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 |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Do you like summer? Residents of Berland do. They especially love eating ice cream in the hot summer. So this summer day a large queue of n Berland residents lined up in front of the ice cream stall. We know that each of them has a certain amount of berland dollars with them. The residents of Berland are nice people, so each person agrees to swap places with the person right behind him for just 1 dollar. More formally, if person a stands just behind person b, then person a can pay person b 1 dollar, then a and b get swapped. Of course, if person a has zero dollars, he can not swap places with person b.
Residents of Berland are strange people. In particular, they get upset when there is someone with a strictly smaller sum of money in the line in front of them.
Can you help the residents of Berland form such order in the line so that they were all happy? A happy resident is the one who stands first in the line or the one in front of who another resident stands with not less number of dollars. Note that the people of Berland are people of honor and they agree to swap places only in the manner described above.
Input
The first line contains integer n (1 ≤ n ≤ 200 000) — the number of residents who stand in the line.
The second line contains n space-separated integers ai (0 ≤ ai ≤ 109), where ai is the number of Berland dollars of a man standing on the i-th position in the line. The positions are numbered starting from the end of the line.
Output
If it is impossible to make all the residents happy, print ":(" without the quotes. Otherwise, print in the single line n space-separated integers, the i-th of them must be equal to the number of money of the person on position i in the new line. If there are multiple answers, print any of them.
Examples
Input
2
11 8
Output
9 10
Input
5
10 9 7 10 6
Output
:(
Input
3
12 3 3
Output
4 4 10
Note
In the first sample two residents should swap places, after that the first resident has 10 dollars and he is at the head of the line and the second resident will have 9 coins and he will be at the end of the line.
In the second sample it is impossible to achieve the desired result.
In the third sample the first person can swap with the second one, then they will have the following numbers of dollars: 4 11 3, then the second person (in the new line) swaps with the third one, and the resulting numbers of dollars will equal to: 4 4 10. In this line everybody will be happy.
Submitted Solution:
```
def main():
n = int(input())
a = [int(x) for x in input().split(" ")]
for i in range(len(a)):
a[i] += i
a.sort()
for i in range(1, len(a)):
a[i] -= i
if a[i] < a[i - 1]:
print(":(")
return
for i in range(len(a)):
a[i] = str(a[i])
print(' '.join(a))
if __name__ == '__main__':
main()
``` | instruction | 0 | 48,343 | 10 | 96,686 |
Yes | output | 1 | 48,343 | 10 | 96,687 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Do you like summer? Residents of Berland do. They especially love eating ice cream in the hot summer. So this summer day a large queue of n Berland residents lined up in front of the ice cream stall. We know that each of them has a certain amount of berland dollars with them. The residents of Berland are nice people, so each person agrees to swap places with the person right behind him for just 1 dollar. More formally, if person a stands just behind person b, then person a can pay person b 1 dollar, then a and b get swapped. Of course, if person a has zero dollars, he can not swap places with person b.
Residents of Berland are strange people. In particular, they get upset when there is someone with a strictly smaller sum of money in the line in front of them.
Can you help the residents of Berland form such order in the line so that they were all happy? A happy resident is the one who stands first in the line or the one in front of who another resident stands with not less number of dollars. Note that the people of Berland are people of honor and they agree to swap places only in the manner described above.
Input
The first line contains integer n (1 ≤ n ≤ 200 000) — the number of residents who stand in the line.
The second line contains n space-separated integers ai (0 ≤ ai ≤ 109), where ai is the number of Berland dollars of a man standing on the i-th position in the line. The positions are numbered starting from the end of the line.
Output
If it is impossible to make all the residents happy, print ":(" without the quotes. Otherwise, print in the single line n space-separated integers, the i-th of them must be equal to the number of money of the person on position i in the new line. If there are multiple answers, print any of them.
Examples
Input
2
11 8
Output
9 10
Input
5
10 9 7 10 6
Output
:(
Input
3
12 3 3
Output
4 4 10
Note
In the first sample two residents should swap places, after that the first resident has 10 dollars and he is at the head of the line and the second resident will have 9 coins and he will be at the end of the line.
In the second sample it is impossible to achieve the desired result.
In the third sample the first person can swap with the second one, then they will have the following numbers of dollars: 4 11 3, then the second person (in the new line) swaps with the third one, and the resulting numbers of dollars will equal to: 4 4 10. In this line everybody will be happy.
Submitted Solution:
```
n=int(input())
fila=list(map(int, input().split()))
break_ = False
for i in range(n):
fila[i] += i
fila = sorted(list(set(fila)))
if len(fila) < n:
print(":(")
else:
for j in range(n):
fila[j] -= j
print(" ".join(map(str, fila)))
``` | instruction | 0 | 48,344 | 10 | 96,688 |
Yes | output | 1 | 48,344 | 10 | 96,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Do you like summer? Residents of Berland do. They especially love eating ice cream in the hot summer. So this summer day a large queue of n Berland residents lined up in front of the ice cream stall. We know that each of them has a certain amount of berland dollars with them. The residents of Berland are nice people, so each person agrees to swap places with the person right behind him for just 1 dollar. More formally, if person a stands just behind person b, then person a can pay person b 1 dollar, then a and b get swapped. Of course, if person a has zero dollars, he can not swap places with person b.
Residents of Berland are strange people. In particular, they get upset when there is someone with a strictly smaller sum of money in the line in front of them.
Can you help the residents of Berland form such order in the line so that they were all happy? A happy resident is the one who stands first in the line or the one in front of who another resident stands with not less number of dollars. Note that the people of Berland are people of honor and they agree to swap places only in the manner described above.
Input
The first line contains integer n (1 ≤ n ≤ 200 000) — the number of residents who stand in the line.
The second line contains n space-separated integers ai (0 ≤ ai ≤ 109), where ai is the number of Berland dollars of a man standing on the i-th position in the line. The positions are numbered starting from the end of the line.
Output
If it is impossible to make all the residents happy, print ":(" without the quotes. Otherwise, print in the single line n space-separated integers, the i-th of them must be equal to the number of money of the person on position i in the new line. If there are multiple answers, print any of them.
Examples
Input
2
11 8
Output
9 10
Input
5
10 9 7 10 6
Output
:(
Input
3
12 3 3
Output
4 4 10
Note
In the first sample two residents should swap places, after that the first resident has 10 dollars and he is at the head of the line and the second resident will have 9 coins and he will be at the end of the line.
In the second sample it is impossible to achieve the desired result.
In the third sample the first person can swap with the second one, then they will have the following numbers of dollars: 4 11 3, then the second person (in the new line) swaps with the third one, and the resulting numbers of dollars will equal to: 4 4 10. In this line everybody will be happy.
Submitted Solution:
```
N = int(input())
nums = list(map(int, input().split()))
for i in range(len(nums)):
nums[i] += i
nums.sort()
for i in range(len(nums)):
nums[i] -= i
if i and nums[i] < nums[i-1]:
print(":(")
exit()
print(' '.join(list(map(str, nums))))
``` | instruction | 0 | 48,345 | 10 | 96,690 |
Yes | output | 1 | 48,345 | 10 | 96,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Do you like summer? Residents of Berland do. They especially love eating ice cream in the hot summer. So this summer day a large queue of n Berland residents lined up in front of the ice cream stall. We know that each of them has a certain amount of berland dollars with them. The residents of Berland are nice people, so each person agrees to swap places with the person right behind him for just 1 dollar. More formally, if person a stands just behind person b, then person a can pay person b 1 dollar, then a and b get swapped. Of course, if person a has zero dollars, he can not swap places with person b.
Residents of Berland are strange people. In particular, they get upset when there is someone with a strictly smaller sum of money in the line in front of them.
Can you help the residents of Berland form such order in the line so that they were all happy? A happy resident is the one who stands first in the line or the one in front of who another resident stands with not less number of dollars. Note that the people of Berland are people of honor and they agree to swap places only in the manner described above.
Input
The first line contains integer n (1 ≤ n ≤ 200 000) — the number of residents who stand in the line.
The second line contains n space-separated integers ai (0 ≤ ai ≤ 109), where ai is the number of Berland dollars of a man standing on the i-th position in the line. The positions are numbered starting from the end of the line.
Output
If it is impossible to make all the residents happy, print ":(" without the quotes. Otherwise, print in the single line n space-separated integers, the i-th of them must be equal to the number of money of the person on position i in the new line. If there are multiple answers, print any of them.
Examples
Input
2
11 8
Output
9 10
Input
5
10 9 7 10 6
Output
:(
Input
3
12 3 3
Output
4 4 10
Note
In the first sample two residents should swap places, after that the first resident has 10 dollars and he is at the head of the line and the second resident will have 9 coins and he will be at the end of the line.
In the second sample it is impossible to achieve the desired result.
In the third sample the first person can swap with the second one, then they will have the following numbers of dollars: 4 11 3, then the second person (in the new line) swaps with the third one, and the resulting numbers of dollars will equal to: 4 4 10. In this line everybody will be happy.
Submitted Solution:
```
def och():
n = int(input())
mas = list(map(int,input().split()))
vag = 0
for i in range(n - 1):
if mas[i] > mas[i+1]:
vag = mas[i] - 1
mas[i] = mas[i+1] + 1
mas[i+1] =vag
for i in range(n):
if mas[i] > mas[i]:
print(':(')
exit()
print(' '.join(map(str,mas)))
och()
``` | instruction | 0 | 48,346 | 10 | 96,692 |
No | output | 1 | 48,346 | 10 | 96,693 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Do you like summer? Residents of Berland do. They especially love eating ice cream in the hot summer. So this summer day a large queue of n Berland residents lined up in front of the ice cream stall. We know that each of them has a certain amount of berland dollars with them. The residents of Berland are nice people, so each person agrees to swap places with the person right behind him for just 1 dollar. More formally, if person a stands just behind person b, then person a can pay person b 1 dollar, then a and b get swapped. Of course, if person a has zero dollars, he can not swap places with person b.
Residents of Berland are strange people. In particular, they get upset when there is someone with a strictly smaller sum of money in the line in front of them.
Can you help the residents of Berland form such order in the line so that they were all happy? A happy resident is the one who stands first in the line or the one in front of who another resident stands with not less number of dollars. Note that the people of Berland are people of honor and they agree to swap places only in the manner described above.
Input
The first line contains integer n (1 ≤ n ≤ 200 000) — the number of residents who stand in the line.
The second line contains n space-separated integers ai (0 ≤ ai ≤ 109), where ai is the number of Berland dollars of a man standing on the i-th position in the line. The positions are numbered starting from the end of the line.
Output
If it is impossible to make all the residents happy, print ":(" without the quotes. Otherwise, print in the single line n space-separated integers, the i-th of them must be equal to the number of money of the person on position i in the new line. If there are multiple answers, print any of them.
Examples
Input
2
11 8
Output
9 10
Input
5
10 9 7 10 6
Output
:(
Input
3
12 3 3
Output
4 4 10
Note
In the first sample two residents should swap places, after that the first resident has 10 dollars and he is at the head of the line and the second resident will have 9 coins and he will be at the end of the line.
In the second sample it is impossible to achieve the desired result.
In the third sample the first person can swap with the second one, then they will have the following numbers of dollars: 4 11 3, then the second person (in the new line) swaps with the third one, and the resulting numbers of dollars will equal to: 4 4 10. In this line everybody will be happy.
Submitted Solution:
```
n = int(input())
A = list(map(int, input().split()))
for i in range(n):
A[i] = [A[i] - n + 1 + i , i]
print(A)
A.sort()
for i, a in enumerate(A):
A[i][0] = A[i][0] + A[i][1] - i + n - 1 - A[i][1]
f = 1
for i in range(1, n):
if A[i] < A[i-1]:
f = 0
break
if not f:
print(":(")
else:
for a in A:
print(a[0], end= ' ')
``` | instruction | 0 | 48,347 | 10 | 96,694 |
No | output | 1 | 48,347 | 10 | 96,695 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Do you like summer? Residents of Berland do. They especially love eating ice cream in the hot summer. So this summer day a large queue of n Berland residents lined up in front of the ice cream stall. We know that each of them has a certain amount of berland dollars with them. The residents of Berland are nice people, so each person agrees to swap places with the person right behind him for just 1 dollar. More formally, if person a stands just behind person b, then person a can pay person b 1 dollar, then a and b get swapped. Of course, if person a has zero dollars, he can not swap places with person b.
Residents of Berland are strange people. In particular, they get upset when there is someone with a strictly smaller sum of money in the line in front of them.
Can you help the residents of Berland form such order in the line so that they were all happy? A happy resident is the one who stands first in the line or the one in front of who another resident stands with not less number of dollars. Note that the people of Berland are people of honor and they agree to swap places only in the manner described above.
Input
The first line contains integer n (1 ≤ n ≤ 200 000) — the number of residents who stand in the line.
The second line contains n space-separated integers ai (0 ≤ ai ≤ 109), where ai is the number of Berland dollars of a man standing on the i-th position in the line. The positions are numbered starting from the end of the line.
Output
If it is impossible to make all the residents happy, print ":(" without the quotes. Otherwise, print in the single line n space-separated integers, the i-th of them must be equal to the number of money of the person on position i in the new line. If there are multiple answers, print any of them.
Examples
Input
2
11 8
Output
9 10
Input
5
10 9 7 10 6
Output
:(
Input
3
12 3 3
Output
4 4 10
Note
In the first sample two residents should swap places, after that the first resident has 10 dollars and he is at the head of the line and the second resident will have 9 coins and he will be at the end of the line.
In the second sample it is impossible to achieve the desired result.
In the third sample the first person can swap with the second one, then they will have the following numbers of dollars: 4 11 3, then the second person (in the new line) swaps with the third one, and the resulting numbers of dollars will equal to: 4 4 10. In this line everybody will be happy.
Submitted Solution:
```
__author__ = 'Adam'
import sys
n = sys.stdin.readline()
a = sys.stdin.readline().split(' ')
a = [int(i) for i in a]
i = 0
b = sorted(a)
if a == b:
print(' '.join([str(i) for i in a]))
else:
for j in range(len(a)-1):
flag = True
for i in range(len(a)-1-j):
#print('q')
if a[i] - a[i+1] == 1:
break
elif a[i] > a[i+1]:
a[i+1], a[i] = a[i]-1, a[i+1]+1
flag = False
if flag:
break
if a == b:
print(' '.join([str(i) for i in a]))
else:
print(':(')
``` | instruction | 0 | 48,348 | 10 | 96,696 |
No | output | 1 | 48,348 | 10 | 96,697 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Do you like summer? Residents of Berland do. They especially love eating ice cream in the hot summer. So this summer day a large queue of n Berland residents lined up in front of the ice cream stall. We know that each of them has a certain amount of berland dollars with them. The residents of Berland are nice people, so each person agrees to swap places with the person right behind him for just 1 dollar. More formally, if person a stands just behind person b, then person a can pay person b 1 dollar, then a and b get swapped. Of course, if person a has zero dollars, he can not swap places with person b.
Residents of Berland are strange people. In particular, they get upset when there is someone with a strictly smaller sum of money in the line in front of them.
Can you help the residents of Berland form such order in the line so that they were all happy? A happy resident is the one who stands first in the line or the one in front of who another resident stands with not less number of dollars. Note that the people of Berland are people of honor and they agree to swap places only in the manner described above.
Input
The first line contains integer n (1 ≤ n ≤ 200 000) — the number of residents who stand in the line.
The second line contains n space-separated integers ai (0 ≤ ai ≤ 109), where ai is the number of Berland dollars of a man standing on the i-th position in the line. The positions are numbered starting from the end of the line.
Output
If it is impossible to make all the residents happy, print ":(" without the quotes. Otherwise, print in the single line n space-separated integers, the i-th of them must be equal to the number of money of the person on position i in the new line. If there are multiple answers, print any of them.
Examples
Input
2
11 8
Output
9 10
Input
5
10 9 7 10 6
Output
:(
Input
3
12 3 3
Output
4 4 10
Note
In the first sample two residents should swap places, after that the first resident has 10 dollars and he is at the head of the line and the second resident will have 9 coins and he will be at the end of the line.
In the second sample it is impossible to achieve the desired result.
In the third sample the first person can swap with the second one, then they will have the following numbers of dollars: 4 11 3, then the second person (in the new line) swaps with the third one, and the resulting numbers of dollars will equal to: 4 4 10. In this line everybody will be happy.
Submitted Solution:
```
n = int(input())
a = [int(x) for x in input().split(" ")]
for i in range(len(a)):
a[i] += i
a.sort()
for i in range(1, len(a)):
a[i] -= i
if a[i] < a[i - 1]:
print(":(")
break
for i in range(len(a)):
a[i] = str(a[i])
print(' '.join(a))
``` | instruction | 0 | 48,349 | 10 | 96,698 |
No | output | 1 | 48,349 | 10 | 96,699 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.
Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.
In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it's impossible.
Input
First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.
Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.
Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.
Output
If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).
Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.
Any of numbers x and y can be equal 0.
Examples
Input
7
2
3
Output
YES
2 1
Input
100
25
10
Output
YES
0 10
Input
15
4
8
Output
NO
Input
9960594
2551
2557
Output
YES
1951 1949
Note
In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles.
In second example Vasya can spend exactly n burles multiple ways:
* buy two bottles of Ber-Cola and five Bars bars;
* buy four bottles of Ber-Cola and don't buy Bars bars;
* don't buy Ber-Cola and buy 10 Bars bars.
In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly n burles. | instruction | 0 | 48,498 | 10 | 96,996 |
Tags: brute force, implementation, number theory
Correct Solution:
```
def read():
return [int(v) for v in input().split()]
def main():
n = read()[0]
a = read()[0]
b = read()[0]
for x in range(0, n // a + 2):
y = int((n - a * x) / b)
if y >= 0 and a * x + b * y == n:
print('YES\n{} {}'.format(x, y))
return
print('NO')
if __name__ == '__main__':
main()
``` | output | 1 | 48,498 | 10 | 96,997 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.
Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.
In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it's impossible.
Input
First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.
Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.
Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.
Output
If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).
Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.
Any of numbers x and y can be equal 0.
Examples
Input
7
2
3
Output
YES
2 1
Input
100
25
10
Output
YES
0 10
Input
15
4
8
Output
NO
Input
9960594
2551
2557
Output
YES
1951 1949
Note
In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles.
In second example Vasya can spend exactly n burles multiple ways:
* buy two bottles of Ber-Cola and five Bars bars;
* buy four bottles of Ber-Cola and don't buy Bars bars;
* don't buy Ber-Cola and buy 10 Bars bars.
In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly n burles. | instruction | 0 | 48,499 | 10 | 96,998 |
Tags: brute force, implementation, number theory
Correct Solution:
```
import sys,math,string,bisect
input=sys.stdin.readline
from collections import deque
L=lambda : list(map(int,input().split()))
Ls=lambda : list(input().split())
M=lambda : map(int,input().split())
I=lambda :int(input())
n=I()
a=I()
b=I()
if(n%math.gcd(a,b)):
print("NO")
else:
if(n%b==0):
print("YES")
print(0,n//b)
exit()
i=0
while(n>0):
if(n%a==0):
print("YES")
print(n//a,i)
exit()
n-=b
i+=1
print("NO")
``` | output | 1 | 48,499 | 10 | 96,999 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.
Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.
In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it's impossible.
Input
First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.
Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.
Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.
Output
If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).
Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.
Any of numbers x and y can be equal 0.
Examples
Input
7
2
3
Output
YES
2 1
Input
100
25
10
Output
YES
0 10
Input
15
4
8
Output
NO
Input
9960594
2551
2557
Output
YES
1951 1949
Note
In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles.
In second example Vasya can spend exactly n burles multiple ways:
* buy two bottles of Ber-Cola and five Bars bars;
* buy four bottles of Ber-Cola and don't buy Bars bars;
* don't buy Ber-Cola and buy 10 Bars bars.
In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly n burles. | instruction | 0 | 48,500 | 10 | 97,000 |
Tags: brute force, implementation, number theory
Correct Solution:
```
'''input
10000000
3
999999
'''
n, a, b = int(input()), int(input()), int(input())
for x in range(n//a+1):
if (n - x*a) % b == 0:
print("YES")
print(x, (n- x*a) // b)
break
else:
print("NO")
``` | output | 1 | 48,500 | 10 | 97,001 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.
Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.
In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it's impossible.
Input
First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.
Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.
Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.
Output
If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).
Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.
Any of numbers x and y can be equal 0.
Examples
Input
7
2
3
Output
YES
2 1
Input
100
25
10
Output
YES
0 10
Input
15
4
8
Output
NO
Input
9960594
2551
2557
Output
YES
1951 1949
Note
In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles.
In second example Vasya can spend exactly n burles multiple ways:
* buy two bottles of Ber-Cola and five Bars bars;
* buy four bottles of Ber-Cola and don't buy Bars bars;
* don't buy Ber-Cola and buy 10 Bars bars.
In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly n burles. | instruction | 0 | 48,501 | 10 | 97,002 |
Tags: brute force, implementation, number theory
Correct Solution:
```
mod = 1000000007
ii = lambda : int(input())
si = lambda : input()
dgl = lambda : list(map(int, input()))
f = lambda : map(int, input().split())
il = lambda : list(map(int, input().split()))
ls = lambda : list(input())
n=ii()
n1=n
n2=n
a=ii()
b=ii()
c1, c2, c3, c4=0, 0, 0, 0
while n>=b:
if n%a==0:
c1+=n//a
break
n-=b
c2+=1
while n1>=a:
if n1%b==0:
c4=n1//b
break
n1-=a
c3+=1
if (c1+c2)>=(c3+c4) and c1*a+c2*b==n2:
print('YES')
print(c1,c2)
elif c3*a+c4*b==n2:
print('YES')
print(c3,c4)
else:
print('NO')
``` | output | 1 | 48,501 | 10 | 97,003 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.
Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.
In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it's impossible.
Input
First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.
Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.
Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.
Output
If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).
Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.
Any of numbers x and y can be equal 0.
Examples
Input
7
2
3
Output
YES
2 1
Input
100
25
10
Output
YES
0 10
Input
15
4
8
Output
NO
Input
9960594
2551
2557
Output
YES
1951 1949
Note
In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles.
In second example Vasya can spend exactly n burles multiple ways:
* buy two bottles of Ber-Cola and five Bars bars;
* buy four bottles of Ber-Cola and don't buy Bars bars;
* don't buy Ber-Cola and buy 10 Bars bars.
In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly n burles. | instruction | 0 | 48,502 | 10 | 97,004 |
Tags: brute force, implementation, number theory
Correct Solution:
```
import bisect
import collections
import copy
import functools
import heapq
import itertools
import math
import random
import re
import sys
import time
import string
from typing import List, Mapping
sys.setrecursionlimit(999999)
n = int(input())
a = int(input())
b = int(input())
if n<min(a,b) or n%math.gcd(a,b):
print("NO")
exit()
if n%a==0:
print("YES")
print(n//a,0)
exit()
if n%b==0:
print("YES")
print(0,n//b)
exit()
for i in range((n//a)+1):
if n-i*a<b:
break
if (n-i*a)%b==0:
print("YES")
print(i,(n-i*a)//b)
exit()
print("NO")
``` | output | 1 | 48,502 | 10 | 97,005 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.
Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.
In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it's impossible.
Input
First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.
Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.
Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.
Output
If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).
Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.
Any of numbers x and y can be equal 0.
Examples
Input
7
2
3
Output
YES
2 1
Input
100
25
10
Output
YES
0 10
Input
15
4
8
Output
NO
Input
9960594
2551
2557
Output
YES
1951 1949
Note
In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles.
In second example Vasya can spend exactly n burles multiple ways:
* buy two bottles of Ber-Cola and five Bars bars;
* buy four bottles of Ber-Cola and don't buy Bars bars;
* don't buy Ber-Cola and buy 10 Bars bars.
In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly n burles. | instruction | 0 | 48,503 | 10 | 97,006 |
Tags: brute force, implementation, number theory
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sat Dec 16 14:43:34 2017
@author: ms
"""
def main():
n = int(input())
a = int(input())
b = int(input())
if(n%a == 0):
print('YES')
print(int(n/a), 0)
return
if(n%b == 0):
print('YES')
print(0, int(n/b))
return
if(a%2 == 0 and b%2 == 0):
if (n%2 == 1):
print('NO')
return
mx = max(a,b)
mn = min(a,b)
new = 0
found = 0
i = 0
while(n>=0):
if(n%mn == 0):
found = 1
new = n
break
n -= mx
i += 1
if found:
print('YES')
if (mx == a):
print(i, int(new/b))
if (mx == b):
print(int(new/a), i)
else:
print('NO')
main()
``` | output | 1 | 48,503 | 10 | 97,007 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.
Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.
In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it's impossible.
Input
First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.
Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.
Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.
Output
If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).
Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.
Any of numbers x and y can be equal 0.
Examples
Input
7
2
3
Output
YES
2 1
Input
100
25
10
Output
YES
0 10
Input
15
4
8
Output
NO
Input
9960594
2551
2557
Output
YES
1951 1949
Note
In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles.
In second example Vasya can spend exactly n burles multiple ways:
* buy two bottles of Ber-Cola and five Bars bars;
* buy four bottles of Ber-Cola and don't buy Bars bars;
* don't buy Ber-Cola and buy 10 Bars bars.
In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly n burles. | instruction | 0 | 48,504 | 10 | 97,008 |
Tags: brute force, implementation, number theory
Correct Solution:
```
from sys import stdin
n = int(stdin.readline())
a = int(stdin.readline())
b = int(stdin.readline())
def get_anwser(n, a, b):
answer = ('NO', -1, -1)
for i in range(0, 10**7 + 1):
if (n - a*i) >= 0 and (n - a*i) % b == 0:
return f'YES\n{i} {(n - a * i) // b}'
return 'NO'
print(get_anwser(n,a,b))
``` | output | 1 | 48,504 | 10 | 97,009 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.
Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.
In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it's impossible.
Input
First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.
Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.
Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.
Output
If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).
Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.
Any of numbers x and y can be equal 0.
Examples
Input
7
2
3
Output
YES
2 1
Input
100
25
10
Output
YES
0 10
Input
15
4
8
Output
NO
Input
9960594
2551
2557
Output
YES
1951 1949
Note
In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles.
In second example Vasya can spend exactly n burles multiple ways:
* buy two bottles of Ber-Cola and five Bars bars;
* buy four bottles of Ber-Cola and don't buy Bars bars;
* don't buy Ber-Cola and buy 10 Bars bars.
In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly n burles. | instruction | 0 | 48,505 | 10 | 97,010 |
Tags: brute force, implementation, number theory
Correct Solution:
```
n, a, b = [int(input()) for i in range(3)]
if a > b:
x = 0
y = -1
while x * a <= n:
if (n - x * a ) % b == 0:
y = (n - x * a) // b
break
x += 1
print("YES\n{} {}".format(x, y) if y != -1 else "NO")
else:
x = -1
y = 0
while y * b <= n:
if (n - y * b ) % a == 0:
x = (n - y * b) // a
break
y += 1
print("YES\n{} {}".format(x, y) if x != -1 else "NO")
``` | output | 1 | 48,505 | 10 | 97,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.
Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.
In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it's impossible.
Input
First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.
Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.
Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.
Output
If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).
Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.
Any of numbers x and y can be equal 0.
Examples
Input
7
2
3
Output
YES
2 1
Input
100
25
10
Output
YES
0 10
Input
15
4
8
Output
NO
Input
9960594
2551
2557
Output
YES
1951 1949
Note
In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles.
In second example Vasya can spend exactly n burles multiple ways:
* buy two bottles of Ber-Cola and five Bars bars;
* buy four bottles of Ber-Cola and don't buy Bars bars;
* don't buy Ber-Cola and buy 10 Bars bars.
In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly n burles.
Submitted Solution:
```
n=int(input())
a=int(input())
b=int(input())
if a==b and n%a!=0:
print('NO')
exit()
is_swapped=b>a
if is_swapped:
a,b=b,a
for i in range(n//a+1):
if (n-a*i)%b==0:
print('YES')
x=i
y=(n-a*i)//b
if is_swapped:
x,y=y,x
print(x,y)
exit()
print('NO')
``` | instruction | 0 | 48,506 | 10 | 97,012 |
Yes | output | 1 | 48,506 | 10 | 97,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.
Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.
In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it's impossible.
Input
First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.
Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.
Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.
Output
If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).
Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.
Any of numbers x and y can be equal 0.
Examples
Input
7
2
3
Output
YES
2 1
Input
100
25
10
Output
YES
0 10
Input
15
4
8
Output
NO
Input
9960594
2551
2557
Output
YES
1951 1949
Note
In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles.
In second example Vasya can spend exactly n burles multiple ways:
* buy two bottles of Ber-Cola and five Bars bars;
* buy four bottles of Ber-Cola and don't buy Bars bars;
* don't buy Ber-Cola and buy 10 Bars bars.
In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly n burles.
Submitted Solution:
```
n=int(input())
a=int(input())
b=int(input())
x=0
flag=0
while x<=n:
if (n-x)%b==0:
print('YES')
print(x//a,(n-x)//b)
flag=1
break
else:
x+=a
if flag==0:
print('NO')
``` | instruction | 0 | 48,507 | 10 | 97,014 |
Yes | output | 1 | 48,507 | 10 | 97,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.
Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.
In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it's impossible.
Input
First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.
Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.
Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.
Output
If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).
Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.
Any of numbers x and y can be equal 0.
Examples
Input
7
2
3
Output
YES
2 1
Input
100
25
10
Output
YES
0 10
Input
15
4
8
Output
NO
Input
9960594
2551
2557
Output
YES
1951 1949
Note
In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles.
In second example Vasya can spend exactly n burles multiple ways:
* buy two bottles of Ber-Cola and five Bars bars;
* buy four bottles of Ber-Cola and don't buy Bars bars;
* don't buy Ber-Cola and buy 10 Bars bars.
In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly n burles.
Submitted Solution:
```
if __name__ == '__main__':
n = int(input())
x = int(input())
y = int(input())
wins = 0
i = 0
while n>=(x*i):
if (n-(x*i))%y==0:
wins = 1
break
i+=1
if wins == 1:
print('YES')
print(i,(n-(x*i))//y)
else:
print('NO')
``` | instruction | 0 | 48,508 | 10 | 97,016 |
Yes | output | 1 | 48,508 | 10 | 97,017 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.
Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.
In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it's impossible.
Input
First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.
Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.
Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.
Output
If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).
Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.
Any of numbers x and y can be equal 0.
Examples
Input
7
2
3
Output
YES
2 1
Input
100
25
10
Output
YES
0 10
Input
15
4
8
Output
NO
Input
9960594
2551
2557
Output
YES
1951 1949
Note
In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles.
In second example Vasya can spend exactly n burles multiple ways:
* buy two bottles of Ber-Cola and five Bars bars;
* buy four bottles of Ber-Cola and don't buy Bars bars;
* don't buy Ber-Cola and buy 10 Bars bars.
In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly n burles.
Submitted Solution:
```
import sys
import math as mt
import bisect
input=sys.stdin.readline
#t=int(input())
t=1
def solve():
if a==b==1:
print("YES")
print(n,0)
return
else:
maxi=max(a,b)
i=0
while i<=n:
if (n-i)%b==0:
print("YES")
print(i//a,(n-i)//b)
return
i+=a
print("NO")
for _ in range(t):
n=int(input())
a=int(input())
b=int(input())
#n,k=map(int,input().split())
#x,y,k=map(int,input().split())
#n,h=(map(int,input().split()))
#l=list(map(int,input().split()))
#l2=list(map(int,input().split()))
(solve())
``` | instruction | 0 | 48,509 | 10 | 97,018 |
Yes | output | 1 | 48,509 | 10 | 97,019 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.
Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.
In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it's impossible.
Input
First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.
Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.
Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.
Output
If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).
Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.
Any of numbers x and y can be equal 0.
Examples
Input
7
2
3
Output
YES
2 1
Input
100
25
10
Output
YES
0 10
Input
15
4
8
Output
NO
Input
9960594
2551
2557
Output
YES
1951 1949
Note
In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles.
In second example Vasya can spend exactly n burles multiple ways:
* buy two bottles of Ber-Cola and five Bars bars;
* buy four bottles of Ber-Cola and don't buy Bars bars;
* don't buy Ber-Cola and buy 10 Bars bars.
In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly n burles.
Submitted Solution:
```
a=int(input())
b=int(input())
c=int(input())
d=True
m=0
l=False
while(d):
if((a-(b*m))%c==0):
l=True
break
m=m+1
if(m>a):
break
if(l):
print("Yes")
print(m,int((a-(b*m))/c))
if(l==False):
print("NO")
``` | instruction | 0 | 48,510 | 10 | 97,020 |
No | output | 1 | 48,510 | 10 | 97,021 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.
Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.
In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it's impossible.
Input
First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.
Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.
Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.
Output
If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).
Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.
Any of numbers x and y can be equal 0.
Examples
Input
7
2
3
Output
YES
2 1
Input
100
25
10
Output
YES
0 10
Input
15
4
8
Output
NO
Input
9960594
2551
2557
Output
YES
1951 1949
Note
In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles.
In second example Vasya can spend exactly n burles multiple ways:
* buy two bottles of Ber-Cola and five Bars bars;
* buy four bottles of Ber-Cola and don't buy Bars bars;
* don't buy Ber-Cola and buy 10 Bars bars.
In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly n burles.
Submitted Solution:
```
n = int(input())
a = int(input())
b = int(input())
def gcd(a, b):
if a * b == 0:
return a + b
return gcd(b, a % b)
d = gcd(a, b)
if n % d != 0:
print("NO")
else:
n, a, b = n//d, a//d, b//d
if a >= b:
for x in range(b + 1):
if (n - a * x) % b == 0:
print("YES")
print(x, (n - a * x) // b)
break
else:
print("NO")
else:
for y in range(a + 1):
if (n - b * y) % a == 0:
print("YES")
print((n - b * y) // a, y)
break
else:
print("NO")
``` | instruction | 0 | 48,511 | 10 | 97,022 |
No | output | 1 | 48,511 | 10 | 97,023 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.
Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.
In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it's impossible.
Input
First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.
Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.
Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.
Output
If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).
Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.
Any of numbers x and y can be equal 0.
Examples
Input
7
2
3
Output
YES
2 1
Input
100
25
10
Output
YES
0 10
Input
15
4
8
Output
NO
Input
9960594
2551
2557
Output
YES
1951 1949
Note
In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles.
In second example Vasya can spend exactly n burles multiple ways:
* buy two bottles of Ber-Cola and five Bars bars;
* buy four bottles of Ber-Cola and don't buy Bars bars;
* don't buy Ber-Cola and buy 10 Bars bars.
In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly n burles.
Submitted Solution:
```
n = int(input())
a = int(input())
b = int(input())
x = 0
y = 0
impossible = False
if a == 1:
print("YES")
print(n, 0)
exit()
if b == 1:
print("YES")
print(0, n)
exit()
while True:
ax = a * x
nax = n - ax
y = nax/b
if (y % 1 == 0) or (ax == n):
print("YES")
print(x,int(y))
exit()
x += 1
if n > 9999999:
if ((n % 2) != (a % 2)) and ((n % 2) != (b % 2)):
impossible = True
if (n == 10000000) and (10/a % 1 != 0) and (10/b % 1 != 0):
impossible = True
if (y < 0) or (impossible):
print("NO")
exit()
``` | instruction | 0 | 48,512 | 10 | 97,024 |
No | output | 1 | 48,512 | 10 | 97,025 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has n burles. One bottle of Ber-Cola costs a burles and one Bars bar costs b burles. He can buy any non-negative integer number of bottles of Ber-Cola and any non-negative integer number of Bars bars.
Find out if it's possible to buy some amount of bottles of Ber-Cola and Bars bars and spend exactly n burles.
In other words, you should find two non-negative integers x and y such that Vasya can buy x bottles of Ber-Cola and y Bars bars and x·a + y·b = n or tell that it's impossible.
Input
First line contains single integer n (1 ≤ n ≤ 10 000 000) — amount of money, that Vasya has.
Second line contains single integer a (1 ≤ a ≤ 10 000 000) — cost of one bottle of Ber-Cola.
Third line contains single integer b (1 ≤ b ≤ 10 000 000) — cost of one Bars bar.
Output
If Vasya can't buy Bars and Ber-Cola in such a way to spend exactly n burles print «NO» (without quotes).
Otherwise in first line print «YES» (without quotes). In second line print two non-negative integers x and y — number of bottles of Ber-Cola and number of Bars bars Vasya should buy in order to spend exactly n burles, i.e. x·a + y·b = n. If there are multiple answers print any of them.
Any of numbers x and y can be equal 0.
Examples
Input
7
2
3
Output
YES
2 1
Input
100
25
10
Output
YES
0 10
Input
15
4
8
Output
NO
Input
9960594
2551
2557
Output
YES
1951 1949
Note
In first example Vasya can buy two bottles of Ber-Cola and one Bars bar. He will spend exactly 2·2 + 1·3 = 7 burles.
In second example Vasya can spend exactly n burles multiple ways:
* buy two bottles of Ber-Cola and five Bars bars;
* buy four bottles of Ber-Cola and don't buy Bars bars;
* don't buy Ber-Cola and buy 10 Bars bars.
In third example it's impossible to but Ber-Cola and Bars bars in order to spend exactly n burles.
Submitted Solution:
```
# B. Proper Nutrition
n=int(input())
a=int(input())
b=int(input())
for i in range(n//2+1):
if i%a==0 and (n-i)%b==0:
print("YES")
print(i//a,(n-i)//b)
break
elif i%b==0 and (n-i)%a==0:
print("YES")
print(i//b,(n-i)//a)
break
else:
print("NO")
``` | instruction | 0 | 48,513 | 10 | 97,026 |
No | output | 1 | 48,513 | 10 | 97,027 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n TV shows you want to watch. Suppose the whole time is split into equal parts called "minutes". The i-th of the shows is going from l_i-th to r_i-th minute, both ends inclusive.
You need a TV to watch a TV show and you can't watch two TV shows which air at the same time on the same TV, so it is possible you will need multiple TVs in some minutes. For example, if segments [l_i, r_i] and [l_j, r_j] intersect, then shows i and j can't be watched simultaneously on one TV.
Once you start watching a show on some TV it is not possible to "move" it to another TV (since it would be too distracting), or to watch another show on the same TV until this show ends.
There is a TV Rental shop near you. It rents a TV for x rupees, and charges y (y < x) rupees for every extra minute you keep the TV. So in order to rent a TV for minutes [a; b] you will need to pay x + y ⋅ (b - a).
You can assume, that taking and returning of the TV doesn't take any time and doesn't distract from watching other TV shows. Find the minimum possible cost to view all shows. Since this value could be too large, print it modulo 10^9 + 7.
Input
The first line contains integers n, x and y (1 ≤ n ≤ 10^5, 1 ≤ y < x ≤ 10^9) — the number of TV shows, the cost to rent a TV for the first minute and the cost to rent a TV for every subsequent minute.
Each of the next n lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10^9) denoting the start and the end minute of the i-th TV show.
Output
Print exactly one integer — the minimum cost to view all the shows taken modulo 10^9 + 7.
Examples
Input
5 4 3
1 2
4 10
2 4
10 11
5 9
Output
60
Input
6 3 2
8 20
6 22
4 15
20 28
17 25
20 27
Output
142
Input
2 1000000000 2
1 2
2 3
Output
999999997
Note
In the first example, the optimal strategy would be to rent 3 TVs to watch:
* Show [1, 2] on the first TV,
* Show [4, 10] on the second TV,
* Shows [2, 4], [5, 9], [10, 11] on the third TV.
This way the cost for the first TV is 4 + 3 ⋅ (2 - 1) = 7, for the second is 4 + 3 ⋅ (10 - 4) = 22 and for the third is 4 + 3 ⋅ (11 - 2) = 31, which gives 60 int total.
In the second example, it is optimal watch each show on a new TV.
In third example, it is optimal to watch both shows on a new TV. Note that the answer is to be printed modulo 10^9 + 7. | instruction | 0 | 48,847 | 10 | 97,694 |
Tags: data structures, greedy, implementation, sortings
Correct Solution:
```
from bisect import bisect_right
n, x, y = map(int, input().split(' '))
s=[0]*n
e=[0]*n
v=[0]*n
c=0
for i in range(n):
s[i],e[i]=map(int, input().split(' '))
c+=x+(e[i]-s[i])*y
s.sort()
e.sort()
for i in range(n-2,-1,-1):
k=bisect_right(s,e[i])
while (k < n) and (v[k]==1) and (s[k]-e[i]) * y < x :
k+=1
if k==n:
continue
if (s[k]-e[i]) * y < x :
v[k] = 1
c+=(s[k]-e[i])*y-x
print(c%(10**9+7))
``` | output | 1 | 48,847 | 10 | 97,695 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n TV shows you want to watch. Suppose the whole time is split into equal parts called "minutes". The i-th of the shows is going from l_i-th to r_i-th minute, both ends inclusive.
You need a TV to watch a TV show and you can't watch two TV shows which air at the same time on the same TV, so it is possible you will need multiple TVs in some minutes. For example, if segments [l_i, r_i] and [l_j, r_j] intersect, then shows i and j can't be watched simultaneously on one TV.
Once you start watching a show on some TV it is not possible to "move" it to another TV (since it would be too distracting), or to watch another show on the same TV until this show ends.
There is a TV Rental shop near you. It rents a TV for x rupees, and charges y (y < x) rupees for every extra minute you keep the TV. So in order to rent a TV for minutes [a; b] you will need to pay x + y ⋅ (b - a).
You can assume, that taking and returning of the TV doesn't take any time and doesn't distract from watching other TV shows. Find the minimum possible cost to view all shows. Since this value could be too large, print it modulo 10^9 + 7.
Input
The first line contains integers n, x and y (1 ≤ n ≤ 10^5, 1 ≤ y < x ≤ 10^9) — the number of TV shows, the cost to rent a TV for the first minute and the cost to rent a TV for every subsequent minute.
Each of the next n lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10^9) denoting the start and the end minute of the i-th TV show.
Output
Print exactly one integer — the minimum cost to view all the shows taken modulo 10^9 + 7.
Examples
Input
5 4 3
1 2
4 10
2 4
10 11
5 9
Output
60
Input
6 3 2
8 20
6 22
4 15
20 28
17 25
20 27
Output
142
Input
2 1000000000 2
1 2
2 3
Output
999999997
Note
In the first example, the optimal strategy would be to rent 3 TVs to watch:
* Show [1, 2] on the first TV,
* Show [4, 10] on the second TV,
* Shows [2, 4], [5, 9], [10, 11] on the third TV.
This way the cost for the first TV is 4 + 3 ⋅ (2 - 1) = 7, for the second is 4 + 3 ⋅ (10 - 4) = 22 and for the third is 4 + 3 ⋅ (11 - 2) = 31, which gives 60 int total.
In the second example, it is optimal watch each show on a new TV.
In third example, it is optimal to watch both shows on a new TV. Note that the answer is to be printed modulo 10^9 + 7. | instruction | 0 | 48,848 | 10 | 97,696 |
Tags: data structures, greedy, implementation, sortings
Correct Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
import sys
from io import BytesIO, IOBase
from heapq import *
def main():
n,x,y=map(int,input().split())
a,b=[],[]
for i in range(n):
u,v=map(int,input().split())
a.append((u,1))
a.append((v,-1))
a.sort(key=lambda x:x[0]*10000000000-x[1])
mod=10**9+7
t,z,ans=1,1,x
for i in range(1,len(a)):
z+=a[i][1]
if z<t:
ans=(ans+t*(a[i][0]-a[i-1][0])*y)%mod
heappush(b,-a[i][0])
else:
if b:
if x<(a[i][0]+b[0])*y:
ans=(ans+t*(a[i][0]-a[i-1][0])*y+x)%mod
else:
ans=(ans+t*(a[i][0]-a[i-1][0])*y+(a[i][0]+b[0])*y)%mod
heappop(b)
else:
ans = (ans + t * (a[i][0] - a[i - 1][0]) * y + x) % mod
t=z
print(ans)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | output | 1 | 48,848 | 10 | 97,697 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n TV shows you want to watch. Suppose the whole time is split into equal parts called "minutes". The i-th of the shows is going from l_i-th to r_i-th minute, both ends inclusive.
You need a TV to watch a TV show and you can't watch two TV shows which air at the same time on the same TV, so it is possible you will need multiple TVs in some minutes. For example, if segments [l_i, r_i] and [l_j, r_j] intersect, then shows i and j can't be watched simultaneously on one TV.
Once you start watching a show on some TV it is not possible to "move" it to another TV (since it would be too distracting), or to watch another show on the same TV until this show ends.
There is a TV Rental shop near you. It rents a TV for x rupees, and charges y (y < x) rupees for every extra minute you keep the TV. So in order to rent a TV for minutes [a; b] you will need to pay x + y ⋅ (b - a).
You can assume, that taking and returning of the TV doesn't take any time and doesn't distract from watching other TV shows. Find the minimum possible cost to view all shows. Since this value could be too large, print it modulo 10^9 + 7.
Input
The first line contains integers n, x and y (1 ≤ n ≤ 10^5, 1 ≤ y < x ≤ 10^9) — the number of TV shows, the cost to rent a TV for the first minute and the cost to rent a TV for every subsequent minute.
Each of the next n lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10^9) denoting the start and the end minute of the i-th TV show.
Output
Print exactly one integer — the minimum cost to view all the shows taken modulo 10^9 + 7.
Examples
Input
5 4 3
1 2
4 10
2 4
10 11
5 9
Output
60
Input
6 3 2
8 20
6 22
4 15
20 28
17 25
20 27
Output
142
Input
2 1000000000 2
1 2
2 3
Output
999999997
Note
In the first example, the optimal strategy would be to rent 3 TVs to watch:
* Show [1, 2] on the first TV,
* Show [4, 10] on the second TV,
* Shows [2, 4], [5, 9], [10, 11] on the third TV.
This way the cost for the first TV is 4 + 3 ⋅ (2 - 1) = 7, for the second is 4 + 3 ⋅ (10 - 4) = 22 and for the third is 4 + 3 ⋅ (11 - 2) = 31, which gives 60 int total.
In the second example, it is optimal watch each show on a new TV.
In third example, it is optimal to watch both shows on a new TV. Note that the answer is to be printed modulo 10^9 + 7. | instruction | 0 | 48,849 | 10 | 97,698 |
Tags: data structures, greedy, implementation, sortings
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
import threading
#sys.setrecursionlimit(300000)
#threading.stack_size(10**8)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------------------------------------------------------------
#mod = 9223372036854775807
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a+b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
class SegmentTree1:
def __init__(self, data, default=10**6, func=lambda a, b: min(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
MOD=10**9+7
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
mod=10**9+7
omod=998244353
#-------------------------------------------------------------------------
prime = [True for i in range(10)]
pp=[0]*10
def SieveOfEratosthenes(n=10):
p = 2
c=0
while (p * p <= n):
if (prime[p] == True):
c+=1
for i in range(p, n+1, p):
pp[i]+=1
prime[i] = False
p += 1
#---------------------------------Binary Search------------------------------------------
def binarySearch(arr, n, key):
left = 0
right = n-1
mid = 0
res=arr[n-1]
while (left <= right):
mid = (right + left)//2
if (arr[mid] >= key):
res=arr[mid]
right = mid-1
else:
left = mid + 1
return res
def binarySearch1(arr, n, key):
left = 0
right = n-1
mid = 0
res=arr[0]
while (left <= right):
mid = (right + left)//2
if (arr[mid] > key):
right = mid-1
else:
res=arr[mid]
left = mid + 1
return res
#---------------------------------running code------------------------------------------
import heapq
n,x,y=map(int,input().split())
start=defaultdict(int)
end=defaultdict(int)
arr=[]
a=set()
h=[]
heapq.heapify(h)
res=0
for i in range (n):
a1,a2=map(int,input().split())
start[a1]+=1
end[a2]+=1
res+=y*(a2-a1)
res%=mod
arr.append([a1,a2])
a.add(a1)
a.add(a2)
a=list(a)
a.sort()
for t in a:
for i in range (start[t]):
if not h:
res+=x
else:
prev=-heapq.heappop(h)
d=y*(t-prev)
if d<x:
res+=d
else:
heapq.heappush(h, -prev)
res+=x
res%=mod
for i in range (end[t]):
heapq.heappush(h, -t)
print(res)
``` | output | 1 | 48,849 | 10 | 97,699 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n TV shows you want to watch. Suppose the whole time is split into equal parts called "minutes". The i-th of the shows is going from l_i-th to r_i-th minute, both ends inclusive.
You need a TV to watch a TV show and you can't watch two TV shows which air at the same time on the same TV, so it is possible you will need multiple TVs in some minutes. For example, if segments [l_i, r_i] and [l_j, r_j] intersect, then shows i and j can't be watched simultaneously on one TV.
Once you start watching a show on some TV it is not possible to "move" it to another TV (since it would be too distracting), or to watch another show on the same TV until this show ends.
There is a TV Rental shop near you. It rents a TV for x rupees, and charges y (y < x) rupees for every extra minute you keep the TV. So in order to rent a TV for minutes [a; b] you will need to pay x + y ⋅ (b - a).
You can assume, that taking and returning of the TV doesn't take any time and doesn't distract from watching other TV shows. Find the minimum possible cost to view all shows. Since this value could be too large, print it modulo 10^9 + 7.
Input
The first line contains integers n, x and y (1 ≤ n ≤ 10^5, 1 ≤ y < x ≤ 10^9) — the number of TV shows, the cost to rent a TV for the first minute and the cost to rent a TV for every subsequent minute.
Each of the next n lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10^9) denoting the start and the end minute of the i-th TV show.
Output
Print exactly one integer — the minimum cost to view all the shows taken modulo 10^9 + 7.
Examples
Input
5 4 3
1 2
4 10
2 4
10 11
5 9
Output
60
Input
6 3 2
8 20
6 22
4 15
20 28
17 25
20 27
Output
142
Input
2 1000000000 2
1 2
2 3
Output
999999997
Note
In the first example, the optimal strategy would be to rent 3 TVs to watch:
* Show [1, 2] on the first TV,
* Show [4, 10] on the second TV,
* Shows [2, 4], [5, 9], [10, 11] on the third TV.
This way the cost for the first TV is 4 + 3 ⋅ (2 - 1) = 7, for the second is 4 + 3 ⋅ (10 - 4) = 22 and for the third is 4 + 3 ⋅ (11 - 2) = 31, which gives 60 int total.
In the second example, it is optimal watch each show on a new TV.
In third example, it is optimal to watch both shows on a new TV. Note that the answer is to be printed modulo 10^9 + 7. | instruction | 0 | 48,850 | 10 | 97,700 |
Tags: data structures, greedy, implementation, sortings
Correct Solution:
```
from bisect import bisect_right,bisect_left
n,c_tv,c_es=map(int,input().split())
start=[]
end=[]
add=0
for _ in range(n):
l,r=map(int,input().split())
add+=(r-l)
start.append(l)
end.append(r)
start.sort()
end.sort()
ans=add*c_es+n*c_tv
M=10**9+7
v=[0]*(n+1)
for i in range(n):
indx=bisect_left(end,start[i])-1
k=indx
while k>=0 and (start[i]-end[k])*c_es<c_tv and v[k]==1:
k-=1
if k==-1:
continue
if (start[i]-end[k])*c_es<c_tv:
ans-=c_tv-(start[i]-end[k])*c_es
v[k]=1
print(ans%M)
``` | output | 1 | 48,850 | 10 | 97,701 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n TV shows you want to watch. Suppose the whole time is split into equal parts called "minutes". The i-th of the shows is going from l_i-th to r_i-th minute, both ends inclusive.
You need a TV to watch a TV show and you can't watch two TV shows which air at the same time on the same TV, so it is possible you will need multiple TVs in some minutes. For example, if segments [l_i, r_i] and [l_j, r_j] intersect, then shows i and j can't be watched simultaneously on one TV.
Once you start watching a show on some TV it is not possible to "move" it to another TV (since it would be too distracting), or to watch another show on the same TV until this show ends.
There is a TV Rental shop near you. It rents a TV for x rupees, and charges y (y < x) rupees for every extra minute you keep the TV. So in order to rent a TV for minutes [a; b] you will need to pay x + y ⋅ (b - a).
You can assume, that taking and returning of the TV doesn't take any time and doesn't distract from watching other TV shows. Find the minimum possible cost to view all shows. Since this value could be too large, print it modulo 10^9 + 7.
Input
The first line contains integers n, x and y (1 ≤ n ≤ 10^5, 1 ≤ y < x ≤ 10^9) — the number of TV shows, the cost to rent a TV for the first minute and the cost to rent a TV for every subsequent minute.
Each of the next n lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ 10^9) denoting the start and the end minute of the i-th TV show.
Output
Print exactly one integer — the minimum cost to view all the shows taken modulo 10^9 + 7.
Examples
Input
5 4 3
1 2
4 10
2 4
10 11
5 9
Output
60
Input
6 3 2
8 20
6 22
4 15
20 28
17 25
20 27
Output
142
Input
2 1000000000 2
1 2
2 3
Output
999999997
Note
In the first example, the optimal strategy would be to rent 3 TVs to watch:
* Show [1, 2] on the first TV,
* Show [4, 10] on the second TV,
* Shows [2, 4], [5, 9], [10, 11] on the third TV.
This way the cost for the first TV is 4 + 3 ⋅ (2 - 1) = 7, for the second is 4 + 3 ⋅ (10 - 4) = 22 and for the third is 4 + 3 ⋅ (11 - 2) = 31, which gives 60 int total.
In the second example, it is optimal watch each show on a new TV.
In third example, it is optimal to watch both shows on a new TV. Note that the answer is to be printed modulo 10^9 + 7. | instruction | 0 | 48,851 | 10 | 97,702 |
Tags: data structures, greedy, implementation, sortings
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
##########################################################
#q.sort(key=lambda x:((x[1]-x[0]),-x[0]))
#from collections import Counter
#from fractions import Fraction
#s=iter(input())
# for _ in range(int(input())):
#from collections import deque
# n=int(input())
# n,k=map(int,input().split())
# arr=list(map(int,input().split()))
#ls=list(map(int,input().split()))
#for in range(m):
from bisect import bisect_right
n, x, y = map(int, input().split(' '))
s=[0]*n
e=[0]*n
v=[0]*n
c=0
for i in range(n):
s[i],e[i]=map(int, input().split(' '))
c+=x+(e[i]-s[i])*y
s.sort()
e.sort()
for i in range(n-2,-1,-1):
k=bisect_right(s,e[i])
while (k < n) and (v[k]==1) and (s[k]-e[i]) * y < x :
k+=1
if k==n:
continue
if (s[k]-e[i]) * y < x :
v[k] = 1
c+=(s[k]-e[i])*y-x
print(c%(10**9+7))
``` | output | 1 | 48,851 | 10 | 97,703 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems.
There are m types of discounts. We assume that the discounts are indexed from 1 to m. To use the discount number i, the customer takes a special basket, where he puts exactly qi items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the qi items in the cart.
Maxim now needs to buy n items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well.
Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.
Input
The first line contains integer m (1 ≤ m ≤ 105) — the number of discount types. The second line contains m integers: q1, q2, ..., qm (1 ≤ qi ≤ 105).
The third line contains integer n (1 ≤ n ≤ 105) — the number of items Maxim needs. The fourth line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 104) — the items' prices.
The numbers in the lines are separated by single spaces.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1
2
4
50 50 100 100
Output
200
Input
2
2 3
5
50 50 50 50 50
Output
150
Input
1
1
7
1 1 1 1 1 1 1
Output
3
Note
In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200.
In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150. | instruction | 0 | 49,128 | 10 | 98,256 |
Tags: greedy, sortings
Correct Solution:
```
if __name__ == '__main__':
m = int(input())
carts = [int(x) for x in input().split()]
n = int(input())
cost = [int(x) for x in input().split()]
x = min(carts)
bought = 0
free = 0
cost = sorted(cost, reverse = True)
amt = 0
for i in cost:
if free > 0:
free -= 1
continue
amt += i
bought += 1
if bought == x:
bought = 0
free = 2
print(amt)
``` | output | 1 | 49,128 | 10 | 98,257 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems.
There are m types of discounts. We assume that the discounts are indexed from 1 to m. To use the discount number i, the customer takes a special basket, where he puts exactly qi items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the qi items in the cart.
Maxim now needs to buy n items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well.
Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.
Input
The first line contains integer m (1 ≤ m ≤ 105) — the number of discount types. The second line contains m integers: q1, q2, ..., qm (1 ≤ qi ≤ 105).
The third line contains integer n (1 ≤ n ≤ 105) — the number of items Maxim needs. The fourth line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 104) — the items' prices.
The numbers in the lines are separated by single spaces.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1
2
4
50 50 100 100
Output
200
Input
2
2 3
5
50 50 50 50 50
Output
150
Input
1
1
7
1 1 1 1 1 1 1
Output
3
Note
In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200.
In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150. | instruction | 0 | 49,129 | 10 | 98,258 |
Tags: greedy, sortings
Correct Solution:
```
def main():
input()
q = min(map(int, input().split()))
input()
aa = sorted(map(int, input().split()), reverse=True)
print(sum(aa) - sum(aa[q::q + 2]) - sum(aa[q + 1::q + 2]))
if __name__ == "__main__":
main()
``` | output | 1 | 49,129 | 10 | 98,259 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems.
There are m types of discounts. We assume that the discounts are indexed from 1 to m. To use the discount number i, the customer takes a special basket, where he puts exactly qi items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the qi items in the cart.
Maxim now needs to buy n items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well.
Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.
Input
The first line contains integer m (1 ≤ m ≤ 105) — the number of discount types. The second line contains m integers: q1, q2, ..., qm (1 ≤ qi ≤ 105).
The third line contains integer n (1 ≤ n ≤ 105) — the number of items Maxim needs. The fourth line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 104) — the items' prices.
The numbers in the lines are separated by single spaces.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1
2
4
50 50 100 100
Output
200
Input
2
2 3
5
50 50 50 50 50
Output
150
Input
1
1
7
1 1 1 1 1 1 1
Output
3
Note
In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200.
In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150. | instruction | 0 | 49,130 | 10 | 98,260 |
Tags: greedy, sortings
Correct Solution:
```
import bisect
from itertools import accumulate
import os
import sys
import math
from decimal import *
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def input(): return sys.stdin.readline().rstrip("\r\n")
def factors(n):
fac=[]
while(n%2==0):
fac.append(2)
n=n//2
for i in range(3,int(math.sqrt(n))+2):
while(n%i==0):
fac.append(i)
n=n//i
if n>1:
fac.append(n)
return fac
# ------------------- fast io --------------------]]
n=int(input())
q=sorted(list(map(int,input().split())))
m=int(input())
m=sorted(list(map(int,input().split())))
m=m[::-1]
i=0
j=0
sumi=0
while(j<len(m)):
for k in range(j,min(j+q[i],len(m))):
sumi+=m[k]
j+=1
j+=2
print(sumi)
``` | output | 1 | 49,130 | 10 | 98,261 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems.
There are m types of discounts. We assume that the discounts are indexed from 1 to m. To use the discount number i, the customer takes a special basket, where he puts exactly qi items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the qi items in the cart.
Maxim now needs to buy n items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well.
Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.
Input
The first line contains integer m (1 ≤ m ≤ 105) — the number of discount types. The second line contains m integers: q1, q2, ..., qm (1 ≤ qi ≤ 105).
The third line contains integer n (1 ≤ n ≤ 105) — the number of items Maxim needs. The fourth line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 104) — the items' prices.
The numbers in the lines are separated by single spaces.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1
2
4
50 50 100 100
Output
200
Input
2
2 3
5
50 50 50 50 50
Output
150
Input
1
1
7
1 1 1 1 1 1 1
Output
3
Note
In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200.
In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150. | instruction | 0 | 49,131 | 10 | 98,262 |
Tags: greedy, sortings
Correct Solution:
```
m = int(input())
q = list(map(int, input().split()))
n = int(input())
a = list(map(int, input().split()))
a.sort(reverse=True)
q.sort()
ans = 0
mm = q[0]
i = 0
while i<n:
j = i
if j+mm <= n:
while i < j+mm:
ans+=a[i]
i+=1
i+=1
else:
break
i+=1
if i <n:
ans += sum(a[i:])
print(ans)
``` | output | 1 | 49,131 | 10 | 98,263 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems.
There are m types of discounts. We assume that the discounts are indexed from 1 to m. To use the discount number i, the customer takes a special basket, where he puts exactly qi items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the qi items in the cart.
Maxim now needs to buy n items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well.
Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.
Input
The first line contains integer m (1 ≤ m ≤ 105) — the number of discount types. The second line contains m integers: q1, q2, ..., qm (1 ≤ qi ≤ 105).
The third line contains integer n (1 ≤ n ≤ 105) — the number of items Maxim needs. The fourth line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 104) — the items' prices.
The numbers in the lines are separated by single spaces.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1
2
4
50 50 100 100
Output
200
Input
2
2 3
5
50 50 50 50 50
Output
150
Input
1
1
7
1 1 1 1 1 1 1
Output
3
Note
In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200.
In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150. | instruction | 0 | 49,132 | 10 | 98,264 |
Tags: greedy, sortings
Correct Solution:
```
m=int(input())
q=[int(i) for i in input().split()]
q.sort()
n=int(input())
a=[int(i) for i in input().split()]
a.sort()
ans=0
i=n-1
while(i>=0):
d=0
while(i>=0 and d<q[0]):
ans+=a[i]
i-=1
d+=1
d=0
while(i>=0 and d<2):
d+=1
i-=1
print(ans)
``` | output | 1 | 49,132 | 10 | 98,265 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems.
There are m types of discounts. We assume that the discounts are indexed from 1 to m. To use the discount number i, the customer takes a special basket, where he puts exactly qi items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the qi items in the cart.
Maxim now needs to buy n items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well.
Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.
Input
The first line contains integer m (1 ≤ m ≤ 105) — the number of discount types. The second line contains m integers: q1, q2, ..., qm (1 ≤ qi ≤ 105).
The third line contains integer n (1 ≤ n ≤ 105) — the number of items Maxim needs. The fourth line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 104) — the items' prices.
The numbers in the lines are separated by single spaces.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1
2
4
50 50 100 100
Output
200
Input
2
2 3
5
50 50 50 50 50
Output
150
Input
1
1
7
1 1 1 1 1 1 1
Output
3
Note
In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200.
In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150. | instruction | 0 | 49,133 | 10 | 98,266 |
Tags: greedy, sortings
Correct Solution:
```
#!/usr/bin/env python
from __future__ import division, print_function
import math
import os
import sys
from fractions import *
from sys import *
from decimal import *
from io import BytesIO, IOBase
from itertools import *
from collections import *
# sys.setrecursionlimit(10**5)
M = 10 ** 9 + 7
# print(math.factorial(5))
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
# sys.setrecursionlimit(10**6)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
def inpu(): return int(inp())
# -----------------------------------------------------------------
def regularbracket(t):
p = 0
for i in t:
if i == "(":
p += 1
else:
p -= 1
if p < 0:
return False
else:
if p > 0:
return False
else:
return True
# -------------------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <= key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# ------------------------------reverse string(pallindrome)
def reverse1(string):
pp = ""
for i in string[::-1]:
pp += i
if pp == string:
return True
return False
# --------------------------------reverse list(paindrome)
def reverse2(list1):
l = []
for i in list1[::-1]:
l.append(i)
if l == list1:
return True
return False
def mex(list1):
# list1 = sorted(list1)
p = max(list1) + 1
for i in range(len(list1)):
if list1[i] != i:
p = i
break
return p
def sumofdigits(n):
n = str(n)
s1 = 0
for i in n:
s1 += int(i)
return s1
def perfect_square(n):
s = math.sqrt(n)
if s == int(s):
return True
return False
# -----------------------------roman
def roman_number(x):
if x > 15999:
return
value = [5000, 4000, 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1]
symbol = ["F", "MF", "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"]
roman = ""
i = 0
while x > 0:
div = x // value[i]
x = x % value[i]
while div:
roman += symbol[i]
div -= 1
i += 1
return roman
def soretd(s):
for i in range(1, len(s)):
if s[i - 1] > s[i]:
return False
return True
# print(soretd("1"))
# ---------------------------
def countRhombi(h, w):
ct = 0
for i in range(2, h + 1, 2):
for j in range(2, w + 1, 2):
ct += (h - i + 1) * (w - j + 1)
return ct
def countrhombi2(h, w):
return ((h * h) // 4) * ((w * w) // 4)
# ---------------------------------
def binpow(a, b):
if b == 0:
return 1
else:
res = binpow(a, b // 2)
if b % 2 != 0:
return res * res * a
else:
return res * res
# -------------------------------------------------------
def binpowmodulus(a, b, m):
a %= m
res = 1
while (b > 0):
if (b & 1):
res = res * a % m
a = a * a % m
b >>= 1
return res
# -------------------------------------------------------------
def coprime_to_n(n):
result = n
i = 2
while (i * i <= n):
if (n % i == 0):
while (n % i == 0):
n //= i
result -= result // i
i += 1
if (n > 1):
result -= result // n
return result
# -------------------prime
def prime(x):
if x == 1:
return False
else:
for i in range(2, int(math.sqrt(x)) + 1):
# print(x)
if (x % i == 0):
return False
else:
return True
def luckynumwithequalnumberoffourandseven(x,n,a):
if x >= n and str(x).count("4") == str(x).count("7"):
a.append(x)
else:
if x < 1e12:
luckynumwithequalnumberoffourandseven(x * 10 + 4,n,a)
luckynumwithequalnumberoffourandseven(x * 10 + 7,n,a)
return a
#----------------------
def luckynum(x,l,r,a):
if x>=l and x<=r:
a.append(x)
if x>r:
a.append(x)
return a
if x < 1e10:
luckynum(x * 10 + 4, l,r,a)
luckynum(x * 10 + 7, l,r,a)
return a
def luckynuber(x, n, a):
p = set(str(x))
if len(p) <= 2:
a.append(x)
if x < n:
luckynuber(x + 1, n, a)
return a
# ------------------------------------------------------interactive problems
def interact(type, x):
if type == "r":
inp = input()
return inp.strip()
else:
print(x, flush=True)
# ------------------------------------------------------------------zero at end of factorial of a number
def findTrailingZeros(n):
# Initialize result
count = 0
# Keep dividing n by
# 5 & update Count
while (n >= 5):
n //= 5
count += n
return count
# -----------------------------------------------merge sort
# Python program for implementation of MergeSort
def mergeSort(arr):
if len(arr) > 1:
# Finding the mid of the array
mid = len(arr) // 2
# Dividing the array elements
L = arr[:mid]
# into 2 halves
R = arr[mid:]
# Sorting the first half
mergeSort(L)
# Sorting the second half
mergeSort(R)
i = j = k = 0
# Copy data to temp arrays L[] and R[]
while i < len(L) and j < len(R):
if L[i] < R[j]:
arr[k] = L[i]
i += 1
else:
arr[k] = R[j]
j += 1
k += 1
# Checking if any element was left
while i < len(L):
arr[k] = L[i]
i += 1
k += 1
while j < len(R):
arr[k] = R[j]
j += 1
k += 1
# -----------------------------------------------lucky number with two lucky any digits
res = set()
def solven(p, l, a, b, n): # given number
if p > n or l > 10:
return
if p > 0:
res.add(p)
solven(p * 10 + a, l + 1, a, b, n)
solven(p * 10 + b, l + 1, a, b, n)
# problem
"""
n = int(input())
for a in range(0, 10):
for b in range(0, a):
solve(0, 0)
print(len(res))
"""
# Python3 program to find all subsets
# by backtracking.
# In the array A at every step we have two
# choices for each element either we can
# ignore the element or we can include the
# element in our subset
def subsetsUtil(A, subset, index, d):
print(*subset)
s = sum(subset)
d.append(s)
for i in range(index, len(A)):
# include the A[i] in subset.
subset.append(A[i])
# move onto the next element.
subsetsUtil(A, subset, i + 1, d)
# exclude the A[i] from subset and
# triggers backtracking.
subset.pop(-1)
return d
def subsetSums(arr, l, r, d, sum=0):
if l > r:
d.append(sum)
return
subsetSums(arr, l + 1, r, d, sum + arr[l])
# Subset excluding arr[l]
subsetSums(arr, l + 1, r, d, sum)
return d
def print_factors(x):
factors = []
for i in range(1, x + 1):
if x % i == 0:
factors.append(i)
return (factors)
# -----------------------------------------------
def calc(X, d, ans, D):
# print(X,d)
if len(X) == 0:
return
i = X.index(max(X))
ans[D[max(X)]] = d
Y = X[:i]
Z = X[i + 1:]
calc(Y, d + 1, ans, D)
calc(Z, d + 1, ans, D)
# ---------------------------------------
def factorization(n, l):
c = n
if prime(n) == True:
l.append(n)
return l
for i in range(2, c):
if n == 1:
break
while n % i == 0:
l.append(i)
n = n // i
return l
# endregion------------------------------
def good(b):
l = []
i = 0
while (len(b) != 0):
if b[i] < b[len(b) - 1 - i]:
l.append(b[i])
b.remove(b[i])
else:
l.append(b[len(b) - 1 - i])
b.remove(b[len(b) - 1 - i])
if l == sorted(l):
# print(l)
return True
return False
# arr=[]
# print(good(arr))
def generate(st, s):
if len(s) == 0:
return
# If current string is not already present.
if s not in st:
st.add(s)
# Traverse current string, one by one
# remove every character and recur.
for i in range(len(s)):
t = list(s).copy()
t.remove(s[i])
t = ''.join(t)
generate(st, t)
return
#=--------------------------------------------longest increasing subsequence
def largestincreasingsubsequence(A):
l = [1]*len(A)
sub=[]
for i in range(1,len(l)):
for k in range(i):
if A[k]<A[i]:
sub.append(l[k])
l[i]=1+max(sub,default=0)
return max(l,default=0)
#----------------------------------longest palindromic substring
# Python3 program for the
# above approach
# Function to calculate
# Bitwise OR of sums of
# all subsequences
def findOR(nums, N):
# Stores the prefix
# sum of nums[]
prefix_sum = 0
# Stores the bitwise OR of
# sum of each subsequence
result = 0
# Iterate through array nums[]
for i in range(N):
# Bits set in nums[i] are
# also set in result
result |= nums[i]
# Calculate prefix_sum
prefix_sum += nums[i]
# Bits set in prefix_sum
# are also set in result
result |= prefix_sum
# Return the result
return result
#l=[]
def OR(a, n):
ans = a[0]
for i in range(1, n):
ans |= a[i]
#l.append(ans)
return ans
#print(prime(12345678987766))
"""
def main():
q=inpu()
x = q
v1 = 0
v2 = 0
i = 2
while i * i <= q:
while q % i == 0:
if v1!=0:
v2 = i
else:
v1 = i
q //= i
i += 1
if q - 1!=0:
v2 = q
if v1 * v2 - x!=0:
print(1)
print(v1 * v2)
else:
print(2)
if __name__ == '__main__':
main()
"""
"""
def main():
l,r = sep()
a=[]
luckynum(0,l,r,a)
a.sort()
#print(a)
i=0
ans=0
l-=1
#print(a)
while(True):
if r>a[i]:
ans+=(a[i]*(a[i]-l))
l=a[i]
else:
ans+=(a[i]*(r-l))
break
i+=1
print(ans)
if __name__ == '__main__':
main()
"""
"""
def main():
sqrt = {i * i: i for i in range(1, 1000)}
#print(sqrt)
a, b = sep()
for y in range(1, a):
x2 = a * a - y * y
if x2 in sqrt:
x = sqrt[x2]
if b * y % a == 0 and b * x % a == 0 and b * x // a != y:
print('YES')
print(-x, y)
print(0, 0)
print(b * y // a, b * x // a)
exit()
print('NO')
if __name__ == '__main__':
main()
"""
def main():
m=inpu()
q=lis()
n=inpu()
arr=lis()
q=min(q)
arr.sort(reverse=True)
s=0
cnt=0
i=0
while(i<n):
cnt+=1
s+=arr[i]
#print(cnt,q)
if cnt==q:
i+=2
cnt=0
i+=1
print(s)
if __name__ == '__main__':
main()
``` | output | 1 | 49,133 | 10 | 98,267 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems.
There are m types of discounts. We assume that the discounts are indexed from 1 to m. To use the discount number i, the customer takes a special basket, where he puts exactly qi items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the qi items in the cart.
Maxim now needs to buy n items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well.
Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.
Input
The first line contains integer m (1 ≤ m ≤ 105) — the number of discount types. The second line contains m integers: q1, q2, ..., qm (1 ≤ qi ≤ 105).
The third line contains integer n (1 ≤ n ≤ 105) — the number of items Maxim needs. The fourth line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 104) — the items' prices.
The numbers in the lines are separated by single spaces.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1
2
4
50 50 100 100
Output
200
Input
2
2 3
5
50 50 50 50 50
Output
150
Input
1
1
7
1 1 1 1 1 1 1
Output
3
Note
In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200.
In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150. | instruction | 0 | 49,134 | 10 | 98,268 |
Tags: greedy, sortings
Correct Solution:
```
I=lambda:map(int,input().split())
m,q,n,a,r,k=int(input()),min(I()),int(input())-1,sorted(I()),0,0
while n>-1:
r+=a[n]
k+=1
if k==q:n-=3;k=0
else:n-=1
print(r)
``` | output | 1 | 49,134 | 10 | 98,269 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems.
There are m types of discounts. We assume that the discounts are indexed from 1 to m. To use the discount number i, the customer takes a special basket, where he puts exactly qi items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the qi items in the cart.
Maxim now needs to buy n items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well.
Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.
Input
The first line contains integer m (1 ≤ m ≤ 105) — the number of discount types. The second line contains m integers: q1, q2, ..., qm (1 ≤ qi ≤ 105).
The third line contains integer n (1 ≤ n ≤ 105) — the number of items Maxim needs. The fourth line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 104) — the items' prices.
The numbers in the lines are separated by single spaces.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1
2
4
50 50 100 100
Output
200
Input
2
2 3
5
50 50 50 50 50
Output
150
Input
1
1
7
1 1 1 1 1 1 1
Output
3
Note
In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200.
In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150. | instruction | 0 | 49,135 | 10 | 98,270 |
Tags: greedy, sortings
Correct Solution:
```
import sys
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
input = sys.stdin.readline
m = int(input())
q = list(map(int, input().split()))
n = int(input())
a = list(map(int, input().split()))
discount = min(q)
a.sort(reverse=True)
ans = 0
for i in range(0, n, discount+2):
ans += sum([a[i] for i in range(i, min(i+discount, n))])
print(ans)
``` | output | 1 | 49,135 | 10 | 98,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems.
There are m types of discounts. We assume that the discounts are indexed from 1 to m. To use the discount number i, the customer takes a special basket, where he puts exactly qi items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the qi items in the cart.
Maxim now needs to buy n items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well.
Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.
Input
The first line contains integer m (1 ≤ m ≤ 105) — the number of discount types. The second line contains m integers: q1, q2, ..., qm (1 ≤ qi ≤ 105).
The third line contains integer n (1 ≤ n ≤ 105) — the number of items Maxim needs. The fourth line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 104) — the items' prices.
The numbers in the lines are separated by single spaces.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1
2
4
50 50 100 100
Output
200
Input
2
2 3
5
50 50 50 50 50
Output
150
Input
1
1
7
1 1 1 1 1 1 1
Output
3
Note
In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200.
In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150.
Submitted Solution:
```
m=int(input())
q=list(map(int,input().split()))
n=int(input())
a=list(map(int,input().split()))
q.sort()
a.sort()
j=n-1
i=0
ans=0
while j>=0:
for i in range(q[0]):
ans=ans+a[j]
j=j-1
if j<0:
break
j=j-2
print(ans)
``` | instruction | 0 | 49,136 | 10 | 98,272 |
Yes | output | 1 | 49,136 | 10 | 98,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems.
There are m types of discounts. We assume that the discounts are indexed from 1 to m. To use the discount number i, the customer takes a special basket, where he puts exactly qi items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the qi items in the cart.
Maxim now needs to buy n items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well.
Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.
Input
The first line contains integer m (1 ≤ m ≤ 105) — the number of discount types. The second line contains m integers: q1, q2, ..., qm (1 ≤ qi ≤ 105).
The third line contains integer n (1 ≤ n ≤ 105) — the number of items Maxim needs. The fourth line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 104) — the items' prices.
The numbers in the lines are separated by single spaces.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1
2
4
50 50 100 100
Output
200
Input
2
2 3
5
50 50 50 50 50
Output
150
Input
1
1
7
1 1 1 1 1 1 1
Output
3
Note
In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200.
In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150.
Submitted Solution:
```
m=int(input())
q=list(map(int,input().split()))
n=int(input())
a=list(map(int,input().split()))
sk=min(q)
a.sort()
j=1
ans=0
for i in range(n-1,-1,-1):
l=sk-j
if l==-1:
pass
elif l==-2:
j=0
else:
ans+=a[i]
j+=1
print(ans)
``` | instruction | 0 | 49,137 | 10 | 98,274 |
Yes | output | 1 | 49,137 | 10 | 98,275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems.
There are m types of discounts. We assume that the discounts are indexed from 1 to m. To use the discount number i, the customer takes a special basket, where he puts exactly qi items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the qi items in the cart.
Maxim now needs to buy n items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well.
Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.
Input
The first line contains integer m (1 ≤ m ≤ 105) — the number of discount types. The second line contains m integers: q1, q2, ..., qm (1 ≤ qi ≤ 105).
The third line contains integer n (1 ≤ n ≤ 105) — the number of items Maxim needs. The fourth line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 104) — the items' prices.
The numbers in the lines are separated by single spaces.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1
2
4
50 50 100 100
Output
200
Input
2
2 3
5
50 50 50 50 50
Output
150
Input
1
1
7
1 1 1 1 1 1 1
Output
3
Note
In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200.
In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150.
Submitted Solution:
```
import sys
import math as mt
input=sys.stdin.buffer.readline
import math as mt
#t=int(input())
t=1
for __ in range(t):
#n,m=map(int,input().split())
m=int(input())
d=list(map(int,input().split()))
d.sort()
n=int(input())
p=list(map(int,input().split()))
p.sort(reverse=True)
ans=0
i=0
while i<n:
#print(111,i)
if (d[0]+i)<n:
for j in range(d[0]):
ans+=p[i]
i+=1
#print(222,i)
i+=1
else:
ans+=p[i]
i+=1
#print(i,ans)
print(ans)
``` | instruction | 0 | 49,138 | 10 | 98,276 |
Yes | output | 1 | 49,138 | 10 | 98,277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems.
There are m types of discounts. We assume that the discounts are indexed from 1 to m. To use the discount number i, the customer takes a special basket, where he puts exactly qi items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the qi items in the cart.
Maxim now needs to buy n items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well.
Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.
Input
The first line contains integer m (1 ≤ m ≤ 105) — the number of discount types. The second line contains m integers: q1, q2, ..., qm (1 ≤ qi ≤ 105).
The third line contains integer n (1 ≤ n ≤ 105) — the number of items Maxim needs. The fourth line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 104) — the items' prices.
The numbers in the lines are separated by single spaces.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1
2
4
50 50 100 100
Output
200
Input
2
2 3
5
50 50 50 50 50
Output
150
Input
1
1
7
1 1 1 1 1 1 1
Output
3
Note
In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200.
In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150.
Submitted Solution:
```
a=input()
b=sorted([int(i) for i in input().split()])[0]
c=int(input())
d=sorted([int(i) for i in input().split()])
p=0
while len(d):
for i in range(0, min(len(d), b)):
p+=d.pop()
for i in range(0, min(len(d), 2)):
a=d.pop()
print(p)
``` | instruction | 0 | 49,139 | 10 | 98,278 |
Yes | output | 1 | 49,139 | 10 | 98,279 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems.
There are m types of discounts. We assume that the discounts are indexed from 1 to m. To use the discount number i, the customer takes a special basket, where he puts exactly qi items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the qi items in the cart.
Maxim now needs to buy n items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well.
Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.
Input
The first line contains integer m (1 ≤ m ≤ 105) — the number of discount types. The second line contains m integers: q1, q2, ..., qm (1 ≤ qi ≤ 105).
The third line contains integer n (1 ≤ n ≤ 105) — the number of items Maxim needs. The fourth line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 104) — the items' prices.
The numbers in the lines are separated by single spaces.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1
2
4
50 50 100 100
Output
200
Input
2
2 3
5
50 50 50 50 50
Output
150
Input
1
1
7
1 1 1 1 1 1 1
Output
3
Note
In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200.
In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150.
Submitted Solution:
```
m1 = input()
m2 = input().split()
n1 = input()
n2 = input().split()
price = 0
a = int(n1)
k = True
sorted(n2)
sorted(m2)
while k:
for i in range(0,int(m2[0])):
if int(m2[0]) < a:
price += int(n2[a-1])
a -= 1
a -= 2
if a <= int(m2[0]):
for j in range(0, a):
price += int(n2[a-1])
a -= 1
k = False
print(price)
``` | instruction | 0 | 49,140 | 10 | 98,280 |
No | output | 1 | 49,140 | 10 | 98,281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems.
There are m types of discounts. We assume that the discounts are indexed from 1 to m. To use the discount number i, the customer takes a special basket, where he puts exactly qi items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the qi items in the cart.
Maxim now needs to buy n items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well.
Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.
Input
The first line contains integer m (1 ≤ m ≤ 105) — the number of discount types. The second line contains m integers: q1, q2, ..., qm (1 ≤ qi ≤ 105).
The third line contains integer n (1 ≤ n ≤ 105) — the number of items Maxim needs. The fourth line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 104) — the items' prices.
The numbers in the lines are separated by single spaces.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1
2
4
50 50 100 100
Output
200
Input
2
2 3
5
50 50 50 50 50
Output
150
Input
1
1
7
1 1 1 1 1 1 1
Output
3
Note
In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200.
In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150.
Submitted Solution:
```
from math import ceil
input()
d=[int(i) for i in input().split()]
c=int(input())
b=[int(i) for i in input().split()]
b.sort()
#d.sort()
take=2
p=min(d)
s=0
left=c%(take+p)
times=ceil(c/(take+p))
s+=sum(b[:left])
print()
for i in range(times):
s+=sum(b[i*(p+take)+left+take:i*(p+take)+left+i+p+take])
print(i*(p+take)+left+take,i*(p+take)+left+i+p+take)
print(s)
``` | instruction | 0 | 49,141 | 10 | 98,282 |
No | output | 1 | 49,141 | 10 | 98,283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems.
There are m types of discounts. We assume that the discounts are indexed from 1 to m. To use the discount number i, the customer takes a special basket, where he puts exactly qi items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the qi items in the cart.
Maxim now needs to buy n items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well.
Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.
Input
The first line contains integer m (1 ≤ m ≤ 105) — the number of discount types. The second line contains m integers: q1, q2, ..., qm (1 ≤ qi ≤ 105).
The third line contains integer n (1 ≤ n ≤ 105) — the number of items Maxim needs. The fourth line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 104) — the items' prices.
The numbers in the lines are separated by single spaces.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1
2
4
50 50 100 100
Output
200
Input
2
2 3
5
50 50 50 50 50
Output
150
Input
1
1
7
1 1 1 1 1 1 1
Output
3
Note
In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200.
In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150.
Submitted Solution:
```
import math
v = int(input())
nums = list(map(int,input().split()))
if min(nums) > v:
print(-1)
else:
l = math.floor(v/min(nums))
mv = None
for i in range(len(nums)):
if nums[i] == min(nums):
mv = i+1
ans = [""]*l
for i in range(l):
ans[i] = str(mv)
rem = v - min(nums)*l
start=0
while rem > 0:
cur=0
for i in range(len(nums)):
if i+1 > mv and rem+min(nums) >= nums[i]:
ans[start] = str(i+1)
cur = nums[i]
start+=1
rem -= cur
print("".join(ans))
``` | instruction | 0 | 49,142 | 10 | 98,284 |
No | output | 1 | 49,142 | 10 | 98,285 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems.
There are m types of discounts. We assume that the discounts are indexed from 1 to m. To use the discount number i, the customer takes a special basket, where he puts exactly qi items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the qi items in the cart.
Maxim now needs to buy n items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well.
Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.
Input
The first line contains integer m (1 ≤ m ≤ 105) — the number of discount types. The second line contains m integers: q1, q2, ..., qm (1 ≤ qi ≤ 105).
The third line contains integer n (1 ≤ n ≤ 105) — the number of items Maxim needs. The fourth line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 104) — the items' prices.
The numbers in the lines are separated by single spaces.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1
2
4
50 50 100 100
Output
200
Input
2
2 3
5
50 50 50 50 50
Output
150
Input
1
1
7
1 1 1 1 1 1 1
Output
3
Note
In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200.
In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
m=int(input())
b=list(map(int,input().split()))
k=max(a)
s=0
for i in range(m):
if(i<k):
d=max(b)
s+=d
b.remove(d)
print(s)
``` | instruction | 0 | 49,143 | 10 | 98,286 |
No | output | 1 | 49,143 | 10 | 98,287 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online.
The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order.
Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 ≤ j ≤ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer.
When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating.
Your task is to calculate the total time it takes for Ayush to process all the orders.
You can assume that the market has endless stock.
Input
The first line contains three integers n, m and k (1 ≤ n, k ≤ 100, 1 ≤ m ≤ k) — the number of users, the number of items each user wants to buy and the total number of items at the market.
The next line contains k distinct integers pl (1 ≤ pl ≤ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k.
Each of the next n lines contains m distinct integers aij (1 ≤ aij ≤ k) — the order of the i-th person.
Output
Print the only integer t — the total time needed for Ayush to process all the orders.
Example
Input
2 2 5
3 4 1 2 5
1 5
3 1
Output
14
Note
Customer 1 wants the items 1 and 5.
pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5].
pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2].
Time taken for the first customer is 3 + 5 = 8.
Customer 2 wants the items 3 and 1.
pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2].
pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2].
Time taken for the second customer is 3 + 3 = 6.
Total time is 8 + 6 = 14.
Formally pos(x) is the index of x in the current row. | instruction | 0 | 49,284 | 10 | 98,568 |
Tags: brute force
Correct Solution:
```
#import sys
#sys.stdin = open('in', 'r')
#n = int(input())
#a = [int(x) for x in input().split()]
n,m,k = map(int, input().split())
p = [int(x) for x in input().split()]
inds = [-1 for i in range(k + 1)]
for i in range(k):
inds[p[i]] = i
res = 0
for i in range(n):
a = [int(x) for x in input().split()]
for aij in a:
pos = inds[aij]
res += pos + 1
while pos > 0:
inds[p[pos-1]] += 1
p[pos] = p[pos-1]
pos -= 1
p[0] = aij
inds[aij]=0
print(res)
``` | output | 1 | 49,284 | 10 | 98,569 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online.
The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order.
Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 ≤ j ≤ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer.
When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating.
Your task is to calculate the total time it takes for Ayush to process all the orders.
You can assume that the market has endless stock.
Input
The first line contains three integers n, m and k (1 ≤ n, k ≤ 100, 1 ≤ m ≤ k) — the number of users, the number of items each user wants to buy and the total number of items at the market.
The next line contains k distinct integers pl (1 ≤ pl ≤ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k.
Each of the next n lines contains m distinct integers aij (1 ≤ aij ≤ k) — the order of the i-th person.
Output
Print the only integer t — the total time needed for Ayush to process all the orders.
Example
Input
2 2 5
3 4 1 2 5
1 5
3 1
Output
14
Note
Customer 1 wants the items 1 and 5.
pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5].
pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2].
Time taken for the first customer is 3 + 5 = 8.
Customer 2 wants the items 3 and 1.
pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2].
pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2].
Time taken for the second customer is 3 + 3 = 6.
Total time is 8 + 6 = 14.
Formally pos(x) is the index of x in the current row. | instruction | 0 | 49,285 | 10 | 98,570 |
Tags: brute force
Correct Solution:
```
n, m, k = map(int, input().split())
v = list(map(int, input().split()))
place = 0
s = 0
for l in range(n):
a = list(map(int, input().split()))
for i in a:
for j in range(k):
if v[j] == i:
place = j
break
s += place+1
while place > 0:
v[place] = v[place-1]
place -= 1
v[0] = i
print(s)
``` | output | 1 | 49,285 | 10 | 98,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online.
The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order.
Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 ≤ j ≤ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer.
When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating.
Your task is to calculate the total time it takes for Ayush to process all the orders.
You can assume that the market has endless stock.
Input
The first line contains three integers n, m and k (1 ≤ n, k ≤ 100, 1 ≤ m ≤ k) — the number of users, the number of items each user wants to buy and the total number of items at the market.
The next line contains k distinct integers pl (1 ≤ pl ≤ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k.
Each of the next n lines contains m distinct integers aij (1 ≤ aij ≤ k) — the order of the i-th person.
Output
Print the only integer t — the total time needed for Ayush to process all the orders.
Example
Input
2 2 5
3 4 1 2 5
1 5
3 1
Output
14
Note
Customer 1 wants the items 1 and 5.
pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5].
pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2].
Time taken for the first customer is 3 + 5 = 8.
Customer 2 wants the items 3 and 1.
pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2].
pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2].
Time taken for the second customer is 3 + 3 = 6.
Total time is 8 + 6 = 14.
Formally pos(x) is the index of x in the current row. | instruction | 0 | 49,286 | 10 | 98,572 |
Tags: brute force
Correct Solution:
```
[n,m,k]=[int(i) for i in input().split()]
a=[0]*(k+1)
temp=[int(i) for i in input().split()]
for i in range(0,k):
a[temp[i]]=i+1
answer=0
for z in range(0,n):
temp=[int(i) for i in input().split()]
for i in temp:
answer+=a[i]
for l in range(1,k+1):
if a[l]<a[i]: a[l]+=1
a[i]=1
print(answer)
``` | output | 1 | 49,286 | 10 | 98,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online.
The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order.
Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 ≤ j ≤ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer.
When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating.
Your task is to calculate the total time it takes for Ayush to process all the orders.
You can assume that the market has endless stock.
Input
The first line contains three integers n, m and k (1 ≤ n, k ≤ 100, 1 ≤ m ≤ k) — the number of users, the number of items each user wants to buy and the total number of items at the market.
The next line contains k distinct integers pl (1 ≤ pl ≤ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k.
Each of the next n lines contains m distinct integers aij (1 ≤ aij ≤ k) — the order of the i-th person.
Output
Print the only integer t — the total time needed for Ayush to process all the orders.
Example
Input
2 2 5
3 4 1 2 5
1 5
3 1
Output
14
Note
Customer 1 wants the items 1 and 5.
pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5].
pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2].
Time taken for the first customer is 3 + 5 = 8.
Customer 2 wants the items 3 and 1.
pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2].
pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2].
Time taken for the second customer is 3 + 3 = 6.
Total time is 8 + 6 = 14.
Formally pos(x) is the index of x in the current row. | instruction | 0 | 49,287 | 10 | 98,574 |
Tags: brute force
Correct Solution:
```
#### IMPORTANT LIBRARY ####
############################
### DO NOT USE import random --> 250ms to load the library
############################
### In case of extra libraries: https://github.com/cheran-senthil/PyRival
######################
####### IMPORT #######
######################
from functools import cmp_to_key
from collections import deque, Counter
from heapq import heappush, heappop
from math import log, ceil
######################
#### STANDARD I/O ####
######################
import sys
import os
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def print(*args, **kwargs):
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
def inp():
return sys.stdin.readline().rstrip("\r\n") # for fast input
def ii():
return int(inp())
def si():
return str(inp())
def li(lag = 0):
l = list(map(int, inp().split()))
if lag != 0:
for i in range(len(l)):
l[i] += lag
return l
def mi(lag = 0):
matrix = list()
for i in range(n):
matrix.append(li(lag))
return matrix
def lsi(): #string list
return list(map(str, inp().split()))
def print_list(lista, space = " "):
print(space.join(map(str, lista)))
######################
### BISECT METHODS ###
######################
def bisect_left(a, x):
"""i tale che a[i] >= x e a[i-1] < x"""
left = 0
right = len(a)
while left < right:
mid = (left+right)//2
if a[mid] < x:
left = mid+1
else:
right = mid
return left
def bisect_right(a, x):
"""i tale che a[i] > x e a[i-1] <= x"""
left = 0
right = len(a)
while left < right:
mid = (left+right)//2
if a[mid] > x:
right = mid
else:
left = mid+1
return left
def bisect_elements(a, x):
"""elementi pari a x nell'árray sortato"""
return bisect_right(a, x) - bisect_left(a, x)
######################
### MOD OPERATION ####
######################
MOD = 10**9 + 7
maxN = 5
FACT = [0] * maxN
INV_FACT = [0] * maxN
def add(x, y):
return (x+y) % MOD
def multiply(x, y):
return (x*y) % MOD
def power(x, y):
if y == 0:
return 1
elif y % 2:
return multiply(x, power(x, y-1))
else:
a = power(x, y//2)
return multiply(a, a)
def inverse(x):
return power(x, MOD-2)
def divide(x, y):
return multiply(x, inverse(y))
def allFactorials():
FACT[0] = 1
for i in range(1, maxN):
FACT[i] = multiply(i, FACT[i-1])
def inverseFactorials():
n = len(INV_FACT)
INV_FACT[n-1] = inverse(FACT[n-1])
for i in range(n-2, -1, -1):
INV_FACT[i] = multiply(INV_FACT[i+1], i+1)
def coeffBinom(n, k):
if n < k:
return 0
return multiply(FACT[n], multiply(INV_FACT[k], INV_FACT[n-k]))
######################
#### GRAPH ALGOS #####
######################
# ZERO BASED GRAPH
def create_graph(n, m, undirected = 1, unweighted = 1):
graph = [[] for i in range(n)]
if unweighted:
for i in range(m):
[x, y] = li(lag = -1)
graph[x].append(y)
if undirected:
graph[y].append(x)
else:
for i in range(m):
[x, y, w] = li(lag = -1)
w += 1
graph[x].append([y,w])
if undirected:
graph[y].append([x,w])
return graph
def create_tree(n, unweighted = 1):
children = [[] for i in range(n)]
if unweighted:
for i in range(n-1):
[x, y] = li(lag = -1)
children[x].append(y)
children[y].append(x)
else:
for i in range(n-1):
[x, y, w] = li(lag = -1)
w += 1
children[x].append([y, w])
children[y].append([x, w])
return children
def dist(tree, n, A, B = -1):
s = [[A, 0]]
massimo, massimo_nodo = 0, 0
distanza = -1
v = [-1] * n
while s:
el, dis = s.pop()
if dis > massimo:
massimo = dis
massimo_nodo = el
if el == B:
distanza = dis
for child in tree[el]:
if v[child] == -1:
v[child] = 1
s.append([child, dis+1])
return massimo, massimo_nodo, distanza
def diameter(tree):
_, foglia, _ = dist(tree, n, 0)
diam, _, _ = dist(tree, n, foglia)
return diam
def dfs(graph, n, A):
v = [-1] * n
s = [[A, 0]]
v[A] = 0
while s:
el, dis = s.pop()
for child in graph[el]:
if v[child] == -1:
v[child] = dis + 1
s.append([child, dis + 1])
return v #visited: -1 if not visited, otherwise v[B] is the distance in terms of edges
def bfs(graph, n, A):
v = [-1] * n
s = deque()
s.append([A, 0])
v[A] = 0
while s:
el, dis = s.popleft()
for child in graph[el]:
if v[child] == -1:
v[child] = dis + 1
s.append([child, dis + 1])
return v #visited: -1 if not visited, otherwise v[B] is the distance in terms of edges
#FROM A GIVEN ROOT, RECOVER THE STRUCTURE
def parents_children_root_unrooted_tree(tree, n, root = 0):
q = deque()
visited = [0] * n
parent = [-1] * n
children = [[] for i in range(n)]
q.append(root)
while q:
all_done = 1
visited[q[0]] = 1
for child in tree[q[0]]:
if not visited[child]:
all_done = 0
q.appendleft(child)
if all_done:
for child in tree[q[0]]:
if parent[child] == -1:
parent[q[0]] = child
children[child].append(q[0])
q.popleft()
return parent, children
# CALCULATING LONGEST PATH FOR ALL THE NODES
def all_longest_path_passing_from_node(parent, children, n):
q = deque()
visited = [len(children[i]) for i in range(n)]
downwards = [[0,0] for i in range(n)]
upward = [1] * n
longest_path = [1] * n
for i in range(n):
if not visited[i]:
q.append(i)
downwards[i] = [1,0]
while q:
node = q.popleft()
if parent[node] != -1:
visited[parent[node]] -= 1
if not visited[parent[node]]:
q.append(parent[node])
else:
root = node
for child in children[node]:
downwards[node] = sorted([downwards[node][0], downwards[node][1], downwards[child][0] + 1], reverse = True)[0:2]
s = [node]
while s:
node = s.pop()
if parent[node] != -1:
if downwards[parent[node]][0] == downwards[node][0] + 1:
upward[node] = 1 + max(upward[parent[node]], downwards[parent[node]][1])
else:
upward[node] = 1 + max(upward[parent[node]], downwards[parent[node]][0])
longest_path[node] = downwards[node][0] + downwards[node][1] + upward[node] - min([downwards[node][0], downwards[node][1], upward[node]]) - 1
for child in children[node]:
s.append(child)
return longest_path
### TBD SUCCESSOR GRAPH 7.5
### TBD TREE QUERIES 10.2 da 2 a 4
### TBD ADVANCED TREE 10.3
### TBD GRAPHS AND MATRICES 11.3.3 e 11.4.3 e 11.5.3 (ON GAMES)
######################
## END OF LIBRARIES ##
######################
n,m,k = li()
s = li()
o = []
for c in range(n):
t = li()
for e in t:
o.append(e)
cost = 0
for el in o:
for i, els in enumerate(s):
if els == el:
cost += i + 1
I = i
s = [s[I]] + s[:I] + s[I+1:]
print(cost)
``` | output | 1 | 49,287 | 10 | 98,575 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online.
The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order.
Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 ≤ j ≤ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer.
When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating.
Your task is to calculate the total time it takes for Ayush to process all the orders.
You can assume that the market has endless stock.
Input
The first line contains three integers n, m and k (1 ≤ n, k ≤ 100, 1 ≤ m ≤ k) — the number of users, the number of items each user wants to buy and the total number of items at the market.
The next line contains k distinct integers pl (1 ≤ pl ≤ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k.
Each of the next n lines contains m distinct integers aij (1 ≤ aij ≤ k) — the order of the i-th person.
Output
Print the only integer t — the total time needed for Ayush to process all the orders.
Example
Input
2 2 5
3 4 1 2 5
1 5
3 1
Output
14
Note
Customer 1 wants the items 1 and 5.
pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5].
pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2].
Time taken for the first customer is 3 + 5 = 8.
Customer 2 wants the items 3 and 1.
pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2].
pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2].
Time taken for the second customer is 3 + 3 = 6.
Total time is 8 + 6 = 14.
Formally pos(x) is the index of x in the current row. | instruction | 0 | 49,288 | 10 | 98,576 |
Tags: brute force
Correct Solution:
```
ln=input().split(" ")
n=int(ln[0])
m=int(ln[1])
k=int(ln[2])
x=0
pila=input().split(" ")
for i in range(n):
cus=input().split(" ")
for j in range(m):
ind=pila.index(cus[j])
x=x+ind+1
pila.insert(0,pila.pop(ind))
print(x)
``` | output | 1 | 49,288 | 10 | 98,577 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ayush is a cashier at the shopping center. Recently his department has started a ''click and collect" service which allows users to shop online.
The store contains k items. n customers have already used the above service. Each user paid for m items. Let aij denote the j-th item in the i-th person's order.
Due to the space limitations all the items are arranged in one single row. When Ayush receives the i-th order he will find one by one all the items aij (1 ≤ j ≤ m) in the row. Let pos(x) denote the position of the item x in the row at the moment of its collection. Then Ayush takes time equal to pos(ai1) + pos(ai2) + ... + pos(aim) for the i-th customer.
When Ayush accesses the x-th element he keeps a new stock in the front of the row and takes away the x-th element. Thus the values are updating.
Your task is to calculate the total time it takes for Ayush to process all the orders.
You can assume that the market has endless stock.
Input
The first line contains three integers n, m and k (1 ≤ n, k ≤ 100, 1 ≤ m ≤ k) — the number of users, the number of items each user wants to buy and the total number of items at the market.
The next line contains k distinct integers pl (1 ≤ pl ≤ k) denoting the initial positions of the items in the store. The items are numbered with integers from 1 to k.
Each of the next n lines contains m distinct integers aij (1 ≤ aij ≤ k) — the order of the i-th person.
Output
Print the only integer t — the total time needed for Ayush to process all the orders.
Example
Input
2 2 5
3 4 1 2 5
1 5
3 1
Output
14
Note
Customer 1 wants the items 1 and 5.
pos(1) = 3, so the new positions are: [1, 3, 4, 2, 5].
pos(5) = 5, so the new positions are: [5, 1, 3, 4, 2].
Time taken for the first customer is 3 + 5 = 8.
Customer 2 wants the items 3 and 1.
pos(3) = 3, so the new positions are: [3, 5, 1, 4, 2].
pos(1) = 3, so the new positions are: [1, 3, 5, 4, 2].
Time taken for the second customer is 3 + 3 = 6.
Total time is 8 + 6 = 14.
Formally pos(x) is the index of x in the current row. | instruction | 0 | 49,289 | 10 | 98,578 |
Tags: brute force
Correct Solution:
```
number = int(input().split()[0])
#orders = int(input())
array = input().split()
sum = 0
for i in range(0,number):
items = input().split()
for j in items:
buffer = []
sum = sum + int(array.index(j)) +1
buffer.append(j)
array.remove(j)
array = buffer + array
print(sum)
``` | output | 1 | 49,289 | 10 | 98,579 |
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