message stringlengths 2 30.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 237 109k | cluster float64 10 10 | __index_level_0__ int64 474 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Alexey, a merry Berland entrant, got sick of the gray reality and he zealously wants to go to university. There are a lot of universities nowadays, so Alexey is getting lost in the diversity β he has not yet decided what profession he wants to get. At school, he had bad grades in all subjects, and it's only thanks to wealthy parents that he was able to obtain the graduation certificate.
The situation is complicated by the fact that each high education institution has the determined amount of voluntary donations, paid by the new students for admission β ni berubleys. He cannot pay more than ni, because then the difference between the paid amount and ni can be regarded as a bribe!
Each rector is wearing the distinctive uniform of his university. Therefore, the uniform's pockets cannot contain coins of denomination more than ri. The rector also does not carry coins of denomination less than li in his pocket β because if everyone pays him with so small coins, they gather a lot of weight and the pocket tears. Therefore, a donation can be paid only by coins of denomination x berubleys, where li β€ x β€ ri (Berland uses coins of any positive integer denomination). Alexey can use the coins of different denominations and he can use the coins of the same denomination any number of times. When Alexey was first confronted with such orders, he was puzzled because it turned out that not all universities can accept him! Alexey is very afraid of going into the army (even though he had long wanted to get the green uniform, but his dad says that the army bullies will beat his son and he cannot pay to ensure the boy's safety). So, Alexey wants to know for sure which universities he can enter so that he could quickly choose his alma mater.
Thanks to the parents, Alexey is not limited in money and we can assume that he has an unlimited number of coins of each type.
In other words, you are given t requests, each of them contains numbers ni, li, ri. For each query you need to answer, whether it is possible to gather the sum of exactly ni berubleys using only coins with an integer denomination from li to ri berubleys. You can use coins of different denominations. Coins of each denomination can be used any number of times.
Input
The first line contains the number of universities t, (1 β€ t β€ 1000) Each of the next t lines contain three space-separated integers: ni, li, ri (1 β€ ni, li, ri β€ 109; li β€ ri).
Output
For each query print on a single line: either "Yes", if Alexey can enter the university, or "No" otherwise.
Examples
Input
2
5 2 3
6 4 5
Output
Yes
No
Note
You can pay the donation to the first university with two coins: one of denomination 2 and one of denomination 3 berubleys. The donation to the second university cannot be paid. | instruction | 0 | 106,497 | 10 | 212,994 |
Tags: constructive algorithms, implementation, math
Correct Solution:
```
class CodeforcesTask397BSolution:
def __init__(self):
self.result = ''
self.t = 0
self.queries = []
def read_input(self):
self.t = int(input())
for _ in range(self.t):
self.queries.append([int(x) for x in input().split(" ")])
def process_task(self):
res = []
for query in self.queries:
k = query[0] // query[1]
res.append("Yes" if k * query[2] >= query[0] else "No")
self.result = "\n".join(res)
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask397BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | output | 1 | 106,497 | 10 | 212,995 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alexey, a merry Berland entrant, got sick of the gray reality and he zealously wants to go to university. There are a lot of universities nowadays, so Alexey is getting lost in the diversity β he has not yet decided what profession he wants to get. At school, he had bad grades in all subjects, and it's only thanks to wealthy parents that he was able to obtain the graduation certificate.
The situation is complicated by the fact that each high education institution has the determined amount of voluntary donations, paid by the new students for admission β ni berubleys. He cannot pay more than ni, because then the difference between the paid amount and ni can be regarded as a bribe!
Each rector is wearing the distinctive uniform of his university. Therefore, the uniform's pockets cannot contain coins of denomination more than ri. The rector also does not carry coins of denomination less than li in his pocket β because if everyone pays him with so small coins, they gather a lot of weight and the pocket tears. Therefore, a donation can be paid only by coins of denomination x berubleys, where li β€ x β€ ri (Berland uses coins of any positive integer denomination). Alexey can use the coins of different denominations and he can use the coins of the same denomination any number of times. When Alexey was first confronted with such orders, he was puzzled because it turned out that not all universities can accept him! Alexey is very afraid of going into the army (even though he had long wanted to get the green uniform, but his dad says that the army bullies will beat his son and he cannot pay to ensure the boy's safety). So, Alexey wants to know for sure which universities he can enter so that he could quickly choose his alma mater.
Thanks to the parents, Alexey is not limited in money and we can assume that he has an unlimited number of coins of each type.
In other words, you are given t requests, each of them contains numbers ni, li, ri. For each query you need to answer, whether it is possible to gather the sum of exactly ni berubleys using only coins with an integer denomination from li to ri berubleys. You can use coins of different denominations. Coins of each denomination can be used any number of times.
Input
The first line contains the number of universities t, (1 β€ t β€ 1000) Each of the next t lines contain three space-separated integers: ni, li, ri (1 β€ ni, li, ri β€ 109; li β€ ri).
Output
For each query print on a single line: either "Yes", if Alexey can enter the university, or "No" otherwise.
Examples
Input
2
5 2 3
6 4 5
Output
Yes
No
Note
You can pay the donation to the first university with two coins: one of denomination 2 and one of denomination 3 berubleys. The donation to the second university cannot be paid. | instruction | 0 | 106,498 | 10 | 212,996 |
Tags: constructive algorithms, implementation, math
Correct Solution:
```
for i in range(int(input())):
n, l, r = map(int, input().split())
print('No' if (n // l) * r < n else 'Yes')
``` | output | 1 | 106,498 | 10 | 212,997 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alexey, a merry Berland entrant, got sick of the gray reality and he zealously wants to go to university. There are a lot of universities nowadays, so Alexey is getting lost in the diversity β he has not yet decided what profession he wants to get. At school, he had bad grades in all subjects, and it's only thanks to wealthy parents that he was able to obtain the graduation certificate.
The situation is complicated by the fact that each high education institution has the determined amount of voluntary donations, paid by the new students for admission β ni berubleys. He cannot pay more than ni, because then the difference between the paid amount and ni can be regarded as a bribe!
Each rector is wearing the distinctive uniform of his university. Therefore, the uniform's pockets cannot contain coins of denomination more than ri. The rector also does not carry coins of denomination less than li in his pocket β because if everyone pays him with so small coins, they gather a lot of weight and the pocket tears. Therefore, a donation can be paid only by coins of denomination x berubleys, where li β€ x β€ ri (Berland uses coins of any positive integer denomination). Alexey can use the coins of different denominations and he can use the coins of the same denomination any number of times. When Alexey was first confronted with such orders, he was puzzled because it turned out that not all universities can accept him! Alexey is very afraid of going into the army (even though he had long wanted to get the green uniform, but his dad says that the army bullies will beat his son and he cannot pay to ensure the boy's safety). So, Alexey wants to know for sure which universities he can enter so that he could quickly choose his alma mater.
Thanks to the parents, Alexey is not limited in money and we can assume that he has an unlimited number of coins of each type.
In other words, you are given t requests, each of them contains numbers ni, li, ri. For each query you need to answer, whether it is possible to gather the sum of exactly ni berubleys using only coins with an integer denomination from li to ri berubleys. You can use coins of different denominations. Coins of each denomination can be used any number of times.
Input
The first line contains the number of universities t, (1 β€ t β€ 1000) Each of the next t lines contain three space-separated integers: ni, li, ri (1 β€ ni, li, ri β€ 109; li β€ ri).
Output
For each query print on a single line: either "Yes", if Alexey can enter the university, or "No" otherwise.
Examples
Input
2
5 2 3
6 4 5
Output
Yes
No
Note
You can pay the donation to the first university with two coins: one of denomination 2 and one of denomination 3 berubleys. The donation to the second university cannot be paid. | instruction | 0 | 106,499 | 10 | 212,998 |
Tags: constructive algorithms, implementation, math
Correct Solution:
```
def possible(numbers):
if int(numbers[0])//int(numbers[1])!=0 and int(numbers[0])%int(numbers[1])//(int(numbers[0])//int(numbers[1]))<=int(numbers[2])-int(numbers[1]) and (int(numbers[0])%int(numbers[1])%(int(numbers[0])//int(numbers[1])) == 0 or int(numbers[0])%int(numbers[1])%int(numbers[0])//int(numbers[1])+int(numbers[0])%int(numbers[1])%(int(numbers[0])//int(numbers[1]))<=(int(numbers[2])-int(numbers[1]))*(int(numbers[0])//int(numbers[1])-int(numbers[0])//int(numbers[2]))):
return 'Yes'
return 'No'
for i in range(0,int(input())):
numbers=input().split(' ')
if i == -1 and int(numbers[0])!=2 and int(numbers[0])!=912247143:
print(numbers)
print(possible(numbers))
``` | output | 1 | 106,499 | 10 | 212,999 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexey, a merry Berland entrant, got sick of the gray reality and he zealously wants to go to university. There are a lot of universities nowadays, so Alexey is getting lost in the diversity β he has not yet decided what profession he wants to get. At school, he had bad grades in all subjects, and it's only thanks to wealthy parents that he was able to obtain the graduation certificate.
The situation is complicated by the fact that each high education institution has the determined amount of voluntary donations, paid by the new students for admission β ni berubleys. He cannot pay more than ni, because then the difference between the paid amount and ni can be regarded as a bribe!
Each rector is wearing the distinctive uniform of his university. Therefore, the uniform's pockets cannot contain coins of denomination more than ri. The rector also does not carry coins of denomination less than li in his pocket β because if everyone pays him with so small coins, they gather a lot of weight and the pocket tears. Therefore, a donation can be paid only by coins of denomination x berubleys, where li β€ x β€ ri (Berland uses coins of any positive integer denomination). Alexey can use the coins of different denominations and he can use the coins of the same denomination any number of times. When Alexey was first confronted with such orders, he was puzzled because it turned out that not all universities can accept him! Alexey is very afraid of going into the army (even though he had long wanted to get the green uniform, but his dad says that the army bullies will beat his son and he cannot pay to ensure the boy's safety). So, Alexey wants to know for sure which universities he can enter so that he could quickly choose his alma mater.
Thanks to the parents, Alexey is not limited in money and we can assume that he has an unlimited number of coins of each type.
In other words, you are given t requests, each of them contains numbers ni, li, ri. For each query you need to answer, whether it is possible to gather the sum of exactly ni berubleys using only coins with an integer denomination from li to ri berubleys. You can use coins of different denominations. Coins of each denomination can be used any number of times.
Input
The first line contains the number of universities t, (1 β€ t β€ 1000) Each of the next t lines contain three space-separated integers: ni, li, ri (1 β€ ni, li, ri β€ 109; li β€ ri).
Output
For each query print on a single line: either "Yes", if Alexey can enter the university, or "No" otherwise.
Examples
Input
2
5 2 3
6 4 5
Output
Yes
No
Note
You can pay the donation to the first university with two coins: one of denomination 2 and one of denomination 3 berubleys. The donation to the second university cannot be paid.
Submitted Solution:
```
def possible(numbers):
if int(numbers[0])==int(numbers[1]) or int(numbers[0])==int(numbers[2]) or (int(numbers[0])>int(numbers[1]) and int(numbers[0])<int(numbers[2])):
return 'Yes'
elif int(numbers[0])<int(numbers[1]) or int(numbers[0])//int(numbers[1])<int(numbers[1])-int(numbers[2]):
return 'No'
else:
if int(numbers[0])%int(numbers[1])==0 or int(numbers[0])%int(numbers[1])+int(numbers[1])<int(numbers[2]) or (int(numbers[0])%int(numbers[1])+int(numbers[1])*(int(numbers[0])//int(numbers[1])-1)<=int(numbers[2]) and int(numbers[0])%int(numbers[1])+int(numbers[1])*(int(numbers[0])//int(numbers[1])-1)>=int(numbers[1])) or int(numbers[0])%int(numbers[1])//(int(numbers[0])//int(numbers[1]))<int(numbers[2])-int(numbers[1]):
return 'Yes'
return 'No'
for i in range(0,int(input())):
numbers=input().split(' ')
if i == -1 and int(numbers[0])!=2:
print(numbers)
print(possible(numbers))
``` | instruction | 0 | 106,500 | 10 | 213,000 |
Yes | output | 1 | 106,500 | 10 | 213,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexey, a merry Berland entrant, got sick of the gray reality and he zealously wants to go to university. There are a lot of universities nowadays, so Alexey is getting lost in the diversity β he has not yet decided what profession he wants to get. At school, he had bad grades in all subjects, and it's only thanks to wealthy parents that he was able to obtain the graduation certificate.
The situation is complicated by the fact that each high education institution has the determined amount of voluntary donations, paid by the new students for admission β ni berubleys. He cannot pay more than ni, because then the difference between the paid amount and ni can be regarded as a bribe!
Each rector is wearing the distinctive uniform of his university. Therefore, the uniform's pockets cannot contain coins of denomination more than ri. The rector also does not carry coins of denomination less than li in his pocket β because if everyone pays him with so small coins, they gather a lot of weight and the pocket tears. Therefore, a donation can be paid only by coins of denomination x berubleys, where li β€ x β€ ri (Berland uses coins of any positive integer denomination). Alexey can use the coins of different denominations and he can use the coins of the same denomination any number of times. When Alexey was first confronted with such orders, he was puzzled because it turned out that not all universities can accept him! Alexey is very afraid of going into the army (even though he had long wanted to get the green uniform, but his dad says that the army bullies will beat his son and he cannot pay to ensure the boy's safety). So, Alexey wants to know for sure which universities he can enter so that he could quickly choose his alma mater.
Thanks to the parents, Alexey is not limited in money and we can assume that he has an unlimited number of coins of each type.
In other words, you are given t requests, each of them contains numbers ni, li, ri. For each query you need to answer, whether it is possible to gather the sum of exactly ni berubleys using only coins with an integer denomination from li to ri berubleys. You can use coins of different denominations. Coins of each denomination can be used any number of times.
Input
The first line contains the number of universities t, (1 β€ t β€ 1000) Each of the next t lines contain three space-separated integers: ni, li, ri (1 β€ ni, li, ri β€ 109; li β€ ri).
Output
For each query print on a single line: either "Yes", if Alexey can enter the university, or "No" otherwise.
Examples
Input
2
5 2 3
6 4 5
Output
Yes
No
Note
You can pay the donation to the first university with two coins: one of denomination 2 and one of denomination 3 berubleys. The donation to the second university cannot be paid.
Submitted Solution:
```
for i in range(int(input())):
n,l,r = map(int,input().split())
print('No') if (n//l)<n/r else print('Yes')
``` | instruction | 0 | 106,501 | 10 | 213,002 |
Yes | output | 1 | 106,501 | 10 | 213,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexey, a merry Berland entrant, got sick of the gray reality and he zealously wants to go to university. There are a lot of universities nowadays, so Alexey is getting lost in the diversity β he has not yet decided what profession he wants to get. At school, he had bad grades in all subjects, and it's only thanks to wealthy parents that he was able to obtain the graduation certificate.
The situation is complicated by the fact that each high education institution has the determined amount of voluntary donations, paid by the new students for admission β ni berubleys. He cannot pay more than ni, because then the difference between the paid amount and ni can be regarded as a bribe!
Each rector is wearing the distinctive uniform of his university. Therefore, the uniform's pockets cannot contain coins of denomination more than ri. The rector also does not carry coins of denomination less than li in his pocket β because if everyone pays him with so small coins, they gather a lot of weight and the pocket tears. Therefore, a donation can be paid only by coins of denomination x berubleys, where li β€ x β€ ri (Berland uses coins of any positive integer denomination). Alexey can use the coins of different denominations and he can use the coins of the same denomination any number of times. When Alexey was first confronted with such orders, he was puzzled because it turned out that not all universities can accept him! Alexey is very afraid of going into the army (even though he had long wanted to get the green uniform, but his dad says that the army bullies will beat his son and he cannot pay to ensure the boy's safety). So, Alexey wants to know for sure which universities he can enter so that he could quickly choose his alma mater.
Thanks to the parents, Alexey is not limited in money and we can assume that he has an unlimited number of coins of each type.
In other words, you are given t requests, each of them contains numbers ni, li, ri. For each query you need to answer, whether it is possible to gather the sum of exactly ni berubleys using only coins with an integer denomination from li to ri berubleys. You can use coins of different denominations. Coins of each denomination can be used any number of times.
Input
The first line contains the number of universities t, (1 β€ t β€ 1000) Each of the next t lines contain three space-separated integers: ni, li, ri (1 β€ ni, li, ri β€ 109; li β€ ri).
Output
For each query print on a single line: either "Yes", if Alexey can enter the university, or "No" otherwise.
Examples
Input
2
5 2 3
6 4 5
Output
Yes
No
Note
You can pay the donation to the first university with two coins: one of denomination 2 and one of denomination 3 berubleys. The donation to the second university cannot be paid.
Submitted Solution:
```
def possible(numbers):
if int(numbers[0])//int(numbers[1])!=0 and int(numbers[0])%int(numbers[1])//(int(numbers[0])//int(numbers[1]))<=int(numbers[2])-int(numbers[1]) and int(numbers[0])%int(numbers[1])%int(numbers[0])//int(numbers[1])+int(numbers[0])%int(numbers[1])%(int(numbers[0])//int(numbers[1]))<=(int(numbers[2])-int(numbers[1]))*(int(numbers[0])//int(numbers[1])-int(numbers[0])//int(numbers[2])):
return 'Yes'
return 'No'
for i in range(0,int(input())):
numbers=input().split(' ')
if i == -1 and int(numbers[0])!=2 and int(numbers[0])!=912247143:
print(numbers)
print(possible(numbers))
``` | instruction | 0 | 106,502 | 10 | 213,004 |
Yes | output | 1 | 106,502 | 10 | 213,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexey, a merry Berland entrant, got sick of the gray reality and he zealously wants to go to university. There are a lot of universities nowadays, so Alexey is getting lost in the diversity β he has not yet decided what profession he wants to get. At school, he had bad grades in all subjects, and it's only thanks to wealthy parents that he was able to obtain the graduation certificate.
The situation is complicated by the fact that each high education institution has the determined amount of voluntary donations, paid by the new students for admission β ni berubleys. He cannot pay more than ni, because then the difference between the paid amount and ni can be regarded as a bribe!
Each rector is wearing the distinctive uniform of his university. Therefore, the uniform's pockets cannot contain coins of denomination more than ri. The rector also does not carry coins of denomination less than li in his pocket β because if everyone pays him with so small coins, they gather a lot of weight and the pocket tears. Therefore, a donation can be paid only by coins of denomination x berubleys, where li β€ x β€ ri (Berland uses coins of any positive integer denomination). Alexey can use the coins of different denominations and he can use the coins of the same denomination any number of times. When Alexey was first confronted with such orders, he was puzzled because it turned out that not all universities can accept him! Alexey is very afraid of going into the army (even though he had long wanted to get the green uniform, but his dad says that the army bullies will beat his son and he cannot pay to ensure the boy's safety). So, Alexey wants to know for sure which universities he can enter so that he could quickly choose his alma mater.
Thanks to the parents, Alexey is not limited in money and we can assume that he has an unlimited number of coins of each type.
In other words, you are given t requests, each of them contains numbers ni, li, ri. For each query you need to answer, whether it is possible to gather the sum of exactly ni berubleys using only coins with an integer denomination from li to ri berubleys. You can use coins of different denominations. Coins of each denomination can be used any number of times.
Input
The first line contains the number of universities t, (1 β€ t β€ 1000) Each of the next t lines contain three space-separated integers: ni, li, ri (1 β€ ni, li, ri β€ 109; li β€ ri).
Output
For each query print on a single line: either "Yes", if Alexey can enter the university, or "No" otherwise.
Examples
Input
2
5 2 3
6 4 5
Output
Yes
No
Note
You can pay the donation to the first university with two coins: one of denomination 2 and one of denomination 3 berubleys. The donation to the second university cannot be paid.
Submitted Solution:
```
for i in range(int(input())):
n, l, r = map(int, input().split())
print('No' if (n // l) * r < n else 'Yes')
# Made By Mostafa_Khaled
``` | instruction | 0 | 106,503 | 10 | 213,006 |
Yes | output | 1 | 106,503 | 10 | 213,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexey, a merry Berland entrant, got sick of the gray reality and he zealously wants to go to university. There are a lot of universities nowadays, so Alexey is getting lost in the diversity β he has not yet decided what profession he wants to get. At school, he had bad grades in all subjects, and it's only thanks to wealthy parents that he was able to obtain the graduation certificate.
The situation is complicated by the fact that each high education institution has the determined amount of voluntary donations, paid by the new students for admission β ni berubleys. He cannot pay more than ni, because then the difference between the paid amount and ni can be regarded as a bribe!
Each rector is wearing the distinctive uniform of his university. Therefore, the uniform's pockets cannot contain coins of denomination more than ri. The rector also does not carry coins of denomination less than li in his pocket β because if everyone pays him with so small coins, they gather a lot of weight and the pocket tears. Therefore, a donation can be paid only by coins of denomination x berubleys, where li β€ x β€ ri (Berland uses coins of any positive integer denomination). Alexey can use the coins of different denominations and he can use the coins of the same denomination any number of times. When Alexey was first confronted with such orders, he was puzzled because it turned out that not all universities can accept him! Alexey is very afraid of going into the army (even though he had long wanted to get the green uniform, but his dad says that the army bullies will beat his son and he cannot pay to ensure the boy's safety). So, Alexey wants to know for sure which universities he can enter so that he could quickly choose his alma mater.
Thanks to the parents, Alexey is not limited in money and we can assume that he has an unlimited number of coins of each type.
In other words, you are given t requests, each of them contains numbers ni, li, ri. For each query you need to answer, whether it is possible to gather the sum of exactly ni berubleys using only coins with an integer denomination from li to ri berubleys. You can use coins of different denominations. Coins of each denomination can be used any number of times.
Input
The first line contains the number of universities t, (1 β€ t β€ 1000) Each of the next t lines contain three space-separated integers: ni, li, ri (1 β€ ni, li, ri β€ 109; li β€ ri).
Output
For each query print on a single line: either "Yes", if Alexey can enter the university, or "No" otherwise.
Examples
Input
2
5 2 3
6 4 5
Output
Yes
No
Note
You can pay the donation to the first university with two coins: one of denomination 2 and one of denomination 3 berubleys. The donation to the second university cannot be paid.
Submitted Solution:
```
def possible(numbers):
if int(numbers[0])//int(numbers[1])!=0 and int(numbers[0])%int(numbers[1])//(int(numbers[0])//int(numbers[1]))<=int(numbers[2])-int(numbers[1]) and (int(numbers[0])%int(numbers[1])%(int(numbers[0])//int(numbers[1])) == 0 or int(numbers[0])%int(numbers[1])%int(numbers[0])//int(numbers[1])<=int(numbers[0])//int(numbers[1])-int(numbers[0])//int(numbers[1])):
return 'Yes'
return 'No'
for i in range(0,int(input())):
numbers=input().split(' ')
if i == 65 and int(numbers[0])!=614580054 and int(numbers[0])!=912247143:
print(numbers)
print(possible(numbers))
``` | instruction | 0 | 106,504 | 10 | 213,008 |
No | output | 1 | 106,504 | 10 | 213,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexey, a merry Berland entrant, got sick of the gray reality and he zealously wants to go to university. There are a lot of universities nowadays, so Alexey is getting lost in the diversity β he has not yet decided what profession he wants to get. At school, he had bad grades in all subjects, and it's only thanks to wealthy parents that he was able to obtain the graduation certificate.
The situation is complicated by the fact that each high education institution has the determined amount of voluntary donations, paid by the new students for admission β ni berubleys. He cannot pay more than ni, because then the difference between the paid amount and ni can be regarded as a bribe!
Each rector is wearing the distinctive uniform of his university. Therefore, the uniform's pockets cannot contain coins of denomination more than ri. The rector also does not carry coins of denomination less than li in his pocket β because if everyone pays him with so small coins, they gather a lot of weight and the pocket tears. Therefore, a donation can be paid only by coins of denomination x berubleys, where li β€ x β€ ri (Berland uses coins of any positive integer denomination). Alexey can use the coins of different denominations and he can use the coins of the same denomination any number of times. When Alexey was first confronted with such orders, he was puzzled because it turned out that not all universities can accept him! Alexey is very afraid of going into the army (even though he had long wanted to get the green uniform, but his dad says that the army bullies will beat his son and he cannot pay to ensure the boy's safety). So, Alexey wants to know for sure which universities he can enter so that he could quickly choose his alma mater.
Thanks to the parents, Alexey is not limited in money and we can assume that he has an unlimited number of coins of each type.
In other words, you are given t requests, each of them contains numbers ni, li, ri. For each query you need to answer, whether it is possible to gather the sum of exactly ni berubleys using only coins with an integer denomination from li to ri berubleys. You can use coins of different denominations. Coins of each denomination can be used any number of times.
Input
The first line contains the number of universities t, (1 β€ t β€ 1000) Each of the next t lines contain three space-separated integers: ni, li, ri (1 β€ ni, li, ri β€ 109; li β€ ri).
Output
For each query print on a single line: either "Yes", if Alexey can enter the university, or "No" otherwise.
Examples
Input
2
5 2 3
6 4 5
Output
Yes
No
Note
You can pay the donation to the first university with two coins: one of denomination 2 and one of denomination 3 berubleys. The donation to the second university cannot be paid.
Submitted Solution:
```
def possible(numbers):
if int(numbers[0])==int(numbers[1]) or int(numbers[0])==int(numbers[2]) or (int(numbers[0])>int(numbers[1]) and int(numbers[0])<int(numbers[2])):
return 'Yes'
elif int(numbers[0])<int(numbers[1]) or int(numbers[0])//int(numbers[1])<int(numbers[1])-int(numbers[2]):
return 'No'
else:
if int(numbers[0])%int(numbers[1])==0 or (int(numbers[0])-int(numbers[2]))%int(numbers[1])+int(numbers[1])<int(numbers[2]):
return 'Yes'
return 'No'
for i in range(0,int(input())):
numbers=input().split(' ')
if i == -1:
print(numbers)
print(possible(numbers))
``` | instruction | 0 | 106,505 | 10 | 213,010 |
No | output | 1 | 106,505 | 10 | 213,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexey, a merry Berland entrant, got sick of the gray reality and he zealously wants to go to university. There are a lot of universities nowadays, so Alexey is getting lost in the diversity β he has not yet decided what profession he wants to get. At school, he had bad grades in all subjects, and it's only thanks to wealthy parents that he was able to obtain the graduation certificate.
The situation is complicated by the fact that each high education institution has the determined amount of voluntary donations, paid by the new students for admission β ni berubleys. He cannot pay more than ni, because then the difference between the paid amount and ni can be regarded as a bribe!
Each rector is wearing the distinctive uniform of his university. Therefore, the uniform's pockets cannot contain coins of denomination more than ri. The rector also does not carry coins of denomination less than li in his pocket β because if everyone pays him with so small coins, they gather a lot of weight and the pocket tears. Therefore, a donation can be paid only by coins of denomination x berubleys, where li β€ x β€ ri (Berland uses coins of any positive integer denomination). Alexey can use the coins of different denominations and he can use the coins of the same denomination any number of times. When Alexey was first confronted with such orders, he was puzzled because it turned out that not all universities can accept him! Alexey is very afraid of going into the army (even though he had long wanted to get the green uniform, but his dad says that the army bullies will beat his son and he cannot pay to ensure the boy's safety). So, Alexey wants to know for sure which universities he can enter so that he could quickly choose his alma mater.
Thanks to the parents, Alexey is not limited in money and we can assume that he has an unlimited number of coins of each type.
In other words, you are given t requests, each of them contains numbers ni, li, ri. For each query you need to answer, whether it is possible to gather the sum of exactly ni berubleys using only coins with an integer denomination from li to ri berubleys. You can use coins of different denominations. Coins of each denomination can be used any number of times.
Input
The first line contains the number of universities t, (1 β€ t β€ 1000) Each of the next t lines contain three space-separated integers: ni, li, ri (1 β€ ni, li, ri β€ 109; li β€ ri).
Output
For each query print on a single line: either "Yes", if Alexey can enter the university, or "No" otherwise.
Examples
Input
2
5 2 3
6 4 5
Output
Yes
No
Note
You can pay the donation to the first university with two coins: one of denomination 2 and one of denomination 3 berubleys. The donation to the second university cannot be paid.
Submitted Solution:
```
def possible(numbers):
if int(numbers[0])<int(numbers[1])*2 or int(numbers[0])<int(numbers[1]):
return 'No'
elif int(numbers[0])<=int(numbers[1]):
return 'Yes'
else:
coins=list(range(int(numbers[1]),int(numbers[2])+1))
for i in coins:
if int(numbers[0])%i==0 or int(numbers[0])%i in coins:
return 'Yes'
return 'No'
for i in range(0,int(input())):
print(possible(input().split(' ')))
``` | instruction | 0 | 106,506 | 10 | 213,012 |
No | output | 1 | 106,506 | 10 | 213,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexey, a merry Berland entrant, got sick of the gray reality and he zealously wants to go to university. There are a lot of universities nowadays, so Alexey is getting lost in the diversity β he has not yet decided what profession he wants to get. At school, he had bad grades in all subjects, and it's only thanks to wealthy parents that he was able to obtain the graduation certificate.
The situation is complicated by the fact that each high education institution has the determined amount of voluntary donations, paid by the new students for admission β ni berubleys. He cannot pay more than ni, because then the difference between the paid amount and ni can be regarded as a bribe!
Each rector is wearing the distinctive uniform of his university. Therefore, the uniform's pockets cannot contain coins of denomination more than ri. The rector also does not carry coins of denomination less than li in his pocket β because if everyone pays him with so small coins, they gather a lot of weight and the pocket tears. Therefore, a donation can be paid only by coins of denomination x berubleys, where li β€ x β€ ri (Berland uses coins of any positive integer denomination). Alexey can use the coins of different denominations and he can use the coins of the same denomination any number of times. When Alexey was first confronted with such orders, he was puzzled because it turned out that not all universities can accept him! Alexey is very afraid of going into the army (even though he had long wanted to get the green uniform, but his dad says that the army bullies will beat his son and he cannot pay to ensure the boy's safety). So, Alexey wants to know for sure which universities he can enter so that he could quickly choose his alma mater.
Thanks to the parents, Alexey is not limited in money and we can assume that he has an unlimited number of coins of each type.
In other words, you are given t requests, each of them contains numbers ni, li, ri. For each query you need to answer, whether it is possible to gather the sum of exactly ni berubleys using only coins with an integer denomination from li to ri berubleys. You can use coins of different denominations. Coins of each denomination can be used any number of times.
Input
The first line contains the number of universities t, (1 β€ t β€ 1000) Each of the next t lines contain three space-separated integers: ni, li, ri (1 β€ ni, li, ri β€ 109; li β€ ri).
Output
For each query print on a single line: either "Yes", if Alexey can enter the university, or "No" otherwise.
Examples
Input
2
5 2 3
6 4 5
Output
Yes
No
Note
You can pay the donation to the first university with two coins: one of denomination 2 and one of denomination 3 berubleys. The donation to the second university cannot be paid.
Submitted Solution:
```
def possible(numbers):
if int(numbers[0])//int(numbers[1])!=0 and int(numbers[0])%int(numbers[1])//(int(numbers[0])//int(numbers[1]))<=int(numbers[2])-int(numbers[1]) and (int(numbers[0])%int(numbers[1])%(int(numbers[0])//int(numbers[1])) == 0 or int(numbers[0])%int(numbers[1])%int(numbers[0])//int(numbers[1])+int(numbers[0])%int(numbers[1])%(int(numbers[0])//int(numbers[1]))<=int(numbers[0])//int(numbers[1])-int(numbers[0])//int(numbers[2])):
return 'Yes'
return 'No'
for i in range(0,int(input())):
numbers=input().split(' ')
if i == -1 and int(numbers[0])!=614580054 and int(numbers[0])!=912247143:
print(numbers)
print(possible(numbers))
``` | instruction | 0 | 106,507 | 10 | 213,014 |
No | output | 1 | 106,507 | 10 | 213,015 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly n Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.
Input
The first and only line contains an integer n (1 β€ n β€ 106) which represents the denomination of the most expensive coin.
Output
Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination n of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.
Examples
Input
10
Output
10 5 1
Input
4
Output
4 2 1
Input
3
Output
3 1 | instruction | 0 | 106,606 | 10 | 213,212 |
Tags: greedy
Correct Solution:
```
n = int(input())
for i in range(n,0,-1):
if n%i==0:
n=i
print(i,end = ' ')
``` | output | 1 | 106,606 | 10 | 213,213 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly n Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.
Input
The first and only line contains an integer n (1 β€ n β€ 106) which represents the denomination of the most expensive coin.
Output
Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination n of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.
Examples
Input
10
Output
10 5 1
Input
4
Output
4 2 1
Input
3
Output
3 1 | instruction | 0 | 106,607 | 10 | 213,214 |
Tags: greedy
Correct Solution:
```
n = int(input())
o = n
e = o
while o != 0:
if n % o == 0 and e % o == 0:
print(o, end = ' ')
e = o
o -= 1
``` | output | 1 | 106,607 | 10 | 213,215 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly n Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.
Input
The first and only line contains an integer n (1 β€ n β€ 106) which represents the denomination of the most expensive coin.
Output
Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination n of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.
Examples
Input
10
Output
10 5 1
Input
4
Output
4 2 1
Input
3
Output
3 1 | instruction | 0 | 106,608 | 10 | 213,216 |
Tags: greedy
Correct Solution:
```
n=int(input())
final=[n]
for i in range(n-1,0,-1):
if n%i==0:
z=0
for j in final:
if j%i!=0:
z=1
break
if z==0:
final.append(i)
print(' '.join(str(x) for x in final))
``` | output | 1 | 106,608 | 10 | 213,217 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly n Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.
Input
The first and only line contains an integer n (1 β€ n β€ 106) which represents the denomination of the most expensive coin.
Output
Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination n of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.
Examples
Input
10
Output
10 5 1
Input
4
Output
4 2 1
Input
3
Output
3 1 | instruction | 0 | 106,609 | 10 | 213,218 |
Tags: greedy
Correct Solution:
```
n = int(input())
l = []
for i in range(1, n):
k = 0
if n % i == 0:
for j in range(len(l)):
if i % l[j] == 0:
k += 1
if k == len(l):
l.append(i)
l.append(n)
l = l[:: - 1]
for item in l:
print(item, end = ' ')
``` | output | 1 | 106,609 | 10 | 213,219 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly n Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.
Input
The first and only line contains an integer n (1 β€ n β€ 106) which represents the denomination of the most expensive coin.
Output
Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination n of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.
Examples
Input
10
Output
10 5 1
Input
4
Output
4 2 1
Input
3
Output
3 1 | instruction | 0 | 106,610 | 10 | 213,220 |
Tags: greedy
Correct Solution:
```
n = int(input())
print(n, end=" ")
i = n // 2
while i:
if not n % i:
n = i
print(n, end=" ")
i -= 1
``` | output | 1 | 106,610 | 10 | 213,221 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly n Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.
Input
The first and only line contains an integer n (1 β€ n β€ 106) which represents the denomination of the most expensive coin.
Output
Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination n of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.
Examples
Input
10
Output
10 5 1
Input
4
Output
4 2 1
Input
3
Output
3 1 | instruction | 0 | 106,611 | 10 | 213,222 |
Tags: greedy
Correct Solution:
```
n = int(input())
if n == 1:
print(1)
exit()
res = [n]
while n > 1:
i = 2
f = 1
while i * i <= n:
if n % i == 0:
res.append(max(i, n // i))
n = res[-1]
f = 0
break
i+= 1
if f:
break
res.append(1)
print(*res)
``` | output | 1 | 106,611 | 10 | 213,223 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly n Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.
Input
The first and only line contains an integer n (1 β€ n β€ 106) which represents the denomination of the most expensive coin.
Output
Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination n of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.
Examples
Input
10
Output
10 5 1
Input
4
Output
4 2 1
Input
3
Output
3 1 | instruction | 0 | 106,612 | 10 | 213,224 |
Tags: greedy
Correct Solution:
```
import math
n = int(input())
coins = []
def getLargestFactor (number):
factors = []
i = 1
while (i < math.floor(math.sqrt(number)) + 1):
if (number % i == 0):
factors.append(i)
factors.append(number/i)
i += 1
factors.sort()
return factors[len(factors) - 2]
temp = n
while (temp != 1):
coins.append(temp)
temp = getLargestFactor(temp)
coins.append(1)
for i in coins:
print(int(i),end = ' ')
``` | output | 1 | 106,612 | 10 | 213,225 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly n Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.
Input
The first and only line contains an integer n (1 β€ n β€ 106) which represents the denomination of the most expensive coin.
Output
Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination n of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.
Examples
Input
10
Output
10 5 1
Input
4
Output
4 2 1
Input
3
Output
3 1 | instruction | 0 | 106,613 | 10 | 213,226 |
Tags: greedy
Correct Solution:
```
n = int(input())
ans = [n]
while n > 1:
d = 2
while n % d:
d += 1
n //= d
ans.append(n)
print(*ans)
``` | output | 1 | 106,613 | 10 | 213,227 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly n Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.
Input
The first and only line contains an integer n (1 β€ n β€ 106) which represents the denomination of the most expensive coin.
Output
Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination n of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.
Examples
Input
10
Output
10 5 1
Input
4
Output
4 2 1
Input
3
Output
3 1
Submitted Solution:
```
a=int(input())
for i in range(a,0,-1):
if a%i==0:
a=i
print(a)
``` | instruction | 0 | 106,614 | 10 | 213,228 |
Yes | output | 1 | 106,614 | 10 | 213,229 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly n Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.
Input
The first and only line contains an integer n (1 β€ n β€ 106) which represents the denomination of the most expensive coin.
Output
Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination n of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.
Examples
Input
10
Output
10 5 1
Input
4
Output
4 2 1
Input
3
Output
3 1
Submitted Solution:
```
def g(n):
sq=int(n**0.5)
for i in range(2,sq+2):
if(n%i==0):
return n//i
return 1
n=int(input())
while(n!=1):
print(n,end=" ")
n=g(n)
print(1)
``` | instruction | 0 | 106,615 | 10 | 213,230 |
Yes | output | 1 | 106,615 | 10 | 213,231 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly n Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.
Input
The first and only line contains an integer n (1 β€ n β€ 106) which represents the denomination of the most expensive coin.
Output
Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination n of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.
Examples
Input
10
Output
10 5 1
Input
4
Output
4 2 1
Input
3
Output
3 1
Submitted Solution:
```
#############################################################################################################
# Autor: Jorge LΓ³pez
# Problema: Coins
#
# Algortmo a seguir:
# 1. Recibimos la moneda
# 2. Ciclo que recoore de coins hasta 0
# 3. Si coins modulo i = 0 entonces:
# coins = i
# 4. Imprmio las coin
# 10 %10 = 0
# 10 %5 = 0
# 5 %1 = 0
#
# Input: 10
#
# Output: 10 5 1
#
##############################################################################################################
#Inicia programa
coin = int(input())
# Recorremos de mayor a menor las moneas posibles
# Recorremos desde coin hasta el 0 con recorrido de -1
for i in range(coin,0,-1):
if (coin % i == 0):
coin = i
print(coin,"",end='')
``` | instruction | 0 | 106,616 | 10 | 213,232 |
Yes | output | 1 | 106,616 | 10 | 213,233 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly n Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.
Input
The first and only line contains an integer n (1 β€ n β€ 106) which represents the denomination of the most expensive coin.
Output
Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination n of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.
Examples
Input
10
Output
10 5 1
Input
4
Output
4 2 1
Input
3
Output
3 1
Submitted Solution:
```
n=int(input())
res=[]
res.append(n)
while(n!=1):
if(n%2==0):
n//=2
res.append(n)
else:
i=3
while(n%i!=0):
i+=2
n//=i
res.append(n)
print(*res)
``` | instruction | 0 | 106,617 | 10 | 213,234 |
Yes | output | 1 | 106,617 | 10 | 213,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly n Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.
Input
The first and only line contains an integer n (1 β€ n β€ 106) which represents the denomination of the most expensive coin.
Output
Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination n of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.
Examples
Input
10
Output
10 5 1
Input
4
Output
4 2 1
Input
3
Output
3 1
Submitted Solution:
```
__author__ = "runekri3"
n = int(input())
is_odd = n % 2
acceptable_coins = [n, 1]
for coin in range(2 + is_odd, n, 2):
if n % coin == 0:
acceptable_coins.append(coin)
acceptable_coins.sort(reverse=True)
print(" ".join(map(str, acceptable_coins)))
``` | instruction | 0 | 106,618 | 10 | 213,236 |
No | output | 1 | 106,618 | 10 | 213,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly n Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.
Input
The first and only line contains an integer n (1 β€ n β€ 106) which represents the denomination of the most expensive coin.
Output
Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination n of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.
Examples
Input
10
Output
10 5 1
Input
4
Output
4 2 1
Input
3
Output
3 1
Submitted Solution:
```
n = int(input())
a = set([1, n])
for p in range(2, n):
if n==p:
break
while n > 1 and n%p==0:
# print (n, p)
a.add(p)
n//=p
if not n:
break
a = list(a)
a.sort(reverse=True)
print(*a)
``` | instruction | 0 | 106,619 | 10 | 213,238 |
No | output | 1 | 106,619 | 10 | 213,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly n Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.
Input
The first and only line contains an integer n (1 β€ n β€ 106) which represents the denomination of the most expensive coin.
Output
Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination n of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.
Examples
Input
10
Output
10 5 1
Input
4
Output
4 2 1
Input
3
Output
3 1
Submitted Solution:
```
import sys
import math
n = int(sys.stdin.readline())
k = [str(n)]
while(n > 1):
if(n % 2 == 0):
n /= 2
k.append(str(int(n)))
elif(n % 3 == 0):
n /= 3
k.append(str(int(n)))
elif(n % 5 == 0):
n /= 5
k.append(str(int(n)))
else:
n = 1
k.append('1')
print(" ".join(k))
``` | instruction | 0 | 106,620 | 10 | 213,240 |
No | output | 1 | 106,620 | 10 | 213,241 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Berland a money reform is being prepared. New coins are being introduced. After long economic calculations was decided that the most expensive coin should possess the denomination of exactly n Berland dollars. Also the following restriction has been introduced for comfort: the denomination of each coin should be divisible by the denomination of any cheaper coin. It is known that among all the possible variants the variant with the largest number of new coins will be chosen. Find this variant. Print in the order of decreasing of the coins' denominations.
Input
The first and only line contains an integer n (1 β€ n β€ 106) which represents the denomination of the most expensive coin.
Output
Print the denominations of all the coins in the order of decreasing. The number of coins must be the largest possible (with the given denomination n of the most expensive coin). Also, the denomination of every coin must be divisible by the denomination of any cheaper coin. Naturally, the denominations of all the coins should be different. If there are several solutins to that problem, print any of them.
Examples
Input
10
Output
10 5 1
Input
4
Output
4 2 1
Input
3
Output
3 1
Submitted Solution:
```
import math
def prime(n):
a=set()
if n%2==0:
while(n%2==0):
a.add(n//2)
n=n//2
p=math.sqrt(n)
p=int(p)
for i in range(3,p+1):
if p%i==0:
a.add(p//i)
p=p//i
a.add(1)
return a
n=int(input())
a=prime(n)
a=list(a)
if n==1:
print(1)
exit()
a.append(n)
a.sort(reverse=True)
print(*a)
``` | instruction | 0 | 106,621 | 10 | 213,242 |
No | output | 1 | 106,621 | 10 | 213,243 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to xi coins.
Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains n integers xi (1 β€ xi β€ 100 000) β prices of the bottles of the drink in the i-th shop.
The third line contains a single integer q (1 β€ q β€ 100 000) β the number of days Vasiliy plans to buy the drink.
Then follow q lines each containing one integer mi (1 β€ mi β€ 109) β the number of coins Vasiliy can spent on the i-th day.
Output
Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.
Example
Input
5
3 10 8 6 11
4
1
10
3
11
Output
0
4
1
5
Note
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop. | instruction | 0 | 106,642 | 10 | 213,284 |
Tags: binary search, dp, implementation
Correct Solution:
```
import bisect
n = int(input())
x = sorted(list(map(int, input().split())))
m = int(input())
for i in range(m):
a = int(input())
a = bisect.bisect_right(x, a)
print(a)
``` | output | 1 | 106,642 | 10 | 213,285 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to xi coins.
Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains n integers xi (1 β€ xi β€ 100 000) β prices of the bottles of the drink in the i-th shop.
The third line contains a single integer q (1 β€ q β€ 100 000) β the number of days Vasiliy plans to buy the drink.
Then follow q lines each containing one integer mi (1 β€ mi β€ 109) β the number of coins Vasiliy can spent on the i-th day.
Output
Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.
Example
Input
5
3 10 8 6 11
4
1
10
3
11
Output
0
4
1
5
Note
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop. | instruction | 0 | 106,643 | 10 | 213,286 |
Tags: binary search, dp, implementation
Correct Solution:
```
n = int(input())
prices = list(map(int, input().split()))
price_occur = [0] * 100001
prefix = [0] * 100001
for p in prices:
price_occur[p] += 1
prefix[0] = price_occur[0]
for i in range(1, 100001):
prefix[i] = prefix[i - 1] + price_occur[i]
q = int(input())
ans = []
for i in range(q):
money = int(input())
if money > 100000:
ans.append(n)
else:
ans.append(prefix[money])
print(*ans, sep='\n')
``` | output | 1 | 106,643 | 10 | 213,287 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to xi coins.
Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains n integers xi (1 β€ xi β€ 100 000) β prices of the bottles of the drink in the i-th shop.
The third line contains a single integer q (1 β€ q β€ 100 000) β the number of days Vasiliy plans to buy the drink.
Then follow q lines each containing one integer mi (1 β€ mi β€ 109) β the number of coins Vasiliy can spent on the i-th day.
Output
Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.
Example
Input
5
3 10 8 6 11
4
1
10
3
11
Output
0
4
1
5
Note
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop. | instruction | 0 | 106,644 | 10 | 213,288 |
Tags: binary search, dp, implementation
Correct Solution:
```
import bisect
n=int(input())
l=list(map(int,input().split()))
l=sorted(l)
for i in range(int(input())) :
print(bisect.bisect_right(l,int(input())))
``` | output | 1 | 106,644 | 10 | 213,289 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to xi coins.
Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains n integers xi (1 β€ xi β€ 100 000) β prices of the bottles of the drink in the i-th shop.
The third line contains a single integer q (1 β€ q β€ 100 000) β the number of days Vasiliy plans to buy the drink.
Then follow q lines each containing one integer mi (1 β€ mi β€ 109) β the number of coins Vasiliy can spent on the i-th day.
Output
Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.
Example
Input
5
3 10 8 6 11
4
1
10
3
11
Output
0
4
1
5
Note
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop. | instruction | 0 | 106,645 | 10 | 213,290 |
Tags: binary search, dp, implementation
Correct Solution:
```
from bisect import bisect
n = int(input())
x = sorted(map(int, input().split()))
q = int(input())
for i in range(q):
m = int(input())
print(bisect(x, m))
``` | output | 1 | 106,645 | 10 | 213,291 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to xi coins.
Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains n integers xi (1 β€ xi β€ 100 000) β prices of the bottles of the drink in the i-th shop.
The third line contains a single integer q (1 β€ q β€ 100 000) β the number of days Vasiliy plans to buy the drink.
Then follow q lines each containing one integer mi (1 β€ mi β€ 109) β the number of coins Vasiliy can spent on the i-th day.
Output
Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.
Example
Input
5
3 10 8 6 11
4
1
10
3
11
Output
0
4
1
5
Note
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop. | instruction | 0 | 106,646 | 10 | 213,292 |
Tags: binary search, dp, implementation
Correct Solution:
```
from bisect import bisect_right
n = int(input())
x = [*map(int, input().split())]
x.sort()
q = int(input())
for i in range(q):
d = int(input())
print(bisect_right(x, d))
``` | output | 1 | 106,646 | 10 | 213,293 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to xi coins.
Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains n integers xi (1 β€ xi β€ 100 000) β prices of the bottles of the drink in the i-th shop.
The third line contains a single integer q (1 β€ q β€ 100 000) β the number of days Vasiliy plans to buy the drink.
Then follow q lines each containing one integer mi (1 β€ mi β€ 109) β the number of coins Vasiliy can spent on the i-th day.
Output
Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.
Example
Input
5
3 10 8 6 11
4
1
10
3
11
Output
0
4
1
5
Note
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop. | instruction | 0 | 106,647 | 10 | 213,294 |
Tags: binary search, dp, implementation
Correct Solution:
```
import bisect
input(); x = sorted(map(int, input().split()))
for i in range(int(input())):
print(bisect.bisect_right(x, int(input())))
``` | output | 1 | 106,647 | 10 | 213,295 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to xi coins.
Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains n integers xi (1 β€ xi β€ 100 000) β prices of the bottles of the drink in the i-th shop.
The third line contains a single integer q (1 β€ q β€ 100 000) β the number of days Vasiliy plans to buy the drink.
Then follow q lines each containing one integer mi (1 β€ mi β€ 109) β the number of coins Vasiliy can spent on the i-th day.
Output
Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.
Example
Input
5
3 10 8 6 11
4
1
10
3
11
Output
0
4
1
5
Note
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop. | instruction | 0 | 106,648 | 10 | 213,296 |
Tags: binary search, dp, implementation
Correct Solution:
```
from bisect import *
def BinarySearch(a, x):
i = bisect(a, x)
if i:
return i
else:
return -1
n=int(input())
x=list(map(int,input().split()))
q=int(input())
x.sort()
#print(*x)
for i in range(q):
a=int(input())
k=BinarySearch(x, a)
if(k==-1):
k=0
print(k)
``` | output | 1 | 106,648 | 10 | 213,297 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to xi coins.
Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains n integers xi (1 β€ xi β€ 100 000) β prices of the bottles of the drink in the i-th shop.
The third line contains a single integer q (1 β€ q β€ 100 000) β the number of days Vasiliy plans to buy the drink.
Then follow q lines each containing one integer mi (1 β€ mi β€ 109) β the number of coins Vasiliy can spent on the i-th day.
Output
Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.
Example
Input
5
3 10 8 6 11
4
1
10
3
11
Output
0
4
1
5
Note
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop. | instruction | 0 | 106,649 | 10 | 213,298 |
Tags: binary search, dp, implementation
Correct Solution:
```
def bst(A, x):
if A[0] > x:
return -1
begin = 0
end = len(A)
while end - begin > 1:
mid = (end + begin)//2
if A[mid] <= x:
begin = mid
else:
end = mid
return begin
n = int(input())
A = [int(i) for i in input().split()]
A = sorted(A)
d = int(input())
D = []
for i in range(d):
D += [int(input())]
ans = []
for i in range(d):
x = D[i]
begin = 0
end = len(A)
while end - begin > 1:
mid = (end + begin) // 2
if A[mid] <= x:
begin = mid
else:
end = mid
oo = begin
if A[0] > x:
oo = -1
ans += [oo+1]
for i in ans:
print(i)
``` | output | 1 | 106,649 | 10 | 213,299 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to xi coins.
Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains n integers xi (1 β€ xi β€ 100 000) β prices of the bottles of the drink in the i-th shop.
The third line contains a single integer q (1 β€ q β€ 100 000) β the number of days Vasiliy plans to buy the drink.
Then follow q lines each containing one integer mi (1 β€ mi β€ 109) β the number of coins Vasiliy can spent on the i-th day.
Output
Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.
Example
Input
5
3 10 8 6 11
4
1
10
3
11
Output
0
4
1
5
Note
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop.
Submitted Solution:
```
from sys import stdin,stdout
nbMagasins=int(stdin.readline())
prix=list(map(int,stdin.readline().split()))
prix.sort()
q=int(stdin.readline())
for i in range(q):
prixMax=int(stdin.readline())
inf=0
sup=nbMagasins-1
while sup>inf:
milieu=(inf+sup)//2
if prix[milieu]>prixMax:
sup=milieu
else:
inf=milieu+1
if prixMax<prix[0]:
print(0)
elif prixMax>=prix[-1]:
print(nbMagasins)
else:
print(inf)
``` | instruction | 0 | 106,650 | 10 | 213,300 |
Yes | output | 1 | 106,650 | 10 | 213,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to xi coins.
Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains n integers xi (1 β€ xi β€ 100 000) β prices of the bottles of the drink in the i-th shop.
The third line contains a single integer q (1 β€ q β€ 100 000) β the number of days Vasiliy plans to buy the drink.
Then follow q lines each containing one integer mi (1 β€ mi β€ 109) β the number of coins Vasiliy can spent on the i-th day.
Output
Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.
Example
Input
5
3 10 8 6 11
4
1
10
3
11
Output
0
4
1
5
Note
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop.
Submitted Solution:
```
def thanos(a, l,r, k):
if l > r:
return 0
else:
mid = (l+r)//2
if a[mid] > k:
return thanos(a, l, mid-1, k)
else:
if mid+1 < len(a):
if a[mid+1] > k:
return mid+1-l
else:
return (mid-l) + thanos(a, mid+1, r, k) + 1
else:
return 1
n = int(input())
x = input().split()
for i in range(len(x)):
x[i] = int(x[i])
x.sort()
q = int(input())
qq = []
while q > 0:
m = int(input())
qq.append(thanos(x, 0, len(x)-1, m))
# print(thanos(x, 0, len(x)-1, m))
q -= 1
print(*qq, sep='\n')
``` | instruction | 0 | 106,651 | 10 | 213,302 |
Yes | output | 1 | 106,651 | 10 | 213,303 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to xi coins.
Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains n integers xi (1 β€ xi β€ 100 000) β prices of the bottles of the drink in the i-th shop.
The third line contains a single integer q (1 β€ q β€ 100 000) β the number of days Vasiliy plans to buy the drink.
Then follow q lines each containing one integer mi (1 β€ mi β€ 109) β the number of coins Vasiliy can spent on the i-th day.
Output
Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.
Example
Input
5
3 10 8 6 11
4
1
10
3
11
Output
0
4
1
5
Note
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop.
Submitted Solution:
```
from bisect import bisect_right
n = int(input())
x = sorted(list(map(int, input().split())))
m = int(input())
for i in range(1, m+1):
count = 0
budget = int(input())
print(bisect_right(x, budget, 0, len(x)))
``` | instruction | 0 | 106,652 | 10 | 213,304 |
Yes | output | 1 | 106,652 | 10 | 213,305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to xi coins.
Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains n integers xi (1 β€ xi β€ 100 000) β prices of the bottles of the drink in the i-th shop.
The third line contains a single integer q (1 β€ q β€ 100 000) β the number of days Vasiliy plans to buy the drink.
Then follow q lines each containing one integer mi (1 β€ mi β€ 109) β the number of coins Vasiliy can spent on the i-th day.
Output
Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.
Example
Input
5
3 10 8 6 11
4
1
10
3
11
Output
0
4
1
5
Note
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop.
Submitted Solution:
```
#!/usr/bin/env python3.5
import sys
import bisect
def read_data():
m = int(next(sys.stdin))
prices = list(map(int, next(sys.stdin).split()))
n = int(next(sys.stdin))
return n, prices
def solve_and_print(n, prices):
prices.sort()
for i in range(n):
s = int(next(sys.stdin))
n = bisect.bisect_right(prices, s)
print(n)
if __name__ == "__main__":
n, prices = read_data()
solve_and_print(n, prices)
``` | instruction | 0 | 106,653 | 10 | 213,306 |
Yes | output | 1 | 106,653 | 10 | 213,307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to xi coins.
Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains n integers xi (1 β€ xi β€ 100 000) β prices of the bottles of the drink in the i-th shop.
The third line contains a single integer q (1 β€ q β€ 100 000) β the number of days Vasiliy plans to buy the drink.
Then follow q lines each containing one integer mi (1 β€ mi β€ 109) β the number of coins Vasiliy can spent on the i-th day.
Output
Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.
Example
Input
5
3 10 8 6 11
4
1
10
3
11
Output
0
4
1
5
Note
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop.
Submitted Solution:
```
n = int(input())
prices = sorted(list(map(int, input().split())))
max_price = max(prices)
botles = [0] * (max_price + 1)
current_idx = 0
val = 0
for i in range(1, max_price + 1):
if i == prices[current_idx]:
current_idx += 1
val += 1
botles[i] = val
q = int(input())
for i in range(q):
m = int(input())
print(botles[m] if m <= max_price else n)
``` | instruction | 0 | 106,654 | 10 | 213,308 |
No | output | 1 | 106,654 | 10 | 213,309 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to xi coins.
Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains n integers xi (1 β€ xi β€ 100 000) β prices of the bottles of the drink in the i-th shop.
The third line contains a single integer q (1 β€ q β€ 100 000) β the number of days Vasiliy plans to buy the drink.
Then follow q lines each containing one integer mi (1 β€ mi β€ 109) β the number of coins Vasiliy can spent on the i-th day.
Output
Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.
Example
Input
5
3 10 8 6 11
4
1
10
3
11
Output
0
4
1
5
Note
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop.
Submitted Solution:
```
n=int(input())
x=[int(X) for X in input().split()]
q=int(input())
m=[]
x.sort()
def binary_search(arr, x):
low = 0
high = len(arr) - 1
mid = 0
while low <= high:
mid = (high + low) // 2
if(mid!=low):
if(x>=arr[mid] and x<arr[mid+1]):
return mid
elif(x<arr[mid] and x>=arr[mid-1]):
return mid-1
elif(x==arr[mid]):
k=mid
while(x==arr[k]):
if(k!=high):
k+=1
else:
break
if(k!=high):
return k-1
else:
return k
# If x is greater, ignore left half
elif arr[mid] < x:
low = mid + 1
# If x is smaller, ignore right half
elif arr[mid] > x:
high = mid - 1
elif(low==high):
if(x>=arr[low]):
return low
else:
return low-1
else:
if(x>=arr[mid] and x<arr[mid+1]):
return mid
elif(x==arr[mid]):
k=mid
while(x==arr[k]):
if(k!=high):
k+=1
else:
break
if(k!=high):
return k-1
else:
return k
# If x is greater, ignore left half
elif arr[mid] < x:
low = mid + 1
# If x is smaller, ignore right half
elif arr[mid] > x:
high = mid - 1
# If we reach here, then the element was not present
return -1
for i in range(q):
a=int(input())
b=binary_search(x, a)+1
print(b)
``` | instruction | 0 | 106,655 | 10 | 213,310 |
No | output | 1 | 106,655 | 10 | 213,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to xi coins.
Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains n integers xi (1 β€ xi β€ 100 000) β prices of the bottles of the drink in the i-th shop.
The third line contains a single integer q (1 β€ q β€ 100 000) β the number of days Vasiliy plans to buy the drink.
Then follow q lines each containing one integer mi (1 β€ mi β€ 109) β the number of coins Vasiliy can spent on the i-th day.
Output
Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.
Example
Input
5
3 10 8 6 11
4
1
10
3
11
Output
0
4
1
5
Note
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop.
Submitted Solution:
```
# Python 3
import bisect
prices = []
n = int(input())
#for price in (int(x) for x in input().split(" ")):
# bisect.insort(prices, price)
price = [int(x) for x in input().split(" ")]
price.sort()
days = int(input())
# if cache[guess] <= coin:
# return True
#
# if prices[guess-1] <= coin:
# if guess not in cache or cache[guess] > coin:
# cache[guess] = coin
#
# return True
#
# return False
cache2 = {}
for _ in range(days):
coins = int(input())
print(bisect.bisect_right(prices, coins))
# if coins in cache2:
# print(cache2[coins])
# continue
#
# low = 0
# high = len(prices)
#
# if not check(coins, low+1):
# print(low)
# cache2[coins] = low
# continue
#
# if check(coins, high):
# print(high)
# cache2[coins] = high
# continue
#
# while(low + 1 < high):
# mid = low + (high - low) // 2
# if (check(coins, mid)):
# low = mid
# else:
# high = mid
#
# if not check(coins, high):
# print(low)
# cache2[coins] = low
# else:
# print(high)
# cache2[coins] = high
``` | instruction | 0 | 106,656 | 10 | 213,312 |
No | output | 1 | 106,656 | 10 | 213,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasiliy likes to rest after a hard work, so you may often meet him in some bar nearby. As all programmers do, he loves the famous drink "Beecola", which can be bought in n different shops in the city. It's known that the price of one bottle in the shop i is equal to xi coins.
Vasiliy plans to buy his favorite drink for q consecutive days. He knows, that on the i-th day he will be able to spent mi coins. Now, for each of the days he want to know in how many different shops he can buy a bottle of "Beecola".
Input
The first line of the input contains a single integer n (1 β€ n β€ 100 000) β the number of shops in the city that sell Vasiliy's favourite drink.
The second line contains n integers xi (1 β€ xi β€ 100 000) β prices of the bottles of the drink in the i-th shop.
The third line contains a single integer q (1 β€ q β€ 100 000) β the number of days Vasiliy plans to buy the drink.
Then follow q lines each containing one integer mi (1 β€ mi β€ 109) β the number of coins Vasiliy can spent on the i-th day.
Output
Print q integers. The i-th of them should be equal to the number of shops where Vasiliy will be able to buy a bottle of the drink on the i-th day.
Example
Input
5
3 10 8 6 11
4
1
10
3
11
Output
0
4
1
5
Note
On the first day, Vasiliy won't be able to buy a drink in any of the shops.
On the second day, Vasiliy can buy a drink in the shops 1, 2, 3 and 4.
On the third day, Vasiliy can buy a drink only in the shop number 1.
Finally, on the last day Vasiliy can buy a drink in any shop.
Submitted Solution:
```
def bs(k, n_list):
l = 0
r = len(n_list) - 1
while l != r:
mid = (l + r) // 2 # - ΠΌΠΈΠ½ΠΈΠΌΠΈΠ·ΠΈΡΠΎΠ²Π°ΡΡ, (r + l + 1) // 2 - ΠΌΠ°ΠΊΡΠΈΠΌΠΈΠ·ΠΈΡΠΎΠ²Π°ΡΡ ΠΎΡΠ²Π΅Ρ
if k > n_list[mid]:
l = mid + 1
else:
r = mid
return l
a = input()
shop_prices = list(map(int, input().split()))
shop_prices.sort()
for day in range(int(input())):
day_money = int(input())
bs_data = bs(day_money, shop_prices)
if shop_prices[bs_data] == day_money:
print(bs_data + 1)
else:
print(bs_data)
``` | instruction | 0 | 106,657 | 10 | 213,314 |
No | output | 1 | 106,657 | 10 | 213,315 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters.
Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times before he does the cut:
* Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan).
* Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen.
Find the minimum amount of money needed before cutting the bar into two parts with the same length.
Constraints
* 2 \leq N \leq 200000
* 1 \leq A_i \leq 2020202020
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_N
Output
Print an integer representing the minimum amount of money needed before cutting the bar into two parts with the same length.
Examples
Input
3
2 4 3
Output
3
Input
12
100 104 102 105 103 103 101 105 104 102 104 101
Output
0 | instruction | 0 | 106,801 | 10 | 213,602 |
"Correct Solution:
```
from itertools import accumulate
n = int(input())
a = map(int, input().split())
acc = tuple(accumulate(a))
ret = min(abs(acc[-1] - x * 2) for x in acc)
print(ret)
``` | output | 1 | 106,801 | 10 | 213,603 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters.
Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times before he does the cut:
* Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan).
* Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen.
Find the minimum amount of money needed before cutting the bar into two parts with the same length.
Constraints
* 2 \leq N \leq 200000
* 1 \leq A_i \leq 2020202020
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_N
Output
Print an integer representing the minimum amount of money needed before cutting the bar into two parts with the same length.
Examples
Input
3
2 4 3
Output
3
Input
12
100 104 102 105 103 103 101 105 104 102 104 101
Output
0 | instruction | 0 | 106,802 | 10 | 213,604 |
"Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
harf=sum(a)/2
l,r,i=0,a[0],0
while r<harf:
i+=1
l=r
r+=a[i]
print(int(min(harf-l,r-harf)*2))
``` | output | 1 | 106,802 | 10 | 213,605 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters.
Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times before he does the cut:
* Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan).
* Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen.
Find the minimum amount of money needed before cutting the bar into two parts with the same length.
Constraints
* 2 \leq N \leq 200000
* 1 \leq A_i \leq 2020202020
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_N
Output
Print an integer representing the minimum amount of money needed before cutting the bar into two parts with the same length.
Examples
Input
3
2 4 3
Output
3
Input
12
100 104 102 105 103 103 101 105 104 102 104 101
Output
0 | instruction | 0 | 106,803 | 10 | 213,606 |
"Correct Solution:
```
N=int(input())
A=list(map(int, input().split()))
S=sum(A)
x=0
ans=S
for a in A:
x+=a
y=abs(S-2*x)
ans=min(ans,y)
print(ans)
``` | output | 1 | 106,803 | 10 | 213,607 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters.
Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times before he does the cut:
* Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan).
* Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen.
Find the minimum amount of money needed before cutting the bar into two parts with the same length.
Constraints
* 2 \leq N \leq 200000
* 1 \leq A_i \leq 2020202020
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_N
Output
Print an integer representing the minimum amount of money needed before cutting the bar into two parts with the same length.
Examples
Input
3
2 4 3
Output
3
Input
12
100 104 102 105 103 103 101 105 104 102 104 101
Output
0 | instruction | 0 | 106,804 | 10 | 213,608 |
"Correct Solution:
```
n,*a=map(int,open(0).read().split())
s=sum(a)
c=0
m=[]
for b in a:c+=b;m+=abs(c+c-s),
print(min(m))
``` | output | 1 | 106,804 | 10 | 213,609 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters.
Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times before he does the cut:
* Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan).
* Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen.
Find the minimum amount of money needed before cutting the bar into two parts with the same length.
Constraints
* 2 \leq N \leq 200000
* 1 \leq A_i \leq 2020202020
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_N
Output
Print an integer representing the minimum amount of money needed before cutting the bar into two parts with the same length.
Examples
Input
3
2 4 3
Output
3
Input
12
100 104 102 105 103 103 101 105 104 102 104 101
Output
0 | instruction | 0 | 106,805 | 10 | 213,610 |
"Correct Solution:
```
import sys
n = int(input())
a = list(map(int, input().split()))
tmp=sys.maxsize
s = sum(a)
t = 0
for i in range(n):
t += a[i]
tmp = min(tmp,abs(s-t*2))
print(tmp)
``` | output | 1 | 106,805 | 10 | 213,611 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters.
Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times before he does the cut:
* Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan).
* Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen.
Find the minimum amount of money needed before cutting the bar into two parts with the same length.
Constraints
* 2 \leq N \leq 200000
* 1 \leq A_i \leq 2020202020
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_N
Output
Print an integer representing the minimum amount of money needed before cutting the bar into two parts with the same length.
Examples
Input
3
2 4 3
Output
3
Input
12
100 104 102 105 103 103 101 105 104 102 104 101
Output
0 | instruction | 0 | 106,806 | 10 | 213,612 |
"Correct Solution:
```
n = int(input())
*a, = map(int, input().split())
s = sum(a)
ans = s
tmp = 0
for i in a[:-1]:
tmp += i
ans = min(ans, abs(tmp*2 - s))
print(ans)
``` | output | 1 | 106,806 | 10 | 213,613 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi, who works at DISCO, is standing before an iron bar. The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters.
Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length. However, this may not be possible as is, so he will do the following operations some number of times before he does the cut:
* Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan).
* Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen.
Find the minimum amount of money needed before cutting the bar into two parts with the same length.
Constraints
* 2 \leq N \leq 200000
* 1 \leq A_i \leq 2020202020
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 A_2 A_3 ... A_N
Output
Print an integer representing the minimum amount of money needed before cutting the bar into two parts with the same length.
Examples
Input
3
2 4 3
Output
3
Input
12
100 104 102 105 103 103 101 105 104 102 104 101
Output
0 | instruction | 0 | 106,807 | 10 | 213,614 |
"Correct Solution:
```
n=int(input())
l=list(map(int,input().split()))
s=sum(l)
m=0
i=-1
while m<s//2:
i+=1
m+=l[i]
print(min(2*m-s,s-2*(m-l[i])))
``` | output | 1 | 106,807 | 10 | 213,615 |
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