 0 20
10
Output
3
Input
2 2
-2000 -2000
3998000 4000000
Output
1
Note
The answer for the first example is 3 because initially the participant could have - 10, 10 or 15 points.
In the second example there is only one correct initial score equaling to 4 002 000.
Submitted Solution:
```
k,n = map(int,input().split())
a = []
b = []
temp = input().split()
s = 0
for i in range(k):
s += int(temp[i])
a.append(s)
list.sort(a)
temp = input().split()
for i in range(n):
b.append(int(temp[i]))
list.sort(b)
count = 0
visit = set()
for i in range(k-n+1):
dif = b[0]-a[0]
if dif not in visit:
visit.add(dif)
add = True
index = 0
for j in range(n):
while index < len(a):
if a[index] == b[j]-dif:
break
elif a[index] > b[j]-dif:
add = False
break
else:
index += 1
if index >= len(a):
add = False
break
if not add:
break
if add:
count += 1
a = a[1:]
print(count)
``` | instruction | 0 | 67,822 | 24 | 135,644 |
Yes | output | 1 | 67,822 | 24 | 135,645 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain number of points (may be negative, i. e. points were subtracted). Initially the participant had some score, and each the marks were one by one added to his score. It is known that the i-th jury member gave ai points.
Polycarp does not remember how many points the participant had before this k marks were given, but he remembers that among the scores announced after each of the k judges rated the participant there were n (n β€ k) values b1, b2, ..., bn (it is guaranteed that all values bj are distinct). It is possible that Polycarp remembers not all of the scores announced, i. e. n < k. Note that the initial score wasn't announced.
Your task is to determine the number of options for the score the participant could have before the judges rated the participant.
Input
The first line contains two integers k and n (1 β€ n β€ k β€ 2 000) β the number of jury members and the number of scores Polycarp remembers.
The second line contains k integers a1, a2, ..., ak ( - 2 000 β€ ai β€ 2 000) β jury's marks in chronological order.
The third line contains n distinct integers b1, b2, ..., bn ( - 4 000 000 β€ bj β€ 4 000 000) β the values of points Polycarp remembers. Note that these values are not necessarily given in chronological order.
Output
Print the number of options for the score the participant could have before the judges rated the participant. If Polycarp messes something up and there is no options, print "0" (without quotes).
Examples
Input
4 1
-5 5 0 20
10
Output
3
Input
2 2
-2000 -2000
3998000 4000000
Output
1
Note
The answer for the first example is 3 because initially the participant could have - 10, 10 or 15 points.
In the second example there is only one correct initial score equaling to 4 002 000.
Submitted Solution:
```
k, n = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
x = set()
sums = [0]
for i in range(k):
sums.append(sums[-1]+a[i])
sums.pop(0)
for i in range(k):
x.add(b[0]-sums[i])
count =0
for s in x:
points = []
for i in range(k):
points.append(s+sums[i])
temp = b.copy()
points.sort()
temp.sort()
for num in points:
if num == temp[0]:
temp.pop(0)
if temp == []: break
if num > temp[0]:
break
if temp == []:
count+= 1
print(count)
``` | instruction | 0 | 67,823 | 24 | 135,646 |
Yes | output | 1 | 67,823 | 24 | 135,647 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp watched TV-show where k jury members one by one rated a participant by adding him a certain number of points (may be negative, i. e. points were subtracted). Initially the participant had some score, and each the marks were one by one added to his score. It is known that the i-th jury member gave ai points.
Polycarp does not remember how many points the participant had before this k marks were given, but he remembers that among the scores announced after each of the k judges rated the participant there were n (n β€ k) values b1, b2, ..., bn (it is guaranteed that all values bj are distinct). It is possible that Polycarp remembers not all of the scores announced, i. e. n < k. Note that the initial score wasn't announced.
Your task is to determine the number of options for the score the participant could have before the judges rated the participant.
Input
The first line contains two integers k and n (1 β€ n β€ k β€ 2 000) β the number of jury members and the number of scores Polycarp remembers.
The second line contains k integers a1, a2, ..., ak ( - 2 000 β€ ai β€ 2 000) β jury's marks in chronological order.
The third line contains n distinct integers b1, b2, ..., bn ( - 4 000 000 β€ bj β€ 4 000 000) β the values of points Polycarp remembers. Note that these values are not necessarily given in chronological order.
Output
Print the number of options for the score the participant could have before the judges rated the participant. If Polycarp messes something up and there is no options, print "0" (without quotes).
Examples
Input
4 1
-5 5 0 20
10
Output
3
Input
2 2
-2000 -2000
3998000 4000000
Output
1
Note
The answer for the first example is 3 because initially the participant could have - 10, 10 or 15 points.
In the second example there is only one correct initial score equaling to 4 002 000.
Submitted Solution:
```
import sys
def solve():
k, n = list(map(int, sys.stdin.readline().split()))
mks = list(map(int, sys.stdin.readline().split()))
pts = list(map(int, sys.stdin.readline().split()))
for i in range(1, k):
mks[i] = mks[i-1] + mks[i]
mks = sorted(mks)
pts = sorted(pts)
vals = set()
for i in range(k - n + 1):
cand = pts[0] - mks[i]
j = 0
off = 0
while j < len(pts):
while i + j + off < k and pts[j] - cand > mks[i + j + off]:
off += 1
if i + j + off < k and pts[j] - mks[i + j + off] == cand:
j += 1
else:
break
if j == len(pts):
vals.add(cand)
print(len(vals))
solve()
``` | instruction | 0 | 67,824 | 24 | 135,648 |
Yes | output | 1 | 67,824 | 24 | 135,649 |
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