message stringlengths 2 30.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 237 109k | cluster float64 10 10 | __index_level_0__ int64 474 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for a rubles, a pack of two copybooks for b rubles, and a pack of three copybooks for c rubles. Alyona already has n copybooks.
What is the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.
Input
The only line contains 4 integers n, a, b, c (1 ≤ n, a, b, c ≤ 109).
Output
Print the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4.
Examples
Input
1 1 3 4
Output
3
Input
6 2 1 1
Output
1
Input
4 4 4 4
Output
0
Input
999999999 1000000000 1000000000 1000000000
Output
1000000000
Note
In the first example Alyona can buy 3 packs of 1 copybook for 3a = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.
In the second example Alyuna can buy a pack of 2 copybooks for b = 1 ruble. She will have 8 copybooks in total.
In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.
In the fourth example Alyona should buy one pack of one copybook. | instruction | 0 | 43,460 | 10 | 86,920 |
Tags: brute force, implementation
Correct Solution:
```
n,a,b,c=map(int,input().split())
xx=(n%4)
if xx==0:
print("0")
elif xx==1:
print(min(a*3,a+b,c))
elif xx==2:
print(min(2*a,b,2*c))
else:
print(min(a,b+c,3*c))
``` | output | 1 | 43,460 | 10 | 86,921 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for a rubles, a pack of two copybooks for b rubles, and a pack of three copybooks for c rubles. Alyona already has n copybooks.
What is the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.
Input
The only line contains 4 integers n, a, b, c (1 ≤ n, a, b, c ≤ 109).
Output
Print the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4.
Examples
Input
1 1 3 4
Output
3
Input
6 2 1 1
Output
1
Input
4 4 4 4
Output
0
Input
999999999 1000000000 1000000000 1000000000
Output
1000000000
Note
In the first example Alyona can buy 3 packs of 1 copybook for 3a = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.
In the second example Alyuna can buy a pack of 2 copybooks for b = 1 ruble. She will have 8 copybooks in total.
In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.
In the fourth example Alyona should buy one pack of one copybook. | instruction | 0 | 43,461 | 10 | 86,922 |
Tags: brute force, implementation
Correct Solution:
```
n,a,b,c=map(int,input().split())
if n%4==0:print(0)
else:
x = 4 - n%4
if x == 1:
print(min(a,b+c,c*3))
if x == 2:
print(min(a*2,b,c*2))
if x == 3:
print(min(a*3,a+b,c))
``` | output | 1 | 43,461 | 10 | 86,923 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for a rubles, a pack of two copybooks for b rubles, and a pack of three copybooks for c rubles. Alyona already has n copybooks.
What is the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.
Input
The only line contains 4 integers n, a, b, c (1 ≤ n, a, b, c ≤ 109).
Output
Print the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4.
Examples
Input
1 1 3 4
Output
3
Input
6 2 1 1
Output
1
Input
4 4 4 4
Output
0
Input
999999999 1000000000 1000000000 1000000000
Output
1000000000
Note
In the first example Alyona can buy 3 packs of 1 copybook for 3a = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.
In the second example Alyuna can buy a pack of 2 copybooks for b = 1 ruble. She will have 8 copybooks in total.
In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.
In the fourth example Alyona should buy one pack of one copybook. | instruction | 0 | 43,462 | 10 | 86,924 |
Tags: brute force, implementation
Correct Solution:
```
n, a, b, c = map(int, input().split())
if n % 4 == 0:
print(0)
if n % 4 == 1:
print(min(3 * a, b + a, c))
if n % 4 == 2:
print(min(2 * a, b, 2 * c))
if n % 4 == 3:
print(min(a, b + c, 3 * c))
``` | output | 1 | 43,462 | 10 | 86,925 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for a rubles, a pack of two copybooks for b rubles, and a pack of three copybooks for c rubles. Alyona already has n copybooks.
What is the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.
Input
The only line contains 4 integers n, a, b, c (1 ≤ n, a, b, c ≤ 109).
Output
Print the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4.
Examples
Input
1 1 3 4
Output
3
Input
6 2 1 1
Output
1
Input
4 4 4 4
Output
0
Input
999999999 1000000000 1000000000 1000000000
Output
1000000000
Note
In the first example Alyona can buy 3 packs of 1 copybook for 3a = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.
In the second example Alyuna can buy a pack of 2 copybooks for b = 1 ruble. She will have 8 copybooks in total.
In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.
In the fourth example Alyona should buy one pack of one copybook. | instruction | 0 | 43,463 | 10 | 86,926 |
Tags: brute force, implementation
Correct Solution:
```
line=[int(x) for x in input().split(' ')]
N=line[0]
a=line[1]
b=line[2]
c=line[3]
prices1=[0,a,b,c]
prices2=[a+b,a+c,b+c,a+b+c]
price=10**9
for i in range(4):
for t in range(1,4):
if (N+i*t)%4==0 and prices1[i]*t<price:
price=prices1[i]*t
for j in range(3,7):
for t in range(1,2):
if (N+j*t)%4==0 and prices2[j-3]*t<price:
price=prices2[j-3]
print(price)
``` | output | 1 | 43,463 | 10 | 86,927 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for a rubles, a pack of two copybooks for b rubles, and a pack of three copybooks for c rubles. Alyona already has n copybooks.
What is the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.
Input
The only line contains 4 integers n, a, b, c (1 ≤ n, a, b, c ≤ 109).
Output
Print the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4.
Examples
Input
1 1 3 4
Output
3
Input
6 2 1 1
Output
1
Input
4 4 4 4
Output
0
Input
999999999 1000000000 1000000000 1000000000
Output
1000000000
Note
In the first example Alyona can buy 3 packs of 1 copybook for 3a = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.
In the second example Alyuna can buy a pack of 2 copybooks for b = 1 ruble. She will have 8 copybooks in total.
In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.
In the fourth example Alyona should buy one pack of one copybook.
Submitted Solution:
```
n, a, b, c = map(int, input().strip().split())
cost = 0 if n%4 == 0 else a * (4 - n % 4)
for i in range(20):
for j in range(20):
for k in range(20):
if (n + i + 2*j + 3*k) % 4 == 0:
cost = min(cost, i*a + j*b + k*c)
print(cost)
``` | instruction | 0 | 43,464 | 10 | 86,928 |
Yes | output | 1 | 43,464 | 10 | 86,929 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for a rubles, a pack of two copybooks for b rubles, and a pack of three copybooks for c rubles. Alyona already has n copybooks.
What is the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.
Input
The only line contains 4 integers n, a, b, c (1 ≤ n, a, b, c ≤ 109).
Output
Print the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4.
Examples
Input
1 1 3 4
Output
3
Input
6 2 1 1
Output
1
Input
4 4 4 4
Output
0
Input
999999999 1000000000 1000000000 1000000000
Output
1000000000
Note
In the first example Alyona can buy 3 packs of 1 copybook for 3a = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.
In the second example Alyuna can buy a pack of 2 copybooks for b = 1 ruble. She will have 8 copybooks in total.
In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.
In the fourth example Alyona should buy one pack of one copybook.
Submitted Solution:
```
n, a, b, c = (int(x) for x in input().split())
nd = 4 - n % 4
if n % 4 == 0:
print(0)
elif nd == 1:
print(min(a, b + c, 3 * c, c + 3 * b))
elif nd == 2:
print(min(2 * a, b, 2 * c))
else:
print(min(3 * a, min(a + b, c)))
``` | instruction | 0 | 43,465 | 10 | 86,930 |
Yes | output | 1 | 43,465 | 10 | 86,931 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for a rubles, a pack of two copybooks for b rubles, and a pack of three copybooks for c rubles. Alyona already has n copybooks.
What is the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.
Input
The only line contains 4 integers n, a, b, c (1 ≤ n, a, b, c ≤ 109).
Output
Print the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4.
Examples
Input
1 1 3 4
Output
3
Input
6 2 1 1
Output
1
Input
4 4 4 4
Output
0
Input
999999999 1000000000 1000000000 1000000000
Output
1000000000
Note
In the first example Alyona can buy 3 packs of 1 copybook for 3a = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.
In the second example Alyuna can buy a pack of 2 copybooks for b = 1 ruble. She will have 8 copybooks in total.
In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.
In the fourth example Alyona should buy one pack of one copybook.
Submitted Solution:
```
def func(n):
if n[0]%4==1:return min(n[3],n[1]*3,n[2]+n[1])
elif n[0]%4==2:return min(n[2],n[3]*2,n[1]*2)
elif n[0]%4==3:return min(n[1],n[2]+n[3],n[3]*3)
else:return 0
print(func(list(map(int,input().split()))))
``` | instruction | 0 | 43,466 | 10 | 86,932 |
Yes | output | 1 | 43,466 | 10 | 86,933 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for a rubles, a pack of two copybooks for b rubles, and a pack of three copybooks for c rubles. Alyona already has n copybooks.
What is the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.
Input
The only line contains 4 integers n, a, b, c (1 ≤ n, a, b, c ≤ 109).
Output
Print the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4.
Examples
Input
1 1 3 4
Output
3
Input
6 2 1 1
Output
1
Input
4 4 4 4
Output
0
Input
999999999 1000000000 1000000000 1000000000
Output
1000000000
Note
In the first example Alyona can buy 3 packs of 1 copybook for 3a = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.
In the second example Alyuna can buy a pack of 2 copybooks for b = 1 ruble. She will have 8 copybooks in total.
In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.
In the fourth example Alyona should buy one pack of one copybook.
Submitted Solution:
```
import sys
INF = sys.maxsize
def solution(n, prices, k):
if k>10:
return INF
if n % 4 == 0:
return 0
else:
ans = min(solution(n + 1, prices,k+1) + prices[0],
solution(n + 2, prices,k+2) + prices[1],
solution(n + 3, prices,k+3) + prices[2])
return ans
def main():
n,a,b,c = map(int, input().split())
prices = [a,b,c]
print(solution(n,prices,0))
main()
``` | instruction | 0 | 43,467 | 10 | 86,934 |
Yes | output | 1 | 43,467 | 10 | 86,935 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for a rubles, a pack of two copybooks for b rubles, and a pack of three copybooks for c rubles. Alyona already has n copybooks.
What is the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.
Input
The only line contains 4 integers n, a, b, c (1 ≤ n, a, b, c ≤ 109).
Output
Print the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4.
Examples
Input
1 1 3 4
Output
3
Input
6 2 1 1
Output
1
Input
4 4 4 4
Output
0
Input
999999999 1000000000 1000000000 1000000000
Output
1000000000
Note
In the first example Alyona can buy 3 packs of 1 copybook for 3a = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.
In the second example Alyuna can buy a pack of 2 copybooks for b = 1 ruble. She will have 8 copybooks in total.
In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.
In the fourth example Alyona should buy one pack of one copybook.
Submitted Solution:
```
n,a,b,c=map(int,input().strip().split(" "))
if(n%4==1):
y=min(3*a,a+b,c)
print(y)
elif(n%4==2):
y=min(2*a,b)
print(y)
elif(n%4==0):
print(0)
else:
print(a)
``` | instruction | 0 | 43,468 | 10 | 86,936 |
No | output | 1 | 43,468 | 10 | 86,937 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for a rubles, a pack of two copybooks for b rubles, and a pack of three copybooks for c rubles. Alyona already has n copybooks.
What is the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.
Input
The only line contains 4 integers n, a, b, c (1 ≤ n, a, b, c ≤ 109).
Output
Print the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4.
Examples
Input
1 1 3 4
Output
3
Input
6 2 1 1
Output
1
Input
4 4 4 4
Output
0
Input
999999999 1000000000 1000000000 1000000000
Output
1000000000
Note
In the first example Alyona can buy 3 packs of 1 copybook for 3a = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.
In the second example Alyuna can buy a pack of 2 copybooks for b = 1 ruble. She will have 8 copybooks in total.
In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.
In the fourth example Alyona should buy one pack of one copybook.
Submitted Solution:
```
n, a, b, c = map(int, input().split())
mod = n % 4
if mod == 0:
print(0)
elif mod == 1:
print(min(3 * a, c))
elif mod == 2:
print(min(2 * a, b, 2 * c))
elif mod == 3:
print(min(a, 3 * c))
``` | instruction | 0 | 43,469 | 10 | 86,938 |
No | output | 1 | 43,469 | 10 | 86,939 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for a rubles, a pack of two copybooks for b rubles, and a pack of three copybooks for c rubles. Alyona already has n copybooks.
What is the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.
Input
The only line contains 4 integers n, a, b, c (1 ≤ n, a, b, c ≤ 109).
Output
Print the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4.
Examples
Input
1 1 3 4
Output
3
Input
6 2 1 1
Output
1
Input
4 4 4 4
Output
0
Input
999999999 1000000000 1000000000 1000000000
Output
1000000000
Note
In the first example Alyona can buy 3 packs of 1 copybook for 3a = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.
In the second example Alyuna can buy a pack of 2 copybooks for b = 1 ruble. She will have 8 copybooks in total.
In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.
In the fourth example Alyona should buy one pack of one copybook.
Submitted Solution:
```
n, a, b, c = (int(x) for x in input().split())
sum, k = 0, 4 - n % 4
if 2 * a > b:
sum = min(k // 2 * b + (k % 2) * a, 2 * c)
elif 3 * a > c and a + b > c:
sum = k // 3 * c + (k % 3) * a
if k % 3 == 0:
sum = min(sum, c * k // 3)
if k % 2 == 0:
sum = min(sum, b * k // 2)
print(min(sum, a * k))
``` | instruction | 0 | 43,470 | 10 | 86,940 |
No | output | 1 | 43,470 | 10 | 86,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little girl Alyona is in a shop to buy some copybooks for school. She study four subjects so she wants to have equal number of copybooks for each of the subjects. There are three types of copybook's packs in the shop: it is possible to buy one copybook for a rubles, a pack of two copybooks for b rubles, and a pack of three copybooks for c rubles. Alyona already has n copybooks.
What is the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4? There are infinitely many packs of any type in the shop. Alyona can buy packs of different type in the same purchase.
Input
The only line contains 4 integers n, a, b, c (1 ≤ n, a, b, c ≤ 109).
Output
Print the minimum amount of rubles she should pay to buy such number of copybooks k that n + k is divisible by 4.
Examples
Input
1 1 3 4
Output
3
Input
6 2 1 1
Output
1
Input
4 4 4 4
Output
0
Input
999999999 1000000000 1000000000 1000000000
Output
1000000000
Note
In the first example Alyona can buy 3 packs of 1 copybook for 3a = 3 rubles in total. After that she will have 4 copybooks which she can split between the subjects equally.
In the second example Alyuna can buy a pack of 2 copybooks for b = 1 ruble. She will have 8 copybooks in total.
In the third example Alyona can split the copybooks she already has between the 4 subject equally, so she doesn't need to buy anything.
In the fourth example Alyona should buy one pack of one copybook.
Submitted Solution:
```
def get_min_rubles(n, l):
if n % 4 == 0:
return 0
a, b, c = l[0], l[1], l[2]
k_a, price_a = 1, 0,
## xa, xa + b, xa + c
while(True):
if (n + k_a) % 4 == 0:
price_a = k_a * a
break
elif (n + k_a + 3) % 4 == 0:
price_a = k_a * a + b
break
elif (n + k_a + 4) % 4 == 0:
price_a = k_a * a + c
break
k_a += 1
## xb, xb + a, xb + c
k_b, price_b = 2, 0
while(True):
if (n + k_b) % 4 == 0:
price_b = (k_b //2) * b
break
elif (n + k_b + 1) % 4 == 0:
price_b = (k_b //2) * b + a
break
elif (n + k_b + 3) % 4 == 0:
price_b = (k_b //2) * b + c
break
k_b += 2
## xc, xc + a, xc + b
k_c, price_c = 3, 0
while(True):
if (n + k_c) % 4 == 0:
price_c = (k_c//3) * c
break
elif (n + k_c + 1) % 4 == 0:
price_c = (k_c//3) * c + a
break
elif (n + k_c + 2) % 4 == 0:
price_c = (k_c//3) * c + b
break
k_c += 3
return min(price_a, price_b, price_c)
prices = list(map(int, input().split()))
print(get_min_rubles(prices[0], prices[1:]))
``` | instruction | 0 | 43,471 | 10 | 86,942 |
No | output | 1 | 43,471 | 10 | 86,943 |
Provide a correct Python 3 solution for this coding contest problem.
You start with the number 0 and you want to reach the number N.
You can change the number, paying a certain amount of coins, with the following operations:
* Multiply the number by 2, paying A coins.
* Multiply the number by 3, paying B coins.
* Multiply the number by 5, paying C coins.
* Increase or decrease the number by 1, paying D coins.
You can perform these operations in arbitrary order and an arbitrary number of times.
What is the minimum number of coins you need to reach N?
You have to solve T testcases.
Constraints
* 1 \le T \le 10
* 1 \le N \le 10^{18}
* 1 \le A, B, C, D \le 10^9
* All numbers N, A, B, C, D are integers.
Input
The input is given from Standard Input. The first line of the input is
T
Then, T lines follow describing the T testcases. Each of the T lines has the format
N A B C D
Output
For each testcase, print the answer on Standard Output followed by a newline.
Example
Input
5
11 1 2 4 8
11 1 2 2 8
32 10 8 5 4
29384293847243 454353412 332423423 934923490 1
900000000000000000 332423423 454353412 934923490 987654321
Output
20
19
26
3821859835
23441258666 | instruction | 0 | 43,560 | 10 | 87,120 |
"Correct Solution:
```
from functools import lru_cache
t=int(input())
for i in range(t):
n,a,b,c,d=map(int,input().split())
@lru_cache(maxsize=None)
def f(n):
if n==0:
return 0
if n==1:
return d
else:
x=n*d
q,r=divmod(n,2)
if r==0:
x=min(x,f(q)+a)
else:
x=min(x,f(q)+a+d,f(q+1)+a+d)
q,r=divmod(n,3)
if r==0:
x=min(x,f(q)+b)
elif r==1:
x=min(x,f(q)+b+d,f(q+1)+b+d*2)
else:
x=min(x,f(q)+b+d*2,f(q+1)+b+d)
q,r=divmod(n,5)
if r==0:
x=min(x,f(q)+c)
elif r==1:
x=min(x,f(q)+c+d,f(q+1)+c+d*4)
elif r==2:
x=min(x,f(q)+c+d*2,f(q+1)+c+d*3)
elif r==3:
x=min(x,f(q)+c+d*3,f(q+1)+c+d*2)
else:
x=min(x,f(q)+c+d*4,f(q+1)+c+d)
return x
print(f(n))
``` | output | 1 | 43,560 | 10 | 87,121 |
Provide a correct Python 3 solution for this coding contest problem.
You start with the number 0 and you want to reach the number N.
You can change the number, paying a certain amount of coins, with the following operations:
* Multiply the number by 2, paying A coins.
* Multiply the number by 3, paying B coins.
* Multiply the number by 5, paying C coins.
* Increase or decrease the number by 1, paying D coins.
You can perform these operations in arbitrary order and an arbitrary number of times.
What is the minimum number of coins you need to reach N?
You have to solve T testcases.
Constraints
* 1 \le T \le 10
* 1 \le N \le 10^{18}
* 1 \le A, B, C, D \le 10^9
* All numbers N, A, B, C, D are integers.
Input
The input is given from Standard Input. The first line of the input is
T
Then, T lines follow describing the T testcases. Each of the T lines has the format
N A B C D
Output
For each testcase, print the answer on Standard Output followed by a newline.
Example
Input
5
11 1 2 4 8
11 1 2 2 8
32 10 8 5 4
29384293847243 454353412 332423423 934923490 1
900000000000000000 332423423 454353412 934923490 987654321
Output
20
19
26
3821859835
23441258666 | instruction | 0 | 43,561 | 10 | 87,122 |
"Correct Solution:
```
t = int(input())
def calc(x):
if x not in dp:
divs = [5, 3, 2]
costs = [c, b, a]
values = lambda div, cost: [calc((x + i) // div) + cost + (sign * x) % div * d for i, sign in [(0, 1), (div - 1, -1)]]
dp[x] = min(*[value for div, cost in zip(divs, costs) for value in values(div, cost)], x * d)
return dp[x]
for _ in range(t):
n, a, b, c, d = map(int, input().split())
dp = {0: 0, 1: d}
print(calc(n))
``` | output | 1 | 43,561 | 10 | 87,123 |
Provide a correct Python 3 solution for this coding contest problem.
You start with the number 0 and you want to reach the number N.
You can change the number, paying a certain amount of coins, with the following operations:
* Multiply the number by 2, paying A coins.
* Multiply the number by 3, paying B coins.
* Multiply the number by 5, paying C coins.
* Increase or decrease the number by 1, paying D coins.
You can perform these operations in arbitrary order and an arbitrary number of times.
What is the minimum number of coins you need to reach N?
You have to solve T testcases.
Constraints
* 1 \le T \le 10
* 1 \le N \le 10^{18}
* 1 \le A, B, C, D \le 10^9
* All numbers N, A, B, C, D are integers.
Input
The input is given from Standard Input. The first line of the input is
T
Then, T lines follow describing the T testcases. Each of the T lines has the format
N A B C D
Output
For each testcase, print the answer on Standard Output followed by a newline.
Example
Input
5
11 1 2 4 8
11 1 2 2 8
32 10 8 5 4
29384293847243 454353412 332423423 934923490 1
900000000000000000 332423423 454353412 934923490 987654321
Output
20
19
26
3821859835
23441258666 | instruction | 0 | 43,562 | 10 | 87,124 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(10000100)
t = int(input())
for _ in range(t):
n, a, b, c, d = map(int, input().split())
used = {}
def solve(t):
if t in used:
return used[t]
if t <= 1:
return t * d
res = d * t
for x, y in zip([2, 3, 5], [a, b, c]):
if t % x == 0:
res = min(res, solve(t // x) + y)
else:
res = min(res, solve(t // x) + y + (t - t // x * x) * d)
res = min(res, solve(t // x + 1) + y + ((t // x + 1) * x - t) * d)
used[t] = res
return res
print(solve(n))
``` | output | 1 | 43,562 | 10 | 87,125 |
Provide a correct Python 3 solution for this coding contest problem.
You start with the number 0 and you want to reach the number N.
You can change the number, paying a certain amount of coins, with the following operations:
* Multiply the number by 2, paying A coins.
* Multiply the number by 3, paying B coins.
* Multiply the number by 5, paying C coins.
* Increase or decrease the number by 1, paying D coins.
You can perform these operations in arbitrary order and an arbitrary number of times.
What is the minimum number of coins you need to reach N?
You have to solve T testcases.
Constraints
* 1 \le T \le 10
* 1 \le N \le 10^{18}
* 1 \le A, B, C, D \le 10^9
* All numbers N, A, B, C, D are integers.
Input
The input is given from Standard Input. The first line of the input is
T
Then, T lines follow describing the T testcases. Each of the T lines has the format
N A B C D
Output
For each testcase, print the answer on Standard Output followed by a newline.
Example
Input
5
11 1 2 4 8
11 1 2 2 8
32 10 8 5 4
29384293847243 454353412 332423423 934923490 1
900000000000000000 332423423 454353412 934923490 987654321
Output
20
19
26
3821859835
23441258666 | instruction | 0 | 43,563 | 10 | 87,126 |
"Correct Solution:
```
from functools import lru_cache
z = 0
n = 0
a = 0
b = 0
c = 0
d = 0
@lru_cache(maxsize=None)
def solve(x):
global a
global b
global c
global d
if x==0:
return 0
if x==1:
return d
ans = d*x
for p in [2,3,5]:
m = x%p
nxt = x//p
A = a
if p == 3:
A = b
if p == 5:
A = c
if nxt < x:
cst = m*d + min(A, (nxt*p-nxt)*d) + solve(nxt)
ans = min(ans, cst)
nxt = (x+p-m)//p
if nxt < x:
cst = min((p-m)*d + A + solve(nxt), (x-nxt)*d + solve(nxt))
ans = min(ans, cst)
#print(x, ans)
return ans
z = int(input())
for _ in range(z):
n, a, b, c, d = map(int, input().split())
print(solve(n))
solve.cache_clear()
``` | output | 1 | 43,563 | 10 | 87,127 |
Provide a correct Python 3 solution for this coding contest problem.
You start with the number 0 and you want to reach the number N.
You can change the number, paying a certain amount of coins, with the following operations:
* Multiply the number by 2, paying A coins.
* Multiply the number by 3, paying B coins.
* Multiply the number by 5, paying C coins.
* Increase or decrease the number by 1, paying D coins.
You can perform these operations in arbitrary order and an arbitrary number of times.
What is the minimum number of coins you need to reach N?
You have to solve T testcases.
Constraints
* 1 \le T \le 10
* 1 \le N \le 10^{18}
* 1 \le A, B, C, D \le 10^9
* All numbers N, A, B, C, D are integers.
Input
The input is given from Standard Input. The first line of the input is
T
Then, T lines follow describing the T testcases. Each of the T lines has the format
N A B C D
Output
For each testcase, print the answer on Standard Output followed by a newline.
Example
Input
5
11 1 2 4 8
11 1 2 2 8
32 10 8 5 4
29384293847243 454353412 332423423 934923490 1
900000000000000000 332423423 454353412 934923490 987654321
Output
20
19
26
3821859835
23441258666 | instruction | 0 | 43,564 | 10 | 87,128 |
"Correct Solution:
```
from functools import lru_cache
import sys
sys.setrecursionlimit(10**7)
T = int(input())
INF = 10**18
for _ in range(T):
N, A, B, C, D = map(int, input().split())
V = set()
@lru_cache(maxsize=None)
def search(now):
if now == 0:
return 0
ret = INF
for div, cost in zip((2, 3, 5), (A, B, C)):
q, r = divmod(now, div)
ret = min(
ret,
search(q) + cost + r * D,
search(q + 1) + cost + (div - r) * D if q + 1 < now else INF,
now * D
)
return ret
print(search(N))
``` | output | 1 | 43,564 | 10 | 87,129 |
Provide a correct Python 3 solution for this coding contest problem.
You start with the number 0 and you want to reach the number N.
You can change the number, paying a certain amount of coins, with the following operations:
* Multiply the number by 2, paying A coins.
* Multiply the number by 3, paying B coins.
* Multiply the number by 5, paying C coins.
* Increase or decrease the number by 1, paying D coins.
You can perform these operations in arbitrary order and an arbitrary number of times.
What is the minimum number of coins you need to reach N?
You have to solve T testcases.
Constraints
* 1 \le T \le 10
* 1 \le N \le 10^{18}
* 1 \le A, B, C, D \le 10^9
* All numbers N, A, B, C, D are integers.
Input
The input is given from Standard Input. The first line of the input is
T
Then, T lines follow describing the T testcases. Each of the T lines has the format
N A B C D
Output
For each testcase, print the answer on Standard Output followed by a newline.
Example
Input
5
11 1 2 4 8
11 1 2 2 8
32 10 8 5 4
29384293847243 454353412 332423423 934923490 1
900000000000000000 332423423 454353412 934923490 987654321
Output
20
19
26
3821859835
23441258666 | instruction | 0 | 43,565 | 10 | 87,130 |
"Correct Solution:
```
T = int(input())
for t in range(T):
N, A, B, C, D = map(int, input().split())
memo = {}
def dist(n):
if n == 0:
return 0
if n == 1:
return D
if n in memo:
return memo[n]
ret = min(
D * n,
D * abs(n - n//5*5) + C + dist(n//5),
D * abs(n - (n+4)//5*5) + C + dist((n+4)//5),
D * abs(n - n//3*3) + B + dist(n//3),
D * abs(n - (n+2)//3*3) + B + dist((n+2)//3),
D * abs(n - n//2*2) + A + dist(n//2),
D * abs(n - (n+1)//2*2) + A + dist((n+1)//2)
)
memo[n] = ret
return ret
print(dist(N))
``` | output | 1 | 43,565 | 10 | 87,131 |
Provide a correct Python 3 solution for this coding contest problem.
You start with the number 0 and you want to reach the number N.
You can change the number, paying a certain amount of coins, with the following operations:
* Multiply the number by 2, paying A coins.
* Multiply the number by 3, paying B coins.
* Multiply the number by 5, paying C coins.
* Increase or decrease the number by 1, paying D coins.
You can perform these operations in arbitrary order and an arbitrary number of times.
What is the minimum number of coins you need to reach N?
You have to solve T testcases.
Constraints
* 1 \le T \le 10
* 1 \le N \le 10^{18}
* 1 \le A, B, C, D \le 10^9
* All numbers N, A, B, C, D are integers.
Input
The input is given from Standard Input. The first line of the input is
T
Then, T lines follow describing the T testcases. Each of the T lines has the format
N A B C D
Output
For each testcase, print the answer on Standard Output followed by a newline.
Example
Input
5
11 1 2 4 8
11 1 2 2 8
32 10 8 5 4
29384293847243 454353412 332423423 934923490 1
900000000000000000 332423423 454353412 934923490 987654321
Output
20
19
26
3821859835
23441258666 | instruction | 0 | 43,566 | 10 | 87,132 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(10 ** 7)
input = sys.stdin.readline
def dfs(n):
if n not in memo:
ans = min(n*d, min(dfs(n//2),dfs((n+1)//2)) + a + d*(n % 2))
if n % 3 <= 1:
ans = min(ans, dfs(n//3) + b + d*(n % 3))
else:
ans = min(ans, dfs((n+2)//3) + b + d*(3 - n % 3))
if n % 5 <= 2:
ans = min(ans, dfs(n//5) + c + d*(n % 5))
else:
ans = min(ans, dfs((n+4)//5) + c + d*(5 - n % 5))
memo[n] = ans
return memo[n]
t = int(input())
for _ in range(t):
n, a, b, c, d = map(int, input().split())
memo = {0: 0, 1: d}
print(dfs(n))
``` | output | 1 | 43,566 | 10 | 87,133 |
Provide a correct Python 3 solution for this coding contest problem.
You start with the number 0 and you want to reach the number N.
You can change the number, paying a certain amount of coins, with the following operations:
* Multiply the number by 2, paying A coins.
* Multiply the number by 3, paying B coins.
* Multiply the number by 5, paying C coins.
* Increase or decrease the number by 1, paying D coins.
You can perform these operations in arbitrary order and an arbitrary number of times.
What is the minimum number of coins you need to reach N?
You have to solve T testcases.
Constraints
* 1 \le T \le 10
* 1 \le N \le 10^{18}
* 1 \le A, B, C, D \le 10^9
* All numbers N, A, B, C, D are integers.
Input
The input is given from Standard Input. The first line of the input is
T
Then, T lines follow describing the T testcases. Each of the T lines has the format
N A B C D
Output
For each testcase, print the answer on Standard Output followed by a newline.
Example
Input
5
11 1 2 4 8
11 1 2 2 8
32 10 8 5 4
29384293847243 454353412 332423423 934923490 1
900000000000000000 332423423 454353412 934923490 987654321
Output
20
19
26
3821859835
23441258666 | instruction | 0 | 43,567 | 10 | 87,134 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(10**9)
from collections import defaultdict
t = int(input())
for _ in range(t):
n,a,b,c,d = map(int,input().split())
dic = defaultdict(int)
dic[0] = 0
dic[1] = d
def calc(x):
if(x==0):
return 0
if(dic[x] != 0):
return dic[x]
tmp = x*d
if(x%2==0):
tmp = min(tmp, calc(x//2) + a)
else:
tmp = min(tmp, calc(x//2) + a + d, calc(x//2 + 1) + a + d)
if(x%3==0):
tmp = min(tmp, calc(x//3) + b)
elif(x%3==1):
tmp = min(tmp, calc(x//3) + b + d)
else:
tmp = min(tmp, calc(1 + x//3) + b + d)
if(x%5==0):
tmp = min(tmp, calc(x//5) + c)
elif(x%5 < 3):
tmp = min(tmp, calc(x//5) + c + d*(x%5))
else:
tmp = min(tmp, calc(1 + x//5) + c + d*(5 - x%5))
dic[x] = tmp
return tmp
print(calc(n))
``` | output | 1 | 43,567 | 10 | 87,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You start with the number 0 and you want to reach the number N.
You can change the number, paying a certain amount of coins, with the following operations:
* Multiply the number by 2, paying A coins.
* Multiply the number by 3, paying B coins.
* Multiply the number by 5, paying C coins.
* Increase or decrease the number by 1, paying D coins.
You can perform these operations in arbitrary order and an arbitrary number of times.
What is the minimum number of coins you need to reach N?
You have to solve T testcases.
Constraints
* 1 \le T \le 10
* 1 \le N \le 10^{18}
* 1 \le A, B, C, D \le 10^9
* All numbers N, A, B, C, D are integers.
Input
The input is given from Standard Input. The first line of the input is
T
Then, T lines follow describing the T testcases. Each of the T lines has the format
N A B C D
Output
For each testcase, print the answer on Standard Output followed by a newline.
Example
Input
5
11 1 2 4 8
11 1 2 2 8
32 10 8 5 4
29384293847243 454353412 332423423 934923490 1
900000000000000000 332423423 454353412 934923490 987654321
Output
20
19
26
3821859835
23441258666
Submitted Solution:
```
from functools import lru_cache
def solve(N, A, B, C, D):
@lru_cache(None)
def f(N):
if N == 0:
return 0
if N == 1:
return D
ret = D * N
q, r = divmod(N, 2)
if r == 0:
ret = min(ret, f(q) + A)
else:
ret = min(ret, f(q) + A + D, f(q + 1) + A + D)
q, r = divmod(N, 3)
if r == 0:
ret = min(ret, f(q) + B)
else:
ret = min(ret, f(q) + B + D*r, f(q + 1) + B + D*(3-r))
q, r = divmod(N, 5)
if r == 0:
ret = min(ret, f(q) + C)
else:
ret = min(ret, f(q) + C + D*r, f(q + 1) + C + D*(5-r))
return ret
return f(N)
T = int(input())
for _ in range(T):
N, A, B, C, D = map(int, input().split())
print(solve(N, A, B, C, D))
``` | instruction | 0 | 43,568 | 10 | 87,136 |
Yes | output | 1 | 43,568 | 10 | 87,137 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You start with the number 0 and you want to reach the number N.
You can change the number, paying a certain amount of coins, with the following operations:
* Multiply the number by 2, paying A coins.
* Multiply the number by 3, paying B coins.
* Multiply the number by 5, paying C coins.
* Increase or decrease the number by 1, paying D coins.
You can perform these operations in arbitrary order and an arbitrary number of times.
What is the minimum number of coins you need to reach N?
You have to solve T testcases.
Constraints
* 1 \le T \le 10
* 1 \le N \le 10^{18}
* 1 \le A, B, C, D \le 10^9
* All numbers N, A, B, C, D are integers.
Input
The input is given from Standard Input. The first line of the input is
T
Then, T lines follow describing the T testcases. Each of the T lines has the format
N A B C D
Output
For each testcase, print the answer on Standard Output followed by a newline.
Example
Input
5
11 1 2 4 8
11 1 2 2 8
32 10 8 5 4
29384293847243 454353412 332423423 934923490 1
900000000000000000 332423423 454353412 934923490 987654321
Output
20
19
26
3821859835
23441258666
Submitted Solution:
```
from functools import *
from math import *
t = int(input())
@lru_cache(maxsize=None)
def sol(n, a, b, c, d):
if n == 1: return d
ans = n * d
cn = (n + 1) // 2
fn = n // 2
if cn: ans = min(ans, sol(cn, a, b, c, d) + abs(cn * 2 - n) * d + a)
if fn: ans = min(ans, sol(fn, a, b, c, d) + abs(fn * 2 - n) * d + a)
cn = (n + 2) // 3
fn = n // 3
if cn: ans = min(ans, sol(cn, a, b, c, d) + abs(cn * 3 - n) * d + b)
if fn: ans = min(ans, sol(fn, a, b, c, d) + abs(fn * 3 - n) * d + b)
cn = (n + 4) // 5
fn = n // 5
if cn: ans = min(ans, sol(cn, a, b, c, d) + abs(cn * 5 - n) * d + c)
if fn: ans = min(ans, sol(fn, a, b, c, d) + abs(fn * 5 - n) * d + c)
# print(str(n) + ", " + str(ans))
return ans
for _ in range(t):
n, a, b, c, d = list(map(int, input().split()))
sol.cache_clear()
print(min(d * n, sol(n, a, b, c, d)))
``` | instruction | 0 | 43,569 | 10 | 87,138 |
Yes | output | 1 | 43,569 | 10 | 87,139 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You start with the number 0 and you want to reach the number N.
You can change the number, paying a certain amount of coins, with the following operations:
* Multiply the number by 2, paying A coins.
* Multiply the number by 3, paying B coins.
* Multiply the number by 5, paying C coins.
* Increase or decrease the number by 1, paying D coins.
You can perform these operations in arbitrary order and an arbitrary number of times.
What is the minimum number of coins you need to reach N?
You have to solve T testcases.
Constraints
* 1 \le T \le 10
* 1 \le N \le 10^{18}
* 1 \le A, B, C, D \le 10^9
* All numbers N, A, B, C, D are integers.
Input
The input is given from Standard Input. The first line of the input is
T
Then, T lines follow describing the T testcases. Each of the T lines has the format
N A B C D
Output
For each testcase, print the answer on Standard Output followed by a newline.
Example
Input
5
11 1 2 4 8
11 1 2 2 8
32 10 8 5 4
29384293847243 454353412 332423423 934923490 1
900000000000000000 332423423 454353412 934923490 987654321
Output
20
19
26
3821859835
23441258666
Submitted Solution:
```
import sys
from functools import lru_cache
sys.setrecursionlimit(10 ** 7)
rl = sys.stdin.readline
def solve():
N, A, B, C, D = map(int, rl().split())
@lru_cache(maxsize=None)
def calc(t):
if t == 0:
return 0
if t == 1:
return D
res = min(
D * t,
D * abs(t - t // 5 * 5) + C + calc(t // 5),
D * abs(t - (t + 4) // 5 * 5) + C + calc((t + 4) // 5),
D * abs(t - t // 3 * 3) + B + calc(t // 3),
D * abs(t - (t + 2) // 3 * 3) + B + calc((t + 2) // 3),
D * abs(t - t // 2 * 2) + A + calc(t // 2),
D * abs(t - (t + 1) // 2 * 2) + A + calc((t + 1) // 2)
)
return res
return calc(N)
if __name__ == '__main__':
T = int(rl())
ans = []
for _ in range(T):
ans.append(solve())
print(*ans, sep='\n')
``` | instruction | 0 | 43,570 | 10 | 87,140 |
Yes | output | 1 | 43,570 | 10 | 87,141 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You start with the number 0 and you want to reach the number N.
You can change the number, paying a certain amount of coins, with the following operations:
* Multiply the number by 2, paying A coins.
* Multiply the number by 3, paying B coins.
* Multiply the number by 5, paying C coins.
* Increase or decrease the number by 1, paying D coins.
You can perform these operations in arbitrary order and an arbitrary number of times.
What is the minimum number of coins you need to reach N?
You have to solve T testcases.
Constraints
* 1 \le T \le 10
* 1 \le N \le 10^{18}
* 1 \le A, B, C, D \le 10^9
* All numbers N, A, B, C, D are integers.
Input
The input is given from Standard Input. The first line of the input is
T
Then, T lines follow describing the T testcases. Each of the T lines has the format
N A B C D
Output
For each testcase, print the answer on Standard Output followed by a newline.
Example
Input
5
11 1 2 4 8
11 1 2 2 8
32 10 8 5 4
29384293847243 454353412 332423423 934923490 1
900000000000000000 332423423 454353412 934923490 987654321
Output
20
19
26
3821859835
23441258666
Submitted Solution:
```
import sys
sys.setrecursionlimit(2147483647)
from functools import lru_cache
@lru_cache(maxsize=None)
def solve(n, a, b, c, d):
if n == 0:
return 0
if n == 1:
return d
res = 10**18
res = min(res, d * n)
res = min(res, d * (n - n // 2 * 2) + a + solve(n // 2, a, b, c, d))
res = min(res, d * (n - n // 3 * 3) + b + solve(n // 3, a, b, c, d))
res = min(res, d * (n - n // 5 * 5) + c + solve(n // 5, a, b, c, d))
res = min(res, d * ((n + 2 - 1) // 2 * 2 - n) + a + solve((n + 2 - 1) // 2, a, b, c, d))
res = min(res, d * ((n + 3 - 1) // 3 * 3 - n) + b + solve((n + 3 - 1) // 3, a, b, c, d))
res = min(res, d * ((n + 5 - 1) // 5 * 5 - n) + c + solve((n + 5 - 1) // 5, a, b, c, d))
return res
T = int(input())
for _ in range(T):
N, A, B, C, D = map(int, input().split())
print(solve(N, A, B, C, D))
``` | instruction | 0 | 43,571 | 10 | 87,142 |
Yes | output | 1 | 43,571 | 10 | 87,143 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You start with the number 0 and you want to reach the number N.
You can change the number, paying a certain amount of coins, with the following operations:
* Multiply the number by 2, paying A coins.
* Multiply the number by 3, paying B coins.
* Multiply the number by 5, paying C coins.
* Increase or decrease the number by 1, paying D coins.
You can perform these operations in arbitrary order and an arbitrary number of times.
What is the minimum number of coins you need to reach N?
You have to solve T testcases.
Constraints
* 1 \le T \le 10
* 1 \le N \le 10^{18}
* 1 \le A, B, C, D \le 10^9
* All numbers N, A, B, C, D are integers.
Input
The input is given from Standard Input. The first line of the input is
T
Then, T lines follow describing the T testcases. Each of the T lines has the format
N A B C D
Output
For each testcase, print the answer on Standard Output followed by a newline.
Example
Input
5
11 1 2 4 8
11 1 2 2 8
32 10 8 5 4
29384293847243 454353412 332423423 934923490 1
900000000000000000 332423423 454353412 934923490 987654321
Output
20
19
26
3821859835
23441258666
Submitted Solution:
```
def dfs(n):
global b,check,d
#ttfは235、chaはabc
#2,3,5の場合それぞれ
for ttf,cha in zip([2,3,5],b):
#-1ずつしていく場合
if check[0]>check[n]+n*d:
check[0]=check[n]+n*d
#最寄りまで-してく場合
x1,x2=n//ttf,n%ttf
if x1 in check:
if check[x1]>check[n]+x2*d+cha:
check[x1]=check[n]+x2*d+cha
dfs(x1)
else:
check[x1]=check[n]+x2*d+cha
dfs(x1)
#最寄りまで+1してく場合
x1,x2=n//ttf+1,ttf-n%ttf
if x1 in check:
if check[x1]>check[n]+x2*d+cha:
check[x1]=check[n]+x2*d+cha
dfs(x1)
else:
check[x1]=check[n]+x2*d+cha
dfs(x1)
t=int(input())
for i in range(t):
N,*b,d=map(int,input().split())
check={N:0,0:N*d}
dfs(N)
print(check[0])
``` | instruction | 0 | 43,572 | 10 | 87,144 |
No | output | 1 | 43,572 | 10 | 87,145 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You start with the number 0 and you want to reach the number N.
You can change the number, paying a certain amount of coins, with the following operations:
* Multiply the number by 2, paying A coins.
* Multiply the number by 3, paying B coins.
* Multiply the number by 5, paying C coins.
* Increase or decrease the number by 1, paying D coins.
You can perform these operations in arbitrary order and an arbitrary number of times.
What is the minimum number of coins you need to reach N?
You have to solve T testcases.
Constraints
* 1 \le T \le 10
* 1 \le N \le 10^{18}
* 1 \le A, B, C, D \le 10^9
* All numbers N, A, B, C, D are integers.
Input
The input is given from Standard Input. The first line of the input is
T
Then, T lines follow describing the T testcases. Each of the T lines has the format
N A B C D
Output
For each testcase, print the answer on Standard Output followed by a newline.
Example
Input
5
11 1 2 4 8
11 1 2 2 8
32 10 8 5 4
29384293847243 454353412 332423423 934923490 1
900000000000000000 332423423 454353412 934923490 987654321
Output
20
19
26
3821859835
23441258666
Submitted Solution:
```
DP = {}
def solve(n, a, b, c, d, did_plus=False):
if n <= 0:
return 0
if n == 1:
return d
if n in DP:
return DP[n]
ans = float('inf')
for i in range(-4, 5):
if did_plus:
if i > 0:
continue
num = n + i
trash = abs(i)*d
if num%5 == 0 and num // 5 < n:
ans = min(ans, solve(num//5, a, b, c, d, i>=0) + min(c, d*(num-(num//5))) + trash)
if num%3 == 0 and num // 3 < n:
ans = min(ans, solve(num//3, a, b, c, d, i>=0) + min(b, d*(num-(num//3))) + trash)
if num%2 == 0 and num // 2 < n:
ans = min(ans, solve(num//2, a, b, c, d, i>=0) + min(a, d*(num-(num//2))) + trash)
# print(str(n) + ' ' + str(ans) + ' ' + str(num) + ' ' + str(i))
DP[n] = int(ans)
return int(ans)
def solve2():
DP.clear()
ans = float('inf')
n, a, b, c, d = map(int, input().split())
# for i in range(0, 65):
# for j in range(0, 30):
# for k in range(0, 10):
# val = pow(2, i)*pow(3, j)*pow(5, k)
# ans = min(ans, a*i + b*j + c*k + abs(n - val)*(d))
# print(ans)
print(solve(n, a, b, c,d))
t = int(input())
for _ in range(t):
solve2()
``` | instruction | 0 | 43,573 | 10 | 87,146 |
No | output | 1 | 43,573 | 10 | 87,147 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You start with the number 0 and you want to reach the number N.
You can change the number, paying a certain amount of coins, with the following operations:
* Multiply the number by 2, paying A coins.
* Multiply the number by 3, paying B coins.
* Multiply the number by 5, paying C coins.
* Increase or decrease the number by 1, paying D coins.
You can perform these operations in arbitrary order and an arbitrary number of times.
What is the minimum number of coins you need to reach N?
You have to solve T testcases.
Constraints
* 1 \le T \le 10
* 1 \le N \le 10^{18}
* 1 \le A, B, C, D \le 10^9
* All numbers N, A, B, C, D are integers.
Input
The input is given from Standard Input. The first line of the input is
T
Then, T lines follow describing the T testcases. Each of the T lines has the format
N A B C D
Output
For each testcase, print the answer on Standard Output followed by a newline.
Example
Input
5
11 1 2 4 8
11 1 2 2 8
32 10 8 5 4
29384293847243 454353412 332423423 934923490 1
900000000000000000 332423423 454353412 934923490 987654321
Output
20
19
26
3821859835
23441258666
Submitted Solution:
```
import sys
check=dict()
N,a,b,c,d=0,0,0,0,0
#ttfは235、chaはabc
def dfs_sub(n,ttf,cha):
global check,d
if 0 in check:
check[0]=min(check[n]+n*d,check[0])
else:
check[0]=check[n]+n*d
x1,x2=n//ttf,n%ttf
if ttf-x2>=x2:
if x1==0:
if x1 in check:
if check[x1]>check[n]+x2*d:
check[x1]=check[n]+x2*d
else:
check[x1]=check[n]+x2*d
else:
if x1 in check:
if check[x1]>check[n]+x2*d+cha:
check[x1]=check[n]+x2*d+cha
dfs(x1)
else:
check[x1]=check[n]+x2*d+cha
dfs(x1)
x1,x2=n//ttf+1,ttf-n%ttf
if ttf-x2>=x2:
if x1 in check:
if check[x1]>check[n]+x2*d+cha:
check[x1]=check[n]+x2*d+cha
dfs(x1)
else:
check[x1]=check[n]+x2*d+cha
dfs(x1)
def dfs(n):
global check,a,b,c,d
dfs_sub(n,2,a)
dfs_sub(n,3,b)
dfs_sub(n,5,c)
t=int(input())
ans=[0]*t
for i in range(t):
N,a,b,c,d=map(int,input().split())
check={N:0}
dfs(N)
ans[i]=check[0]
for i in range(t):
print(ans[i])
``` | instruction | 0 | 43,574 | 10 | 87,148 |
No | output | 1 | 43,574 | 10 | 87,149 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You start with the number 0 and you want to reach the number N.
You can change the number, paying a certain amount of coins, with the following operations:
* Multiply the number by 2, paying A coins.
* Multiply the number by 3, paying B coins.
* Multiply the number by 5, paying C coins.
* Increase or decrease the number by 1, paying D coins.
You can perform these operations in arbitrary order and an arbitrary number of times.
What is the minimum number of coins you need to reach N?
You have to solve T testcases.
Constraints
* 1 \le T \le 10
* 1 \le N \le 10^{18}
* 1 \le A, B, C, D \le 10^9
* All numbers N, A, B, C, D are integers.
Input
The input is given from Standard Input. The first line of the input is
T
Then, T lines follow describing the T testcases. Each of the T lines has the format
N A B C D
Output
For each testcase, print the answer on Standard Output followed by a newline.
Example
Input
5
11 1 2 4 8
11 1 2 2 8
32 10 8 5 4
29384293847243 454353412 332423423 934923490 1
900000000000000000 332423423 454353412 934923490 987654321
Output
20
19
26
3821859835
23441258666
Submitted Solution:
```
import heapq
from collections import deque
T = int(input())
for t in range(T):
N,A,B,C,D = map(int,input().split())
q = [(0,N)]
heapq.heapify(q)
seen = set()
while q:
cost, n = heapq.heappop(q)
if n in seen:
continue
seen.add(n)
if n == 0:
print(cost)
break
heapq.heappush(q, (cost + N * D, 0))
if n % 2 ==0:
heapq.heappush(q, (cost + A, n // 2))
else: # odd number
heapq.heappush(q, (cost + D + A, (n + 1) // 2))
heapq.heappush(q, (cost + D + A, (n - 1) // 2))
if n % 3 ==0:
heapq.heappush(q, (cost + B, n // 3))
else:
rest = n % 3
heapq.heappush(q, (cost + rest * D + B, (n - rest) // 3))
heapq.heappush(q, (cost + (3-rest) * D + B, (n - rest + 3) // 3))
if n % 5 ==0:
heapq.heappush(q, (cost + C, n // 5))
else:
rest = n % 5
heapq.heappush(q, (cost + rest * D + C, (n - rest) // 5))
heapq.heappush(q, (cost + (5-rest) * D + C, (n - rest + 5) // 5))
``` | instruction | 0 | 43,575 | 10 | 87,150 |
No | output | 1 | 43,575 | 10 | 87,151 |
Provide a correct Python 3 solution for this coding contest problem.
There are two buttons, one of size A and one of size B.
When you press a button of size X, you get X coins and the size of that button decreases by 1.
You will press a button twice. Here, you can press the same button twice, or press both buttons once.
At most how many coins can you get?
Constraints
* All values in input are integers.
* 3 \leq A, B \leq 20
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of coins you can get.
Examples
Input
5 3
Output
9
Input
3 4
Output
7
Input
6 6
Output
12 | instruction | 0 | 43,608 | 10 | 87,216 |
"Correct Solution:
```
a, b = map(int, input().split())
print(max([a+(a-1), a+b, b+(b-1)]))
``` | output | 1 | 43,608 | 10 | 87,217 |
Provide a correct Python 3 solution for this coding contest problem.
There are two buttons, one of size A and one of size B.
When you press a button of size X, you get X coins and the size of that button decreases by 1.
You will press a button twice. Here, you can press the same button twice, or press both buttons once.
At most how many coins can you get?
Constraints
* All values in input are integers.
* 3 \leq A, B \leq 20
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of coins you can get.
Examples
Input
5 3
Output
9
Input
3 4
Output
7
Input
6 6
Output
12 | instruction | 0 | 43,609 | 10 | 87,218 |
"Correct Solution:
```
A, B = map(int, input().split())
print(max(A+B, max(A, B)*2-1))
``` | output | 1 | 43,609 | 10 | 87,219 |
Provide a correct Python 3 solution for this coding contest problem.
There are two buttons, one of size A and one of size B.
When you press a button of size X, you get X coins and the size of that button decreases by 1.
You will press a button twice. Here, you can press the same button twice, or press both buttons once.
At most how many coins can you get?
Constraints
* All values in input are integers.
* 3 \leq A, B \leq 20
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of coins you can get.
Examples
Input
5 3
Output
9
Input
3 4
Output
7
Input
6 6
Output
12 | instruction | 0 | 43,610 | 10 | 87,220 |
"Correct Solution:
```
A,B= [int(x) for x in input().split()]
print(max(A+B,2*A-1,2*B-1))
``` | output | 1 | 43,610 | 10 | 87,221 |
Provide a correct Python 3 solution for this coding contest problem.
There are two buttons, one of size A and one of size B.
When you press a button of size X, you get X coins and the size of that button decreases by 1.
You will press a button twice. Here, you can press the same button twice, or press both buttons once.
At most how many coins can you get?
Constraints
* All values in input are integers.
* 3 \leq A, B \leq 20
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of coins you can get.
Examples
Input
5 3
Output
9
Input
3 4
Output
7
Input
6 6
Output
12 | instruction | 0 | 43,611 | 10 | 87,222 |
"Correct Solution:
```
a,b = map(int,input().split())
lst = [a+a-1,a+b,b+b-1]
print(max(lst))
``` | output | 1 | 43,611 | 10 | 87,223 |
Provide a correct Python 3 solution for this coding contest problem.
There are two buttons, one of size A and one of size B.
When you press a button of size X, you get X coins and the size of that button decreases by 1.
You will press a button twice. Here, you can press the same button twice, or press both buttons once.
At most how many coins can you get?
Constraints
* All values in input are integers.
* 3 \leq A, B \leq 20
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of coins you can get.
Examples
Input
5 3
Output
9
Input
3 4
Output
7
Input
6 6
Output
12 | instruction | 0 | 43,612 | 10 | 87,224 |
"Correct Solution:
```
a,b=list(map(int,input().split()))
print(max(a,b)*2-(a!=b))
``` | output | 1 | 43,612 | 10 | 87,225 |
Provide a correct Python 3 solution for this coding contest problem.
There are two buttons, one of size A and one of size B.
When you press a button of size X, you get X coins and the size of that button decreases by 1.
You will press a button twice. Here, you can press the same button twice, or press both buttons once.
At most how many coins can you get?
Constraints
* All values in input are integers.
* 3 \leq A, B \leq 20
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of coins you can get.
Examples
Input
5 3
Output
9
Input
3 4
Output
7
Input
6 6
Output
12 | instruction | 0 | 43,613 | 10 | 87,226 |
"Correct Solution:
```
#ABC124
a,b = map(int,input().split())
print(max(a+b,a+(a-1),b+(b-1)))
``` | output | 1 | 43,613 | 10 | 87,227 |
Provide a correct Python 3 solution for this coding contest problem.
There are two buttons, one of size A and one of size B.
When you press a button of size X, you get X coins and the size of that button decreases by 1.
You will press a button twice. Here, you can press the same button twice, or press both buttons once.
At most how many coins can you get?
Constraints
* All values in input are integers.
* 3 \leq A, B \leq 20
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of coins you can get.
Examples
Input
5 3
Output
9
Input
3 4
Output
7
Input
6 6
Output
12 | instruction | 0 | 43,614 | 10 | 87,228 |
"Correct Solution:
```
a, b = map(int, input().split())
print(max((a+b), (2*a-1), (2*b-1),))
``` | output | 1 | 43,614 | 10 | 87,229 |
Provide a correct Python 3 solution for this coding contest problem.
There are two buttons, one of size A and one of size B.
When you press a button of size X, you get X coins and the size of that button decreases by 1.
You will press a button twice. Here, you can press the same button twice, or press both buttons once.
At most how many coins can you get?
Constraints
* All values in input are integers.
* 3 \leq A, B \leq 20
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of coins you can get.
Examples
Input
5 3
Output
9
Input
3 4
Output
7
Input
6 6
Output
12 | instruction | 0 | 43,615 | 10 | 87,230 |
"Correct Solution:
```
a,b=(map(int,input().split()))
print(max(a+(a-1),a+b,b+(b-1)))
``` | output | 1 | 43,615 | 10 | 87,231 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two buttons, one of size A and one of size B.
When you press a button of size X, you get X coins and the size of that button decreases by 1.
You will press a button twice. Here, you can press the same button twice, or press both buttons once.
At most how many coins can you get?
Constraints
* All values in input are integers.
* 3 \leq A, B \leq 20
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of coins you can get.
Examples
Input
5 3
Output
9
Input
3 4
Output
7
Input
6 6
Output
12
Submitted Solution:
```
A,B=map(int,input().split())
print(sum(sorted([A,B,A-1,B-1])[-2:]))
``` | instruction | 0 | 43,616 | 10 | 87,232 |
Yes | output | 1 | 43,616 | 10 | 87,233 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two buttons, one of size A and one of size B.
When you press a button of size X, you get X coins and the size of that button decreases by 1.
You will press a button twice. Here, you can press the same button twice, or press both buttons once.
At most how many coins can you get?
Constraints
* All values in input are integers.
* 3 \leq A, B \leq 20
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of coins you can get.
Examples
Input
5 3
Output
9
Input
3 4
Output
7
Input
6 6
Output
12
Submitted Solution:
```
A,B=map(int,input().split())
print(max(A+B,(A*2)-1,(B*2)-1))
``` | instruction | 0 | 43,617 | 10 | 87,234 |
Yes | output | 1 | 43,617 | 10 | 87,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two buttons, one of size A and one of size B.
When you press a button of size X, you get X coins and the size of that button decreases by 1.
You will press a button twice. Here, you can press the same button twice, or press both buttons once.
At most how many coins can you get?
Constraints
* All values in input are integers.
* 3 \leq A, B \leq 20
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of coins you can get.
Examples
Input
5 3
Output
9
Input
3 4
Output
7
Input
6 6
Output
12
Submitted Solution:
```
A,B = list(map(int,input().split()))
print(max((A*2-1,A+B,B*2-1)))
``` | instruction | 0 | 43,618 | 10 | 87,236 |
Yes | output | 1 | 43,618 | 10 | 87,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two buttons, one of size A and one of size B.
When you press a button of size X, you get X coins and the size of that button decreases by 1.
You will press a button twice. Here, you can press the same button twice, or press both buttons once.
At most how many coins can you get?
Constraints
* All values in input are integers.
* 3 \leq A, B \leq 20
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of coins you can get.
Examples
Input
5 3
Output
9
Input
3 4
Output
7
Input
6 6
Output
12
Submitted Solution:
```
A,B =map(int,input().split())
l =[A,A-1,B,B-1]
l.sort()
print(l[2]+l[3])
``` | instruction | 0 | 43,619 | 10 | 87,238 |
Yes | output | 1 | 43,619 | 10 | 87,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two buttons, one of size A and one of size B.
When you press a button of size X, you get X coins and the size of that button decreases by 1.
You will press a button twice. Here, you can press the same button twice, or press both buttons once.
At most how many coins can you get?
Constraints
* All values in input are integers.
* 3 \leq A, B \leq 20
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of coins you can get.
Examples
Input
5 3
Output
9
Input
3 4
Output
7
Input
6 6
Output
12
Submitted Solution:
```
p, q = map(int, input().split())
score = max(p, q) * 2 - 1
print(score)
``` | instruction | 0 | 43,620 | 10 | 87,240 |
No | output | 1 | 43,620 | 10 | 87,241 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two buttons, one of size A and one of size B.
When you press a button of size X, you get X coins and the size of that button decreases by 1.
You will press a button twice. Here, you can press the same button twice, or press both buttons once.
At most how many coins can you get?
Constraints
* All values in input are integers.
* 3 \leq A, B \leq 20
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of coins you can get.
Examples
Input
5 3
Output
9
Input
3 4
Output
7
Input
6 6
Output
12
Submitted Solution:
```
def check_max_coin(a,b):
max_coin = 0
for i in range(1:3):
if a < b :
max_coin += b
else:
max_coin += a
print(max_coin)
a, b = map(int, input().split())
check_max_coin(a,b)
``` | instruction | 0 | 43,621 | 10 | 87,242 |
No | output | 1 | 43,621 | 10 | 87,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two buttons, one of size A and one of size B.
When you press a button of size X, you get X coins and the size of that button decreases by 1.
You will press a button twice. Here, you can press the same button twice, or press both buttons once.
At most how many coins can you get?
Constraints
* All values in input are integers.
* 3 \leq A, B \leq 20
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of coins you can get.
Examples
Input
5 3
Output
9
Input
3 4
Output
7
Input
6 6
Output
12
Submitted Solution:
```
a, b = map(int,input())
ma = max(2*a-1, a+b, 2*b-1)
print(ma)
``` | instruction | 0 | 43,622 | 10 | 87,244 |
No | output | 1 | 43,622 | 10 | 87,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are two buttons, one of size A and one of size B.
When you press a button of size X, you get X coins and the size of that button decreases by 1.
You will press a button twice. Here, you can press the same button twice, or press both buttons once.
At most how many coins can you get?
Constraints
* All values in input are integers.
* 3 \leq A, B \leq 20
Input
Input is given from Standard Input in the following format:
A B
Output
Print the maximum number of coins you can get.
Examples
Input
5 3
Output
9
Input
3 4
Output
7
Input
6 6
Output
12
Submitted Solution:
```
N,K = map(int,input().split())
s = list(map(int,input()))
s.append(2)
start = s[0]
choku = [0]
saka = [0]
number = 1
for i in range(N):
if s[i] == s[i+1]:
number += 1
else:
if s[i] == 0:
choku.append(number)
number = 1
else:
saka.append(number)
number = 1
choku.append(0)
saka.append(0)
while(1):
max_align = 0
m = len(choku)
n = len(saka)
if m <= K:
print(N)
break
if start == 0: #最初が直立の場合
for i in range(m-K+1):
if max_align <= sum(choku[i:i+K]) + sum(saka[i-1:i+K]):
max_align = sum(choku[i:i+K]) + sum(saka[i-1:i+K])
else:
for i in range(m-K+1):
if max_align <= sum(choku[i:i+K]) + sum(saka[i:i+K+1]):
max_align = sum(choku[i:i+K]) + sum(saka[i:i+K+1])
print(max_align)
break
``` | instruction | 0 | 43,623 | 10 | 87,246 |
No | output | 1 | 43,623 | 10 | 87,247 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
William is hosting a party for n of his trader friends. They started a discussion on various currencies they trade, but there's an issue: not all of his trader friends like every currency. They like some currencies, but not others.
For each William's friend i it is known whether he likes currency j. There are m currencies in total. It is also known that a trader may not like more than p currencies.
Because friends need to have some common topic for discussions they need to find the largest by cardinality (possibly empty) subset of currencies, such that there are at least ⌈ n/2 ⌉ friends (rounded up) who like each currency in this subset.
Input
The first line contains three integers n, m and p (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ p ≤ m ≤ 60, 1 ≤ p ≤ 15), which is the number of trader friends, the number of currencies, the maximum number of currencies each friend can like.
Each of the next n lines contain m characters. The j-th character of i-th line is 1 if friend i likes the currency j and 0 otherwise. It is guaranteed that the number of ones in each line does not exceed p.
Output
Print a string of length m, which defines the subset of currencies of the maximum size, which are liked by at least half of all friends. Currencies belonging to this subset must be signified by the character 1.
If there are multiple answers, print any.
Examples
Input
3 4 3
1000
0110
1001
Output
1000
Input
5 5 4
11001
10101
10010
01110
11011
Output
10001
Note
In the first sample test case only the first currency is liked by at least ⌈ 3/2 ⌉ = 2 friends, therefore it's easy to demonstrate that a better answer cannot be found.
In the second sample test case the answer includes 2 currencies and will be liked by friends 1, 2, and 5. For this test case there are other currencies that are liked by at least half of the friends, but using them we cannot achieve a larger subset size. | instruction | 0 | 45,624 | 10 | 91,248 |
Tags: bitmasks, brute force, dp, probabilities
Correct Solution:
```
import sys
import bisect
from bisect import bisect_left as lb
input_=lambda: sys.stdin.readline().strip("\r\n")
from math import log
from math import gcd
from math import atan2,acos
from random import randint
sa=lambda :input_()
sb=lambda:int(input_())
sc=lambda:input_().split()
sd=lambda:list(map(int,input_().split()))
sflo=lambda:list(map(float,input_().split()))
se=lambda:float(input_())
sf=lambda:list(input_())
flsh=lambda: sys.stdout.flush()
#sys.setrecursionlimit(10**6)
mod=10**9+7
mod1=998244353
gp=[]
cost=[]
dp=[]
mx=[]
ans1=[]
ans2=[]
special=[]
specnode=[]
a=0
kthpar=[]
def dfs(root,par):
if par!=-1:
dp[root]=dp[par]+1
for i in range(1,20):
if kthpar[root][i-1]!=-1:
kthpar[root][i]=kthpar[kthpar[root][i-1]][i-1]
for child in gp[root]:
if child==par:continue
kthpar[child][0]=root
dfs(child,root)
ans=0
b=[]
def hnbhai(tc):
n,m,p=sd()
a=[]
for i in range(n):
a.append(sa())
ans=""
mx=-1
for _ in range(40):
ind=(randint(0,10**6))%n
pos=[]
num=0
cnt=0
for i in range(m):
if a[ind][i]=='1':
pos.append(i)
num|=(1<<i)
cnt+=1
dp=[0]*(1<<cnt)
for i in a:
nn=0
for j in range(len(pos)):
if i[pos[j]]=='1':
nn|=1<<j
dp[nn]+=1
for i in range(len(pos)):
for j in range(1<<cnt):
if (j>>i)&1:
dp[j^(1<<i)]+=dp[j]
for i in range(1<<cnt):
if 2*dp[i]>=n and (bin(i)[2:]).count('1')>mx:
temp=["0"]*m
for j in range(len(pos)):
if (i>>j)&1:
temp[pos[j]]='1'
mx=(bin(i)[2:]).count('1')
ans="".join(temp)
print(ans)
for _ in range(1):
hnbhai(_+1)
``` | output | 1 | 45,624 | 10 | 91,249 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
William is hosting a party for n of his trader friends. They started a discussion on various currencies they trade, but there's an issue: not all of his trader friends like every currency. They like some currencies, but not others.
For each William's friend i it is known whether he likes currency j. There are m currencies in total. It is also known that a trader may not like more than p currencies.
Because friends need to have some common topic for discussions they need to find the largest by cardinality (possibly empty) subset of currencies, such that there are at least ⌈ n/2 ⌉ friends (rounded up) who like each currency in this subset.
Input
The first line contains three integers n, m and p (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ p ≤ m ≤ 60, 1 ≤ p ≤ 15), which is the number of trader friends, the number of currencies, the maximum number of currencies each friend can like.
Each of the next n lines contain m characters. The j-th character of i-th line is 1 if friend i likes the currency j and 0 otherwise. It is guaranteed that the number of ones in each line does not exceed p.
Output
Print a string of length m, which defines the subset of currencies of the maximum size, which are liked by at least half of all friends. Currencies belonging to this subset must be signified by the character 1.
If there are multiple answers, print any.
Examples
Input
3 4 3
1000
0110
1001
Output
1000
Input
5 5 4
11001
10101
10010
01110
11011
Output
10001
Note
In the first sample test case only the first currency is liked by at least ⌈ 3/2 ⌉ = 2 friends, therefore it's easy to demonstrate that a better answer cannot be found.
In the second sample test case the answer includes 2 currencies and will be liked by friends 1, 2, and 5. For this test case there are other currencies that are liked by at least half of the friends, but using them we cannot achieve a larger subset size. | instruction | 0 | 45,625 | 10 | 91,250 |
Tags: bitmasks, brute force, dp, probabilities
Correct Solution:
```
import random
# Python program to compute sum of ranges for different range queries
from collections import defaultdict
import math
import os
import sys
from io import BytesIO, IOBase
from types import GeneratorType
from collections import defaultdict
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
n, m, p = map(int, input().split())
b = []
for j in range(n):
s = input()
c = []
for j in s:
c.append(int(j))
b.append(c)
tot=0
ans=["0" for i in range(m)]
atleast=math.ceil(n/2)
for i in range(10):
ind = random.randint(-10 ** 18, 10 ** 18) % n
req = b[ind]
d1 = []
d = defaultdict(lambda:0)
l = 0
for j in range(m):
if req[j] == 1:
d[l] = j
d1.append(j)
l += 1
ma=1<<l
cnt = [0 for i in range(ma)]
for j in range(n):
req = 0
c = 0
for k in d1:
if b[j][k] == 1:
req += (1<<c)
c += 1
cnt[req]+=1
for i in range(l):
for mask in range(ma):
if mask&(1<<i):
cnt[mask^(1<<i)]+=cnt[mask]
for mask in range(ma):
if cnt[mask]>=atleast:
req=["0" for k in range(m)]
total=0
for u in range(16):
if (mask & (1<<u)):
req[d[u]]="1"
total+=1
if total>tot:
tot=total
ans=req
print("".join(ans))
``` | output | 1 | 45,625 | 10 | 91,251 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
William is hosting a party for n of his trader friends. They started a discussion on various currencies they trade, but there's an issue: not all of his trader friends like every currency. They like some currencies, but not others.
For each William's friend i it is known whether he likes currency j. There are m currencies in total. It is also known that a trader may not like more than p currencies.
Because friends need to have some common topic for discussions they need to find the largest by cardinality (possibly empty) subset of currencies, such that there are at least ⌈ n/2 ⌉ friends (rounded up) who like each currency in this subset.
Input
The first line contains three integers n, m and p (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ p ≤ m ≤ 60, 1 ≤ p ≤ 15), which is the number of trader friends, the number of currencies, the maximum number of currencies each friend can like.
Each of the next n lines contain m characters. The j-th character of i-th line is 1 if friend i likes the currency j and 0 otherwise. It is guaranteed that the number of ones in each line does not exceed p.
Output
Print a string of length m, which defines the subset of currencies of the maximum size, which are liked by at least half of all friends. Currencies belonging to this subset must be signified by the character 1.
If there are multiple answers, print any.
Examples
Input
3 4 3
1000
0110
1001
Output
1000
Input
5 5 4
11001
10101
10010
01110
11011
Output
10001
Note
In the first sample test case only the first currency is liked by at least ⌈ 3/2 ⌉ = 2 friends, therefore it's easy to demonstrate that a better answer cannot be found.
In the second sample test case the answer includes 2 currencies and will be liked by friends 1, 2, and 5. For this test case there are other currencies that are liked by at least half of the friends, but using them we cannot achieve a larger subset size. | instruction | 0 | 45,626 | 10 | 91,252 |
Tags: bitmasks, brute force, dp, probabilities
Correct Solution:
```
import random
from math import ceil
import os
import sys
from io import BytesIO, IOBase
# 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")
n,m,p = map(int,input().split())
A = []
for _ in range(n):
A.append(input())
num_iter = 20
mx_bit_cnt = 0
ans = None
def solve(i):
global n,m,p,mx_bit_cnt,ans
I = [idx for idx in range(m) if A[i][idx] == "1"]
L = len(I)
C = [0]*(1<<L)
for j in range(n):
cur = 0
for l in range(L):
if A[j][I[l]] == "1":
cur += (1<<l)
C[cur] += 1
for i in range(L):
for mask in range((1<<L)-1,-1,-1):
if mask & (1<<i):
C[mask^(1<<i)] += C[mask]
for mask in range(1<<L):
bit_cnt = bin(mask).count("1")
if C[mask] < ceil(n/2):
continue
if bit_cnt >= mx_bit_cnt:
mx_bit_cnt = bit_cnt
temp = ["0"]*m
for l in range(L):
if mask & (1<<l):
temp[I[l]] = "1"
ans = "".join(temp)
while num_iter:
num_iter -= 1
i = random.randint(0,n-1)
solve(i)
print(ans)
``` | output | 1 | 45,626 | 10 | 91,253 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
William is hosting a party for n of his trader friends. They started a discussion on various currencies they trade, but there's an issue: not all of his trader friends like every currency. They like some currencies, but not others.
For each William's friend i it is known whether he likes currency j. There are m currencies in total. It is also known that a trader may not like more than p currencies.
Because friends need to have some common topic for discussions they need to find the largest by cardinality (possibly empty) subset of currencies, such that there are at least ⌈ n/2 ⌉ friends (rounded up) who like each currency in this subset.
Input
The first line contains three integers n, m and p (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ p ≤ m ≤ 60, 1 ≤ p ≤ 15), which is the number of trader friends, the number of currencies, the maximum number of currencies each friend can like.
Each of the next n lines contain m characters. The j-th character of i-th line is 1 if friend i likes the currency j and 0 otherwise. It is guaranteed that the number of ones in each line does not exceed p.
Output
Print a string of length m, which defines the subset of currencies of the maximum size, which are liked by at least half of all friends. Currencies belonging to this subset must be signified by the character 1.
If there are multiple answers, print any.
Examples
Input
3 4 3
1000
0110
1001
Output
1000
Input
5 5 4
11001
10101
10010
01110
11011
Output
10001
Note
In the first sample test case only the first currency is liked by at least ⌈ 3/2 ⌉ = 2 friends, therefore it's easy to demonstrate that a better answer cannot be found.
In the second sample test case the answer includes 2 currencies and will be liked by friends 1, 2, and 5. For this test case there are other currencies that are liked by at least half of the friends, but using them we cannot achieve a larger subset size. | instruction | 0 | 45,627 | 10 | 91,254 |
Tags: bitmasks, brute force, dp, probabilities
Correct Solution:
```
import random
import sys
from sys import stdin
def popcnt(x):
ret = 0
while x > 0:
ret += x % 2
x //= 2
return ret
n,m,p = map(int,stdin.readline().split())
s = [stdin.readline()[:-1] for i in range(n)]
ans = 0
anslis = ["0"] * m
for loop in range(20):
v = random.randint(0,n-1)
vbits = []
for i in range(m):
if s[v][i] == "1":
vbits.append(i)
dp = [0] * (2**len(vbits))
vbits.reverse()
for i in range(n):
now = 0
for j in vbits:
now *= 2
if s[i][j] == "1":
now += 1
dp[now] += 1
vbits.reverse()
#print (v,dp)
for j in range(len(vbits)):
for i in range(len(dp)):
if i & 2**j == 0:
dp[i] += dp[i | 2**j]
#print (v,dp)
for i in range(len(dp)):
if dp[i] >= (n+1)//2:
pc = popcnt(i)
if ans < pc:
#print ("change!")
ans = pc
anslis = [0] * m
for j in range(len(vbits)):
if 2**j & i > 0:
anslis[vbits[j]] = 1
print ("".join(map(str,anslis)))
``` | output | 1 | 45,627 | 10 | 91,255 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
William is hosting a party for n of his trader friends. They started a discussion on various currencies they trade, but there's an issue: not all of his trader friends like every currency. They like some currencies, but not others.
For each William's friend i it is known whether he likes currency j. There are m currencies in total. It is also known that a trader may not like more than p currencies.
Because friends need to have some common topic for discussions they need to find the largest by cardinality (possibly empty) subset of currencies, such that there are at least ⌈ n/2 ⌉ friends (rounded up) who like each currency in this subset.
Input
The first line contains three integers n, m and p (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ p ≤ m ≤ 60, 1 ≤ p ≤ 15), which is the number of trader friends, the number of currencies, the maximum number of currencies each friend can like.
Each of the next n lines contain m characters. The j-th character of i-th line is 1 if friend i likes the currency j and 0 otherwise. It is guaranteed that the number of ones in each line does not exceed p.
Output
Print a string of length m, which defines the subset of currencies of the maximum size, which are liked by at least half of all friends. Currencies belonging to this subset must be signified by the character 1.
If there are multiple answers, print any.
Examples
Input
3 4 3
1000
0110
1001
Output
1000
Input
5 5 4
11001
10101
10010
01110
11011
Output
10001
Note
In the first sample test case only the first currency is liked by at least ⌈ 3/2 ⌉ = 2 friends, therefore it's easy to demonstrate that a better answer cannot be found.
In the second sample test case the answer includes 2 currencies and will be liked by friends 1, 2, and 5. For this test case there are other currencies that are liked by at least half of the friends, but using them we cannot achieve a larger subset size. | instruction | 0 | 45,628 | 10 | 91,256 |
Tags: bitmasks, brute force, dp, probabilities
Correct Solution:
```
import random
import sys
from sys import stdin
def popcnt(x):
ret = 0
while x > 0:
ret += x % 2
x //= 2
return ret
n,m,p = map(int,stdin.readline().split())
s = [stdin.readline()[:-1] for i in range(n)]
ans = 0
anslis = ["0"] * m
for loop in range(8):
v = random.randint(0,n-1)
vbits = []
for i in range(m):
if s[v][i] == "1":
vbits.append(i)
dp = [0] * (2**len(vbits))
vbits.reverse()
for i in range(n):
now = 0
for j in vbits:
now *= 2
if s[i][j] == "1":
now += 1
dp[now] += 1
vbits.reverse()
#print (v,dp)
for j in range(len(vbits)):
for i in range(len(dp)):
if i & 2**j == 0:
dp[i] += dp[i | 2**j]
#print (v,dp)
for i in range(len(dp)):
if dp[i] >= (n+1)//2:
pc = popcnt(i)
if ans < pc:
#print ("change!")
ans = pc
anslis = [0] * m
for j in range(len(vbits)):
if 2**j & i > 0:
anslis[vbits[j]] = 1
print ("".join(map(str,anslis)))
``` | output | 1 | 45,628 | 10 | 91,257 |
Provide tags and a correct Python 3 solution for this coding contest problem.
<image>
William is hosting a party for n of his trader friends. They started a discussion on various currencies they trade, but there's an issue: not all of his trader friends like every currency. They like some currencies, but not others.
For each William's friend i it is known whether he likes currency j. There are m currencies in total. It is also known that a trader may not like more than p currencies.
Because friends need to have some common topic for discussions they need to find the largest by cardinality (possibly empty) subset of currencies, such that there are at least ⌈ n/2 ⌉ friends (rounded up) who like each currency in this subset.
Input
The first line contains three integers n, m and p (1 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ p ≤ m ≤ 60, 1 ≤ p ≤ 15), which is the number of trader friends, the number of currencies, the maximum number of currencies each friend can like.
Each of the next n lines contain m characters. The j-th character of i-th line is 1 if friend i likes the currency j and 0 otherwise. It is guaranteed that the number of ones in each line does not exceed p.
Output
Print a string of length m, which defines the subset of currencies of the maximum size, which are liked by at least half of all friends. Currencies belonging to this subset must be signified by the character 1.
If there are multiple answers, print any.
Examples
Input
3 4 3
1000
0110
1001
Output
1000
Input
5 5 4
11001
10101
10010
01110
11011
Output
10001
Note
In the first sample test case only the first currency is liked by at least ⌈ 3/2 ⌉ = 2 friends, therefore it's easy to demonstrate that a better answer cannot be found.
In the second sample test case the answer includes 2 currencies and will be liked by friends 1, 2, and 5. For this test case there are other currencies that are liked by at least half of the friends, but using them we cannot achieve a larger subset size. | instruction | 0 | 45,629 | 10 | 91,258 |
Tags: bitmasks, brute force, dp, probabilities
Correct Solution:
```
import sys
import bisect
from bisect import bisect_left as lb
input_=lambda: sys.stdin.readline().strip("\r\n")
from math import log
from math import gcd
from math import atan2,acos
from random import randint
sa=lambda :input_()
sb=lambda:int(input_())
sc=lambda:input_().split()
sd=lambda:list(map(int,input_().split()))
sflo=lambda:list(map(float,input_().split()))
se=lambda:float(input_())
sf=lambda:list(input_())
flsh=lambda: sys.stdout.flush()
#sys.setrecursionlimit(10**6)
mod=10**9+7
mod1=998244353
gp=[]
cost=[]
dp=[]
mx=[]
ans1=[]
ans2=[]
special=[]
specnode=[]
a=0
kthpar=[]
def dfs(root,par):
if par!=-1:
dp[root]=dp[par]+1
for i in range(1,20):
if kthpar[root][i-1]!=-1:
kthpar[root][i]=kthpar[kthpar[root][i-1]][i-1]
for child in gp[root]:
if child==par:continue
kthpar[child][0]=root
dfs(child,root)
ans=0
b=[]
def hnbhai(tc):
n,m,p=sd()
a=[]
for i in range(n):
a.append(sa())
ans=""
mx=-1
for _ in range(40):
ind=randint(0,n-1)
pos=[]
num=0
cnt=0
for i in range(m):
if a[ind][i]=='1':
pos.append(i)
num|=(1<<i)
cnt+=1
dp=[0]*(1<<cnt)
for i in a:
nn=0
for j in range(len(pos)):
if i[pos[j]]=='1':
nn|=1<<j
dp[nn]+=1
for i in range(len(pos)):
for j in range(1<<cnt):
if (j>>i)&1:
dp[j^(1<<i)]+=dp[j]
for i in range(1<<cnt):
if 2*dp[i]>=n and (bin(i)[2:]).count('1')>mx:
temp=["0"]*m
for j in range(len(pos)):
if (i>>j)&1:
temp[pos[j]]='1'
mx=(bin(i)[2:]).count('1')
ans="".join(temp)
print(ans)
for _ in range(1):
hnbhai(_+1)
``` | output | 1 | 45,629 | 10 | 91,259 |
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