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.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 100) — the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it — the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Submitted Solution:
```
q = int(input())
for i in range(q):
n = int(input())
a = list(map(int, input().split()))
s = sum(a)
print(s / n + (s % n != 0))
``` | instruction | 0 | 13,408 | 10 | 26,816 |
No | output | 1 | 13,408 | 10 | 26,817 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 100) — the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it — the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Submitted Solution:
```
t=int(input())
#l=list()
min1=99999
for q in range(t):
n=int(input())
a=list(map(int,input().split()))
s=sum(a)
for i in range(1,max(a)+1):
if(n*i>=s):
if(min1>i):
min1=i
print(min1)
#print(l)
#x=len(l)
#print(l[x])
``` | instruction | 0 | 13,409 | 10 | 26,818 |
No | output | 1 | 13,409 | 10 | 26,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 100) — the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 ≤ n ≤ 100) — the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it — the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Submitted Solution:
```
import math
quer = input()
quer = int(quer)
for _ in range(1, quer+1):
num = input()
num = int(num)
inp_lst = [int(i) for i in input().split()]
inp_lst.sort()
avg = math.ceil(sum(inp_lst) / num)
#if avg. not present in list, find the immediate larger one than this
ans_lst = sorted(i for i in inp_lst if i >= avg)
print(ans_lst[0])
``` | instruction | 0 | 13,410 | 10 | 26,820 |
No | output | 1 | 13,410 | 10 | 26,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to n (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor of n (not equal to n, of course). For example, if n = 6 then Funt has to pay 3 burles, while for n = 25 he needs to pay 5 and if n = 2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial n in several parts n1 + n2 + ... + nk = n (here k is arbitrary, even k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition ni ≥ 2 should hold for all i from 1 to k.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split n in parts.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.
Output
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
Examples
Input
4
Output
2
Input
27
Output
3
Submitted Solution:
```
n = int(input())
if n == 2:
print(1)
elif n % 2 == 0:
print(2)
else:
i = 2
flag = True
while i * i <= n:
if n % i == 0:
flag = False
break
i +=1
if flag:
print(1)
else:
i = 2
flag = True
while i * i <= n - 2:
if (n - 2) % i == 0:
flag = False
break
i += 1
if flag:
print(2)
else:
print(3)
``` | instruction | 0 | 13,782 | 10 | 27,564 |
Yes | output | 1 | 13,782 | 10 | 27,565 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to n (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor of n (not equal to n, of course). For example, if n = 6 then Funt has to pay 3 burles, while for n = 25 he needs to pay 5 and if n = 2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial n in several parts n1 + n2 + ... + nk = n (here k is arbitrary, even k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition ni ≥ 2 should hold for all i from 1 to k.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split n in parts.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.
Output
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
Examples
Input
4
Output
2
Input
27
Output
3
Submitted Solution:
```
mark = []
p = [1]
import math
def isPrime(n):
batas_atas = int(n ** 0.5)
for i in range(2, batas_atas+1):
if n % i == 0:
return False
return True
def gen_prime(upto):
for i in range(upto+1):
mark.append(True)
for i in range(2, upto+1):
if(mark[i]):
p.append(i)
for j in range(i,upto+1,i):
mark[j] = False
ans = 0
def cari_prima_terdekat(n):
global ans
#print(n)
if(n > 0):
for i in range(n,0,-1):
if(isPrime(i)):
ans += 1
print(i)
cari_prima_terdekat(n - i)
break
n = int(input())
if(isPrime(n)):
print(1)
elif(n % 2 == 0):
print(2)
elif(isPrime(n-2)):
print(2)
else:
print(3)
``` | instruction | 0 | 13,783 | 10 | 27,566 |
Yes | output | 1 | 13,783 | 10 | 27,567 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to n (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor of n (not equal to n, of course). For example, if n = 6 then Funt has to pay 3 burles, while for n = 25 he needs to pay 5 and if n = 2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial n in several parts n1 + n2 + ... + nk = n (here k is arbitrary, even k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition ni ≥ 2 should hold for all i from 1 to k.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split n in parts.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.
Output
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
Examples
Input
4
Output
2
Input
27
Output
3
Submitted Solution:
```
from math import sqrt
def getPrimeFactors(n):
factors = []
while n % 2 == 0:
factors.append(2)
n >>= 1
for i in range(3, int(sqrt(n)) + 1):
while n % i == 0:
factors.append(i)
n = n // i
if n > 2:
factors.append(n)
return factors
n = int(input())
print(len(getPrimeFactors(n)))
``` | instruction | 0 | 13,784 | 10 | 27,568 |
No | output | 1 | 13,784 | 10 | 27,569 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to n (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor of n (not equal to n, of course). For example, if n = 6 then Funt has to pay 3 burles, while for n = 25 he needs to pay 5 and if n = 2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial n in several parts n1 + n2 + ... + nk = n (here k is arbitrary, even k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition ni ≥ 2 should hold for all i from 1 to k.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split n in parts.
Input
The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.
Output
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
Examples
Input
4
Output
2
Input
27
Output
3
Submitted Solution:
```
def isPrime(n):
if (n <= 1):
return False
if (n <= 3):
return True
if (n % 2 == 0 or n % 3 == 0):
return False
i = 5
while (i * i <= n):
if (n % i == 0 or n % (i + 2) == 0):
return False
i = i + 6
return True
n=int(input())
print("1") if isPrime(n) else print("2") if n%2==0 else print("3")
``` | instruction | 0 | 13,787 | 10 | 27,574 |
No | output | 1 | 13,787 | 10 | 27,575 |
Provide a correct Python 3 solution for this coding contest problem.
You are now in a bookshop with your friend Alice to buy a book, "The Winning Strategy for the Programming Koshien Contest,” just released today. As you definitely want to buy it, you are planning to borrow some money from Alice in case the amount you have falls short of the price. If the amount you receive from Alice still fails to meet the price, you have to abandon buying the book this time.
Write a program to calculate the minimum amount of money you need to borrow from Alice given the following three items of data: the money you and Alice have now and the price of the book.
Input
The input is given in the following format.
m f b
A line containing the three amounts of money is given: the amount you have with you now m (0 ≤ m ≤ 10000), the money Alice has now f (0 ≤ f ≤ 10000) and the price of the book b (100 ≤ b ≤ 20000).
Output
Output a line suggesting the minimum amount of money you need to borrow from Alice. Output "NA" if all the money Alice has with him now is not a sufficient amount for you to buy the book.
Examples
Input
1000 3000 3000
Output
2000
Input
5000 3000 4500
Output
0
Input
500 1000 2000
Output
NA | instruction | 0 | 14,053 | 10 | 28,106 |
"Correct Solution:
```
m, f, b = map(int, input().split())
print("NA") if m < b - f else print(max(0, b - m))
``` | output | 1 | 14,053 | 10 | 28,107 |
Provide a correct Python 3 solution for this coding contest problem.
You are now in a bookshop with your friend Alice to buy a book, "The Winning Strategy for the Programming Koshien Contest,” just released today. As you definitely want to buy it, you are planning to borrow some money from Alice in case the amount you have falls short of the price. If the amount you receive from Alice still fails to meet the price, you have to abandon buying the book this time.
Write a program to calculate the minimum amount of money you need to borrow from Alice given the following three items of data: the money you and Alice have now and the price of the book.
Input
The input is given in the following format.
m f b
A line containing the three amounts of money is given: the amount you have with you now m (0 ≤ m ≤ 10000), the money Alice has now f (0 ≤ f ≤ 10000) and the price of the book b (100 ≤ b ≤ 20000).
Output
Output a line suggesting the minimum amount of money you need to borrow from Alice. Output "NA" if all the money Alice has with him now is not a sufficient amount for you to buy the book.
Examples
Input
1000 3000 3000
Output
2000
Input
5000 3000 4500
Output
0
Input
500 1000 2000
Output
NA | instruction | 0 | 14,054 | 10 | 28,108 |
"Correct Solution:
```
m,f,b=map(int,input().split())
if(m+f<b):
print("NA")
else:
if(m>b):
print(0)
else:
print(b-m)
``` | output | 1 | 14,054 | 10 | 28,109 |
Provide a correct Python 3 solution for this coding contest problem.
You are now in a bookshop with your friend Alice to buy a book, "The Winning Strategy for the Programming Koshien Contest,” just released today. As you definitely want to buy it, you are planning to borrow some money from Alice in case the amount you have falls short of the price. If the amount you receive from Alice still fails to meet the price, you have to abandon buying the book this time.
Write a program to calculate the minimum amount of money you need to borrow from Alice given the following three items of data: the money you and Alice have now and the price of the book.
Input
The input is given in the following format.
m f b
A line containing the three amounts of money is given: the amount you have with you now m (0 ≤ m ≤ 10000), the money Alice has now f (0 ≤ f ≤ 10000) and the price of the book b (100 ≤ b ≤ 20000).
Output
Output a line suggesting the minimum amount of money you need to borrow from Alice. Output "NA" if all the money Alice has with him now is not a sufficient amount for you to buy the book.
Examples
Input
1000 3000 3000
Output
2000
Input
5000 3000 4500
Output
0
Input
500 1000 2000
Output
NA | instruction | 0 | 14,055 | 10 | 28,110 |
"Correct Solution:
```
m,f,b=map(int,input().split())
if m>b:
print(0)
elif b-m<=f:
print(b-m)
else:
print("NA")
``` | output | 1 | 14,055 | 10 | 28,111 |
Provide a correct Python 3 solution for this coding contest problem.
You are now in a bookshop with your friend Alice to buy a book, "The Winning Strategy for the Programming Koshien Contest,” just released today. As you definitely want to buy it, you are planning to borrow some money from Alice in case the amount you have falls short of the price. If the amount you receive from Alice still fails to meet the price, you have to abandon buying the book this time.
Write a program to calculate the minimum amount of money you need to borrow from Alice given the following three items of data: the money you and Alice have now and the price of the book.
Input
The input is given in the following format.
m f b
A line containing the three amounts of money is given: the amount you have with you now m (0 ≤ m ≤ 10000), the money Alice has now f (0 ≤ f ≤ 10000) and the price of the book b (100 ≤ b ≤ 20000).
Output
Output a line suggesting the minimum amount of money you need to borrow from Alice. Output "NA" if all the money Alice has with him now is not a sufficient amount for you to buy the book.
Examples
Input
1000 3000 3000
Output
2000
Input
5000 3000 4500
Output
0
Input
500 1000 2000
Output
NA | instruction | 0 | 14,056 | 10 | 28,112 |
"Correct Solution:
```
m, f, b = map(int,input().split())
if m >= b:
print(0)
if m < b:
a = b-m
if b-m <= f:
print(a)
else:
print('NA')
``` | output | 1 | 14,056 | 10 | 28,113 |
Provide a correct Python 3 solution for this coding contest problem.
You are now in a bookshop with your friend Alice to buy a book, "The Winning Strategy for the Programming Koshien Contest,” just released today. As you definitely want to buy it, you are planning to borrow some money from Alice in case the amount you have falls short of the price. If the amount you receive from Alice still fails to meet the price, you have to abandon buying the book this time.
Write a program to calculate the minimum amount of money you need to borrow from Alice given the following three items of data: the money you and Alice have now and the price of the book.
Input
The input is given in the following format.
m f b
A line containing the three amounts of money is given: the amount you have with you now m (0 ≤ m ≤ 10000), the money Alice has now f (0 ≤ f ≤ 10000) and the price of the book b (100 ≤ b ≤ 20000).
Output
Output a line suggesting the minimum amount of money you need to borrow from Alice. Output "NA" if all the money Alice has with him now is not a sufficient amount for you to buy the book.
Examples
Input
1000 3000 3000
Output
2000
Input
5000 3000 4500
Output
0
Input
500 1000 2000
Output
NA | instruction | 0 | 14,057 | 10 | 28,114 |
"Correct Solution:
```
m, f, b = map(int, input().split())
print("NA" if m+f < b else max(0, b-m))
``` | output | 1 | 14,057 | 10 | 28,115 |
Provide a correct Python 3 solution for this coding contest problem.
You are now in a bookshop with your friend Alice to buy a book, "The Winning Strategy for the Programming Koshien Contest,” just released today. As you definitely want to buy it, you are planning to borrow some money from Alice in case the amount you have falls short of the price. If the amount you receive from Alice still fails to meet the price, you have to abandon buying the book this time.
Write a program to calculate the minimum amount of money you need to borrow from Alice given the following three items of data: the money you and Alice have now and the price of the book.
Input
The input is given in the following format.
m f b
A line containing the three amounts of money is given: the amount you have with you now m (0 ≤ m ≤ 10000), the money Alice has now f (0 ≤ f ≤ 10000) and the price of the book b (100 ≤ b ≤ 20000).
Output
Output a line suggesting the minimum amount of money you need to borrow from Alice. Output "NA" if all the money Alice has with him now is not a sufficient amount for you to buy the book.
Examples
Input
1000 3000 3000
Output
2000
Input
5000 3000 4500
Output
0
Input
500 1000 2000
Output
NA | instruction | 0 | 14,058 | 10 | 28,116 |
"Correct Solution:
```
i = input()
m,f,b = map(int,i.split())
if m >= b:
print(0)
elif (m+f) < b:
print('NA')
else:
print(b - m)
``` | output | 1 | 14,058 | 10 | 28,117 |
Provide a correct Python 3 solution for this coding contest problem.
You are now in a bookshop with your friend Alice to buy a book, "The Winning Strategy for the Programming Koshien Contest,” just released today. As you definitely want to buy it, you are planning to borrow some money from Alice in case the amount you have falls short of the price. If the amount you receive from Alice still fails to meet the price, you have to abandon buying the book this time.
Write a program to calculate the minimum amount of money you need to borrow from Alice given the following three items of data: the money you and Alice have now and the price of the book.
Input
The input is given in the following format.
m f b
A line containing the three amounts of money is given: the amount you have with you now m (0 ≤ m ≤ 10000), the money Alice has now f (0 ≤ f ≤ 10000) and the price of the book b (100 ≤ b ≤ 20000).
Output
Output a line suggesting the minimum amount of money you need to borrow from Alice. Output "NA" if all the money Alice has with him now is not a sufficient amount for you to buy the book.
Examples
Input
1000 3000 3000
Output
2000
Input
5000 3000 4500
Output
0
Input
500 1000 2000
Output
NA | instruction | 0 | 14,059 | 10 | 28,118 |
"Correct Solution:
```
m,f,b = map(int, input(). split())
if b <= m:
print('0')
elif b-m <= f:
print(b-m)
else:
print('NA')
``` | output | 1 | 14,059 | 10 | 28,119 |
Provide a correct Python 3 solution for this coding contest problem.
You are now in a bookshop with your friend Alice to buy a book, "The Winning Strategy for the Programming Koshien Contest,” just released today. As you definitely want to buy it, you are planning to borrow some money from Alice in case the amount you have falls short of the price. If the amount you receive from Alice still fails to meet the price, you have to abandon buying the book this time.
Write a program to calculate the minimum amount of money you need to borrow from Alice given the following three items of data: the money you and Alice have now and the price of the book.
Input
The input is given in the following format.
m f b
A line containing the three amounts of money is given: the amount you have with you now m (0 ≤ m ≤ 10000), the money Alice has now f (0 ≤ f ≤ 10000) and the price of the book b (100 ≤ b ≤ 20000).
Output
Output a line suggesting the minimum amount of money you need to borrow from Alice. Output "NA" if all the money Alice has with him now is not a sufficient amount for you to buy the book.
Examples
Input
1000 3000 3000
Output
2000
Input
5000 3000 4500
Output
0
Input
500 1000 2000
Output
NA | instruction | 0 | 14,060 | 10 | 28,120 |
"Correct Solution:
```
m,f,b=map(int,input().split())
x=b-m
if b<=m:
print(0)
elif f>=x and 0<=x:
print(x)
else:
print("NA")
``` | output | 1 | 14,060 | 10 | 28,121 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are now in a bookshop with your friend Alice to buy a book, "The Winning Strategy for the Programming Koshien Contest,” just released today. As you definitely want to buy it, you are planning to borrow some money from Alice in case the amount you have falls short of the price. If the amount you receive from Alice still fails to meet the price, you have to abandon buying the book this time.
Write a program to calculate the minimum amount of money you need to borrow from Alice given the following three items of data: the money you and Alice have now and the price of the book.
Input
The input is given in the following format.
m f b
A line containing the three amounts of money is given: the amount you have with you now m (0 ≤ m ≤ 10000), the money Alice has now f (0 ≤ f ≤ 10000) and the price of the book b (100 ≤ b ≤ 20000).
Output
Output a line suggesting the minimum amount of money you need to borrow from Alice. Output "NA" if all the money Alice has with him now is not a sufficient amount for you to buy the book.
Examples
Input
1000 3000 3000
Output
2000
Input
5000 3000 4500
Output
0
Input
500 1000 2000
Output
NA
Submitted Solution:
```
m, f, b = map(int, input().split())
if m >= b:
print("0")
elif m + f >= b:
print(b - m)
else:
print("NA")
``` | instruction | 0 | 14,061 | 10 | 28,122 |
Yes | output | 1 | 14,061 | 10 | 28,123 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are now in a bookshop with your friend Alice to buy a book, "The Winning Strategy for the Programming Koshien Contest,” just released today. As you definitely want to buy it, you are planning to borrow some money from Alice in case the amount you have falls short of the price. If the amount you receive from Alice still fails to meet the price, you have to abandon buying the book this time.
Write a program to calculate the minimum amount of money you need to borrow from Alice given the following three items of data: the money you and Alice have now and the price of the book.
Input
The input is given in the following format.
m f b
A line containing the three amounts of money is given: the amount you have with you now m (0 ≤ m ≤ 10000), the money Alice has now f (0 ≤ f ≤ 10000) and the price of the book b (100 ≤ b ≤ 20000).
Output
Output a line suggesting the minimum amount of money you need to borrow from Alice. Output "NA" if all the money Alice has with him now is not a sufficient amount for you to buy the book.
Examples
Input
1000 3000 3000
Output
2000
Input
5000 3000 4500
Output
0
Input
500 1000 2000
Output
NA
Submitted Solution:
```
m, f, b = map(int, input().split())
if(m + f < b) :
print("NA")
elif(m >= b) :
print(0)
else :
print(b - m)
``` | instruction | 0 | 14,062 | 10 | 28,124 |
Yes | output | 1 | 14,062 | 10 | 28,125 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are now in a bookshop with your friend Alice to buy a book, "The Winning Strategy for the Programming Koshien Contest,” just released today. As you definitely want to buy it, you are planning to borrow some money from Alice in case the amount you have falls short of the price. If the amount you receive from Alice still fails to meet the price, you have to abandon buying the book this time.
Write a program to calculate the minimum amount of money you need to borrow from Alice given the following three items of data: the money you and Alice have now and the price of the book.
Input
The input is given in the following format.
m f b
A line containing the three amounts of money is given: the amount you have with you now m (0 ≤ m ≤ 10000), the money Alice has now f (0 ≤ f ≤ 10000) and the price of the book b (100 ≤ b ≤ 20000).
Output
Output a line suggesting the minimum amount of money you need to borrow from Alice. Output "NA" if all the money Alice has with him now is not a sufficient amount for you to buy the book.
Examples
Input
1000 3000 3000
Output
2000
Input
5000 3000 4500
Output
0
Input
500 1000 2000
Output
NA
Submitted Solution:
```
m,a,b=map(int,input().split())
if m>=b:
print(0)
elif m+a<b:
print("NA")
else:
print(b-m)
``` | instruction | 0 | 14,063 | 10 | 28,126 |
Yes | output | 1 | 14,063 | 10 | 28,127 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are now in a bookshop with your friend Alice to buy a book, "The Winning Strategy for the Programming Koshien Contest,” just released today. As you definitely want to buy it, you are planning to borrow some money from Alice in case the amount you have falls short of the price. If the amount you receive from Alice still fails to meet the price, you have to abandon buying the book this time.
Write a program to calculate the minimum amount of money you need to borrow from Alice given the following three items of data: the money you and Alice have now and the price of the book.
Input
The input is given in the following format.
m f b
A line containing the three amounts of money is given: the amount you have with you now m (0 ≤ m ≤ 10000), the money Alice has now f (0 ≤ f ≤ 10000) and the price of the book b (100 ≤ b ≤ 20000).
Output
Output a line suggesting the minimum amount of money you need to borrow from Alice. Output "NA" if all the money Alice has with him now is not a sufficient amount for you to buy the book.
Examples
Input
1000 3000 3000
Output
2000
Input
5000 3000 4500
Output
0
Input
500 1000 2000
Output
NA
Submitted Solution:
```
m,f,b = list(map(int, input().split()))
t = b - m
if t>0:
if t<=f: print(t)
else: print("NA")
else: print(0)
``` | instruction | 0 | 14,064 | 10 | 28,128 |
Yes | output | 1 | 14,064 | 10 | 28,129 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are now in a bookshop with your friend Alice to buy a book, "The Winning Strategy for the Programming Koshien Contest,” just released today. As you definitely want to buy it, you are planning to borrow some money from Alice in case the amount you have falls short of the price. If the amount you receive from Alice still fails to meet the price, you have to abandon buying the book this time.
Write a program to calculate the minimum amount of money you need to borrow from Alice given the following three items of data: the money you and Alice have now and the price of the book.
Input
The input is given in the following format.
m f b
A line containing the three amounts of money is given: the amount you have with you now m (0 ≤ m ≤ 10000), the money Alice has now f (0 ≤ f ≤ 10000) and the price of the book b (100 ≤ b ≤ 20000).
Output
Output a line suggesting the minimum amount of money you need to borrow from Alice. Output "NA" if all the money Alice has with him now is not a sufficient amount for you to buy the book.
Examples
Input
1000 3000 3000
Output
2000
Input
5000 3000 4500
Output
0
Input
500 1000 2000
Output
NA
Submitted Solution:
```
m,f,b=map(int,input().split())
ans=0
if m>b:
ans=0
elif m<b:
ans=b-m
elif ans>f:
ans="NA"
print(ans)
``` | instruction | 0 | 14,065 | 10 | 28,130 |
No | output | 1 | 14,065 | 10 | 28,131 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are now in a bookshop with your friend Alice to buy a book, "The Winning Strategy for the Programming Koshien Contest,” just released today. As you definitely want to buy it, you are planning to borrow some money from Alice in case the amount you have falls short of the price. If the amount you receive from Alice still fails to meet the price, you have to abandon buying the book this time.
Write a program to calculate the minimum amount of money you need to borrow from Alice given the following three items of data: the money you and Alice have now and the price of the book.
Input
The input is given in the following format.
m f b
A line containing the three amounts of money is given: the amount you have with you now m (0 ≤ m ≤ 10000), the money Alice has now f (0 ≤ f ≤ 10000) and the price of the book b (100 ≤ b ≤ 20000).
Output
Output a line suggesting the minimum amount of money you need to borrow from Alice. Output "NA" if all the money Alice has with him now is not a sufficient amount for you to buy the book.
Examples
Input
1000 3000 3000
Output
2000
Input
5000 3000 4500
Output
0
Input
500 1000 2000
Output
NA
Submitted Solution:
```
m,f,b=map(int,input().split())
ans=0
if m>b:
ans=0
elif m<b:
ans=b-m
elif m+f<b:
ans="NA"
print(ans)
``` | instruction | 0 | 14,066 | 10 | 28,132 |
No | output | 1 | 14,066 | 10 | 28,133 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1 | instruction | 0 | 14,596 | 10 | 29,192 |
Tags: binary search, sortings
Correct Solution:
```
f = lambda: map(int, input().split())
n, k = f()
t = list(f())
d = {0: 0}
for q in t:
for i in range(1, k + 1): d[q * i] = i
for j in range(int(input())):
a = int(input())
p = [i + d[a - b] for b, i in d.items() if a - b in d]
print(min(p) if p and min(p) <= k else -1)
``` | output | 1 | 14,596 | 10 | 29,193 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1 | instruction | 0 | 14,597 | 10 | 29,194 |
Tags: binary search, sortings
Correct Solution:
```
f = lambda: map(int, input().split())
n, k = f()
t = list(f())
d = {0: 0}
for q in t:
for i in range(1, k + 1): d[q * i] = i
for j in range(int(input())):
a = int(input())
p = [i + d[a - b] for b, i in d.items() if a - b in d]
print(min(p) if p and min(p) <= k else -1)
# Made By Mostafa_Khaled
``` | output | 1 | 14,597 | 10 | 29,195 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1 | instruction | 0 | 14,598 | 10 | 29,196 |
Tags: binary search, sortings
Correct Solution:
```
n, k = map(int, input().split())
a = set(map(int, input().split()))
q = int(input())
# def isIn(x, fm, to):
# if fm >= to:
# return a[fm] == x
# t = a[(fm+to) // 2]
# if t > x:
# return isIn(x, fm, (fm+to) // 2 - 1)
# elif t < x:
# return isIn(x, (fm+to) // 2 + 1, to)
# else:
# return True
for _ in range(q):
x = int(input())
if x in a:
print(1)
continue
found = False
for i in range(2, k + 1):
for j in range(1, i // 2 + 1):
for l in a:
t = x - l * j
if t % (i - j) != 0:
continue
# if isIn(t // (i - j), 0, n - 1):
if t // (i - j) in a:
print(i)
found = True
break
if found:
break
if found:
break
if not found:
print(-1)
``` | output | 1 | 14,598 | 10 | 29,197 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1 | instruction | 0 | 14,599 | 10 | 29,198 |
Tags: binary search, sortings
Correct Solution:
```
n_k = input()
n_k = n_k.split(" ")
n = int(n_k[0])
k = int(n_k[1])
ais = input()
ais = ais.split(" ")
q = int(input())
pares = {}
for a in ais:
a = int(a)
for i in range(k):
p = int((i+1)*a)
if (p not in pares) or (i+1 < pares[p]):
pares[p] = i+1
m = 1000000000
for i in range(q):
x = int(input())
ans = 1000;
minimo = m
for plata, deuda in pares.items():
if plata == x:
if deuda <= k:
if deuda < minimo:
minimo = deuda
else:
r = x-plata
if r in pares:
if deuda+pares[r] < minimo:
if deuda + pares[r] <= k:
minimo = deuda+pares[r]
if minimo == m:
print(-1)
else:
print(minimo)
``` | output | 1 | 14,599 | 10 | 29,199 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1 | instruction | 0 | 14,600 | 10 | 29,200 |
Tags: binary search, sortings
Correct Solution:
```
from collections import defaultdict
n, m = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
d = defaultdict(int)
d[0] = 0
for v in a:
for i in range(1, m + 1):
d[v * i] = i
q = int(input())
for _ in range(q):
x = int(input())
r = m + 1
for k, v in d.items():
y = x - k
if y in d:
r = min(r, v + d[y])
print(r if r <= m else -1)
``` | output | 1 | 14,600 | 10 | 29,201 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1 | instruction | 0 | 14,601 | 10 | 29,202 |
Tags: binary search, sortings
Correct Solution:
```
def find(x):
ret = 100
for k in mp:
if x - k in mp:
ret = min(ret, mp[k] + mp[x - k])
return ret if ret <= m else -1
n, m = (int(x) for x in input().split())
a = [int(x) for x in input().split()]
mp = {}
for i in a:
for j in range(m + 1):
x = i * j
if x in mp:
mp[x] = min(j, mp[x])
else:
mp[x] = j
q = int(input())
for Q in range(q):
x = int(input())
print(find(x))
``` | output | 1 | 14,601 | 10 | 29,203 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1 | instruction | 0 | 14,602 | 10 | 29,204 |
Tags: binary search, sortings
Correct Solution:
```
n_k = input()
n_k = n_k.split(" ")
n = int(n_k[0])
k = int(n_k[1])
ais = input()
ais = ais.split(" ")
q = int(input())
pares = {}
for a in ais:
a = int(a)
for i in range(k):
p = int((i+1)*a)
if (p not in pares) or (i+1 < pares[p]):
pares[p] = i+1
m = 1000000000
for i in range(q):
x = int(input())
ans = 1000;
minimo = m
for plata, deuda in pares.items():
if plata == x:
if deuda <= k:
if deuda < minimo:
minimo = deuda
else:
r = x-plata
if r in pares:
if deuda+pares[r] < minimo:
if deuda + pares[r] <= k:
minimo = deuda+pares[r]
if minimo == m:
print(-1)
else:
print(minimo)
# 1538793706889
``` | output | 1 | 14,602 | 10 | 29,205 |
Provide tags and a correct Python 3 solution for this coding contest problem.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1 | instruction | 0 | 14,603 | 10 | 29,206 |
Tags: binary search, sortings
Correct Solution:
```
n_k = input()
n_k = n_k.split(" ")
n = int(n_k[0])
k = int(n_k[1])
ais = input()
ais = ais.split(" ")
q = int(input())
pares = {}
for a in ais:
a = int(a)
for i in range(k):
p = int((i+1)*a)
if (p not in pares) or (i+1 < pares[p]):
pares[p] = i+1
m = 1000000000
for i in range(q):
x = int(input())
ans = 1000;
minimo = m
for money, bills in pares.items():
if money == x and bills <= k and bills < minimo:
minimo = bills
else:
r = x-money
if r in pares and bills+pares[r] < minimo and bills + pares[r] <= k:
minimo = bills+pares[r]
if minimo == m:
print(-1)
else:
print(minimo)
``` | output | 1 | 14,603 | 10 | 29,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1
Submitted Solution:
```
f = lambda: map(int, input().split())
n, k = f()
t = list(f())
u = [{i * q for q in t} for i in range(1, k + 1)]
def g(d):
v = [{d - q for q in p} for p in u]
for j in range(1, k + 1):
if d in u[j - 1]: return j
for i in range(j // 2):
if u[i].intersection(v[j - i - 2]): return j
return -1
for i in range(int(input())): print(g(int(input())))
``` | instruction | 0 | 14,604 | 10 | 29,208 |
Yes | output | 1 | 14,604 | 10 | 29,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1
Submitted Solution:
```
n,k = [int(s) for s in input().split()]
bills = [int(s) for s in input().split()]
pairs = {}
for bill in bills:
for i in range(k):
possible = (i+1)*bill
if possible not in pairs or i+1 < pairs[possible]:
pairs[possible] = i+1
q = int(input())
mymax = 100000000000000000
for i in range(q):
query = int(input())
minumum = mymax
for money, nbills in pairs.items():
if money == query and nbills <=k and nbills < minumum:
minumum = nbills
else:
rest = query-money
if rest in pairs and nbills+pairs[rest] < minumum and nbills+pairs[rest] <= k:
minumum = nbills+pairs[rest]
if minumum == mymax:
print(-1)
else:
print(minumum)
``` | instruction | 0 | 14,605 | 10 | 29,210 |
Yes | output | 1 | 14,605 | 10 | 29,211 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1
Submitted Solution:
```
n, k = (int(x) for x in input().split())
a = [int(x) for x in input().split()]
q = int(input())
for _ in range(q):
x = int(input())
m = -1
for i in range(n):
for j in range(i+1, n):
#print("trying", x, "with", a[i], a[j])
k1, r = divmod(x, a[j])
#print("result:", k1, r)
if r % a[i] == 0:
k2 = r //a[i]
if k1 + k2 <= k and (m == -1 or k1 + k2 < m):
m = k1 + k2
print(m)
``` | instruction | 0 | 14,606 | 10 | 29,212 |
No | output | 1 | 14,606 | 10 | 29,213 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1
Submitted Solution:
```
n, k= map(int, input().split())
d = list(map(int, input().split()))
d.reverse()
q = int(input())
while q>0:
q -= 1
count = 0
anothercount = 0
amt = int(input())
for i in range(0, n):
if amt < d[i]:
i += 1
else:
count += amt//d[i]
anothercount += 1
amt %= d[i]
if amt != 0 or count > k or anothercount > 2:
print(-1)
else:
print(count)
``` | instruction | 0 | 14,607 | 10 | 29,214 |
No | output | 1 | 14,607 | 10 | 29,215 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1
Submitted Solution:
```
n = list(map(int, input().split()))
k = n[1]
n = n[0]
bills = list(map(int, input().split()))
n_queries = int(input())
# used_bills = set()
#
#
# def check(amount, times, used_bills=None):
# used_bills = used_bills if used_bills else set()
# if times > k:
# return -1
# for bill in list(reversed(bills)):
# if amount - bill >= 0:
# used_bills.add(bill)
# if len(used_bills) > 2:
# return -1
# times += 1
# amount -= bill
# if amount == 0:
# return times
# return check(amount, times, used_bills)
# return -1
#
# times = 0
# for i in range(n_queries):
# print(check(int(input()), times))
def max_bill(amount):
times = 0
for bill in list(reversed(bills)):
if amount - bill >= 0:
times += amount//bill
# print("Times:", times)
# print("Bill:", bill)
if times > k:
return -1
amount -= amount//bill * bill
if amount == 0:
return times
return other_bill(amount, bill, times)
def other_bill(amount, bill, times):
for billy in list(reversed(bills)):
if billy >= bill:
continue
if not (amount/billy).is_integer():
continue
else:
times += amount//billy
# print("Billy:", billy)
if times > k:
return -1
return times
return False
for i in range(n_queries):
test_a = max_bill(int(input()))
print(test_a)
``` | instruction | 0 | 14,608 | 10 | 29,216 |
No | output | 1 | 14,608 | 10 | 29,217 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
ATMs of a well-known bank of a small country are arranged so that they can not give any amount of money requested by the user. Due to the limited size of the bill dispenser (the device that is directly giving money from an ATM) and some peculiarities of the ATM structure, you can get at most k bills from it, and the bills may be of at most two distinct denominations.
For example, if a country uses bills with denominations 10, 50, 100, 500, 1000 and 5000 burles, then at k = 20 such ATM can give sums 100 000 burles and 96 000 burles, but it cannot give sums 99 000 and 101 000 burles.
Let's suppose that the country uses bills of n distinct denominations, and the ATM that you are using has an unlimited number of bills of each type. You know that during the day you will need to withdraw a certain amount of cash q times. You know that when the ATM has multiple ways to give money, it chooses the one which requires the minimum number of bills, or displays an error message if it cannot be done. Determine the result of each of the q of requests for cash withdrawal.
Input
The first line contains two integers n, k (1 ≤ n ≤ 5000, 1 ≤ k ≤ 20).
The next line contains n space-separated integers ai (1 ≤ ai ≤ 107) — the denominations of the bills that are used in the country. Numbers ai follow in the strictly increasing order.
The next line contains integer q (1 ≤ q ≤ 20) — the number of requests for cash withdrawal that you will make.
The next q lines contain numbers xi (1 ≤ xi ≤ 2·108) — the sums of money in burles that you are going to withdraw from the ATM.
Output
For each request for cash withdrawal print on a single line the minimum number of bills it can be done, or print - 1, if it is impossible to get the corresponding sum.
Examples
Input
6 20
10 50 100 500 1000 5000
8
4200
100000
95000
96000
99000
10100
2015
9950
Output
6
20
19
20
-1
3
-1
-1
Input
5 2
1 2 3 5 8
8
1
3
5
7
9
11
13
15
Output
1
1
1
2
2
2
2
-1
Submitted Solution:
```
n = list(map(int, input().split()))
k = n[1]
n = n[0]
bills = list(map(int, input().split()))
n_queries = int(input())
used_bills = set()
def check(amount, times, used_bills=None):
used_bills = used_bills if used_bills else set()
if times > k:
return -1
for bill in list(reversed(bills)):
if amount - bill >= 0:
used_bills.add(bill)
if len(used_bills) > 2:
return -1
times += 1
amount -= bill
if amount == 0:
return times
return check(amount, times, used_bills)
return -1
times = 0
for i in range(n_queries):
print(check(int(input()), times))
``` | instruction | 0 | 14,609 | 10 | 29,218 |
No | output | 1 | 14,609 | 10 | 29,219 |
Provide a correct Python 3 solution for this coding contest problem.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 40
* 1 ≤ vi ≤ 1015
* 1 ≤ wi ≤ 1015
* 1 ≤ W ≤ 1015
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9 | instruction | 0 | 15,030 | 10 | 30,060 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
# http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=DPL_1_H&lang=ja
# 全列挙する関数
def allp(aa):
now = [(0, 0)]
for v, w in aa:
nxt = [(w + w0, v + v0) for w0, v0 in now]
now += nxt
# wでソート
now.sort(key=lambda x: (x[0], -x[1]))
# wが昇順ならvも昇順でないと、無駄な組み合わせがあるということなので
# それを除いて、wとvを別々に返す
pv = -1
ww = []
vv = []
for w, v in now:
if v <= pv: continue
ww.append(w)
vv.append(v)
pv = v
return vv, ww
def main():
n, wn = MI()
vw = [LI() for _ in range(n)]
# 半分全列挙する
vv1, ww1 = allp(vw[:n // 2])
vv2, ww2 = allp(vw[n // 2:])
# 前半は1つずつ動かして、後半はできるだけ動かさない
# wの和が制限を越えたら後半は動かす
ans = 0
j = len(vv2) - 1
for v1, w1 in zip(vv1, ww1):
while w1 + ww2[j] > wn:
if j == 0:
print(ans)
exit()
else:
j -= 1
if v1 + vv2[j] > ans: ans = v1 + vv2[j]
print(ans)
main()
``` | output | 1 | 15,030 | 10 | 30,061 |
Provide a correct Python 3 solution for this coding contest problem.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 40
* 1 ≤ vi ≤ 1015
* 1 ≤ wi ≤ 1015
* 1 ≤ W ≤ 1015
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9 | instruction | 0 | 15,031 | 10 | 30,062 |
"Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
sys.setrecursionlimit(10 ** 7)
def branchAndBound(N, items, lim_w):
items.sort(key=lambda x: x[0] / x[1], reverse=True)
Value = []
Weight = []
Ratio = []
for v, w in items:
Value.append(v)
Weight.append(w)
Ratio.append(v / w)
def upperbound(cur, cur_w, cur_v):
rest_w = lim_w - cur_w
max_v = ub_v = cur_v
for j in range(cur + 1, N):
if Weight[j] <= rest_w:
rest_w -= Weight[j]
max_v += Value[j]
ub_v = max_v
else:
ub_v += Ratio[j] * rest_w
break
return max_v, ub_v
answer_v, ub_v = upperbound(-1, 0, 0)
def dfs(cur, ub_v, cur_w, cur_v):
nonlocal answer_v
if answer_v > ub_v or cur == N:
return
if cur_w + Weight[cur] <= lim_w:
dfs(cur + 1, ub_v, cur_w + Weight[cur], cur_v + Value[cur])
max_v, ub_v = upperbound(cur, cur_w, cur_v)
if max_v > answer_v:
answer_v = max_v
if ub_v > answer_v:
dfs(cur + 1, ub_v, cur_w, cur_v)
dfs(0, ub_v, 0, 0)
return answer_v
def main():
N, lim_w = map(int, input().split())
items = [list(map(int, input().split())) for _ in range(N)]
print(branchAndBound(N, items, lim_w))
if __name__ == "__main__":
main()
``` | output | 1 | 15,031 | 10 | 30,063 |
Provide a correct Python 3 solution for this coding contest problem.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 40
* 1 ≤ vi ≤ 1015
* 1 ≤ wi ≤ 1015
* 1 ≤ W ≤ 1015
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9 | instruction | 0 | 15,032 | 10 | 30,064 |
"Correct Solution:
```
def solve():
from bisect import bisect_right
N, W = map(int, input().split())
a = [tuple(map(int, input().split())) for _ in [0]*N]
def bitdp(items):
n = len(items)
dp = [(0, 0) for _ in [0]*(2**n)]
for bitset in range(1, 2**n):
for item_num, (item_v, item_w) in enumerate(items):
if bitset & 2**item_num:
w, v = dp[bitset-2**item_num]
if w+item_w <= W:
dp[bitset] = (w+item_w, v+item_v)
break
dp.sort()
l, maxv, inf = [(0, 0)], 0, float("inf")
append = l.append
for item, (w, v), (nextw, _) in zip(dp, dp, dp[1:]+[(inf, inf)]):
if w < nextw and maxv < v:
append(item)
maxv = v
return l
dp1 = bitdp(a[:N//2])
dp2 = bitdp(a[N//2:])
inf = float("inf")
result = 0
for w, v in dp1:
total = v + dp2[bisect_right(dp2, (W-w, inf))-1][1]
if result < total:
result = total
print(result)
if __name__ == "__main__":
solve()
``` | output | 1 | 15,032 | 10 | 30,065 |
Provide a correct Python 3 solution for this coding contest problem.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 40
* 1 ≤ vi ≤ 1015
* 1 ≤ wi ≤ 1015
* 1 ≤ W ≤ 1015
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9 | instruction | 0 | 15,033 | 10 | 30,066 |
"Correct Solution:
```
n, W = map(int, input().split())
value = []
weight = []
n1 = n // 2
n2 = n - n1
vw1 = [[0, 0]]
vw2 = [[0, 0]]
for _ in range(n1):
v, w = map(int, input().split())
sz = len(vw1)
for i in range(sz):
p = vw1[i][:]
p[0] += w
p[1] += v
vw1.append(p)
for _ in range(n2):
v, w = map(int, input().split())
sz = len(vw2)
for i in range(sz):
p = vw2[i][:]
p[0] += w
p[1] += v
vw2.append(p)
vw1.sort()
vw2.sort()
maxv = 0
for i in range(2 ** (n - n // 2)):
w, v = vw2[i]
if maxv < v:
maxv = v
else:
vw2[i][1] = maxv
ind1 = 0
ind2 = 2 ** (n - n // 2) - 1
ans = 0
while ind1 < 2 ** (n // 2) and ind2 >= 0:
w1, v1 = vw1[ind1]
w2, v2 = vw2[ind2]
while ind2 > 0 and w1 + w2 > W:
ind2 -= 1
w2, v2 = vw2[ind2]
if w1 + w2 <= W:
ans = max(ans, v1 + v2)
ind1 += 1
print(ans)
``` | output | 1 | 15,033 | 10 | 30,067 |
Provide a correct Python 3 solution for this coding contest problem.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 40
* 1 ≤ vi ≤ 1015
* 1 ≤ wi ≤ 1015
* 1 ≤ W ≤ 1015
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9 | instruction | 0 | 15,034 | 10 | 30,068 |
"Correct Solution:
```
import bisect
n,maxw = map(int , input().split())
data1 = [[0,0]] #data1enu[i]:iのbitが立ってる品物の価値、重さの総和
data2 = [[0,0]]
nn = n // 2
for i in range(nn):
v,w = map(int , input().split())
sz = len(data1)
for j in range(sz):
ppp = data1[j][:]
ppp[0] += w
ppp[1] += v
data1.append(ppp)
for i in range(n-nn):
v,w = map(int , input().split())
sz = len(data2)
for j in range(sz):
ppp = data2[j][:]
ppp[0] += w
ppp[1] += v
data2.append(ppp)
data2.sort()
maxx = 0
ss = len(data2)
for i in range(ss):
maxx = max(maxx , data2[i][1])
data2[i][1] = maxx
ans = 0
ss = len(data1)
for i in range(ss):
ww , vv = data1[i]
if (ww > maxw) : continue
amari = maxw - ww
ans = max (ans , vv + data2[bisect.bisect_right(data2, [amari , float('inf')])-1][1])
print(ans)
``` | output | 1 | 15,034 | 10 | 30,069 |
Provide a correct Python 3 solution for this coding contest problem.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 40
* 1 ≤ vi ≤ 1015
* 1 ≤ wi ≤ 1015
* 1 ≤ W ≤ 1015
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9 | instruction | 0 | 15,035 | 10 | 30,070 |
"Correct Solution:
```
N, W = map(int, input().split())
S = [list(map(int, input().split())) for i in range(N)]
def make(S):
T = {0: 0}
for v, w in S:
T0 = dict(T)
for k, val in T.items():
if k+w > W:
continue
if k+w in T0:
T0[k+w] = max(T0[k+w], val + v)
else:
T0[k+w] = val + v
T = T0
v = 0
R = []
for k in sorted(T):
v = max(v, T[k])
R.append((k, v))
return R
def solve(T0, T1):
T0.sort()
T1.sort(reverse=1)
it = iter(T1); k1, v1 = next(it)
yield 0
for k0, v0 in T0:
while k0 + k1 > W:
k1, v1 = next(it)
yield v0 + v1
T0 = make(S[:N//2])
T1 = make(S[N//2:])
print(max(solve(T0, T1)))
``` | output | 1 | 15,035 | 10 | 30,071 |
Provide a correct Python 3 solution for this coding contest problem.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 40
* 1 ≤ vi ≤ 1015
* 1 ≤ wi ≤ 1015
* 1 ≤ W ≤ 1015
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9 | instruction | 0 | 15,036 | 10 | 30,072 |
"Correct Solution:
```
#!python3
iim = lambda: map(int, input().rstrip().split())
from heapq import heappush, heappop
def resolve():
N, W = iim()
S = [list(iim()) for i in range(N)]
S.sort(key=lambda x: x[0] / x[1], reverse=True)
def ubounds(w, v, i):
for j in range(i, N):
vj, wj = S[j]
if w + wj > W:
return (-v, -v - vj / wj * (W - w))
w += wj
v += vj
return (-v, -v)
v1, v2 = ubounds(0, 0, 0)
q = []
heappush(q, (v2, v1, 0, 0, 0))
vm1 = v1
#print(W, S)
while q:
#print(q)
vq1, vq2, wq, vq, i = heappop(q)
if i == N:
continue
vi, wi = S[i]
if wq + wi <= W:
heappush(q, (vq1, vq1, wq+wi, vq+vi, i+1))
v1, v2 = ubounds(wq, vq, i + 1)
if v2 < vm1:
if v1 < vm1:
vm1 = v1
heappush(q, (v2, v1, wq, vq, i + 1))
print(-vm1)
if __name__ == "__main__":
resolve()
``` | output | 1 | 15,036 | 10 | 30,073 |
Provide a correct Python 3 solution for this coding contest problem.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 40
* 1 ≤ vi ≤ 1015
* 1 ≤ wi ≤ 1015
* 1 ≤ W ≤ 1015
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9 | instruction | 0 | 15,037 | 10 | 30,074 |
"Correct Solution:
```
import sys
from bisect import bisect_right
def meet_in_the_middle(a: list, limit: int) -> tuple:
first_v, first_w, second_v, second_w = [], [], [], []
for items, v_append, w_append in (
(a[:len(a)//2], first_v.append, first_w.append),
(a[len(a)//2:], second_v.append, second_w.append)
):
enumerated = [(0, 0)]
for v, w in items:
enumerated += [(w+_w, v+_v) for _w, _v in enumerated]
enumerated.sort()
enumerated = enumerated[:bisect_right(enumerated, (limit, float("inf")))] + [(limit+1, 0)]
max_v = -1
for (cw, cv), (nw, nv) in zip(enumerated, enumerated[1:]):
if cw < nw and max_v < cv:
v_append(cv)
w_append(cw)
max_v = cv
return (first_v, first_w), (second_v, second_w)
N, W = map(int, input().split())
a = [list(map(int, l.split())) for l in sys.stdin]
(first_v, first_w), (second_v, second_w) = meet_in_the_middle(a, W)
ans = 0
for v, w in zip(first_v, first_w):
i = bisect_right(second_w, W-w)-1
new_v = v + second_v[i]
if ans < new_v:
ans = new_v
print(ans)
``` | output | 1 | 15,037 | 10 | 30,075 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 40
* 1 ≤ vi ≤ 1015
* 1 ≤ wi ≤ 1015
* 1 ≤ W ≤ 1015
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools
import time,random
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
mod2 = 998244353
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def pe(s): return print(str(s), file=sys.stderr)
def JA(a, sep): return sep.join(map(str, a))
def JAA(a, s, t): return s.join(t.join(map(str, b)) for b in a)
def main():
n,W = LI()
vw = [LI() for _ in range(n)]
def rk(vw):
r = collections.defaultdict(int)
r[0] = 0
for v,w in vw:
rl = list(r.items())
for rw,rv in rl:
if rw + w > W:
continue
if r[rw+w] < rv + v:
r[rw+w] = rv + v
t = []
m = -1
for w,v in sorted(r.items()):
if m < v:
m = v
t.append((w,v))
return t
r1 = rk(vw[:n//2])
r2 = rk(vw[n//2:])
i2 = len(r2) - 1
r = 0
for w,v in r1:
while w + r2[i2][0] > W:
i2 -= 1
vv = v + r2[i2][1]
if r < vv:
r = vv
return r
print(main())
``` | instruction | 0 | 15,038 | 10 | 30,076 |
Yes | output | 1 | 15,038 | 10 | 30,077 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 40
* 1 ≤ vi ≤ 1015
* 1 ≤ wi ≤ 1015
* 1 ≤ W ≤ 1015
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9
Submitted Solution:
```
#!usr/bin/env python3
from collections import defaultdict
from collections import deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS():return list(map(list, sys.stdin.readline().split()))
def S(): return list(sys.stdin.readline())[:-1]
def IR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = I()
return l
def LIR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = LI()
return l
def SR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = S()
return l
def LSR(n):
l = [None for i in range(n)]
for i in range(n):l[i] = LS()
return l
sys.setrecursionlimit(1000000)
mod = 1000000007
#1:combinatorial
#1_A
"""
n, m = map(int, input().split())
c = list(map(int, input().split()))
dp = [float("inf") for i in range(n+1)]
dp[0] = 0
for i in range(n):
for j in c:
if i+j <= n:
dp[i+j] = min(dp[i+j], dp[i]+1)
print(dp[n])
"""
#1_B
"""
n,w = map(int,input().split())
b = [list(map(int, input().split())) for i in range(n)]
dp = [0 for i in range(w+1)]
memo = [[] for i in range(w+1)]
for i in range(w):
for j in range(n):
if i+b[j][1] <= w and j not in memo[i]:
if dp[i+b[j][1]] > dp[i]+b[j][0]:
continue
else:
memo[i+b[j][1]] = memo[i]+[j]
dp[i+b[j][1]] = dp[i]+b[j][0]
print(max(dp))
"""
#1_C
"""
n,w = map(int,input().split())
b = [list(map(int, input().split())) for i in range(n)]
dp = [0 for i in range(w+1)]
for i in range(w):
for j in b:
if i+j[1] <= w:
dp[i+j[1]] = max(dp[i+j[1]], dp[i]+j[0])
print(dp[w])
"""
#1_D
"""
import bisect
n = int(input())
s = [int(input()) for i in range(n)]
dp = [float("INF") for i in range(n)]
for i in range(n):
dp[bisect.bisect_left(dp,s[i])] = s[i]
print(bisect.bisect_left(dp,100000000000000))
"""
#1_E
"""
s1 = input()
s2 = input()
n = len(s1)
m = len(s2)
dp = [[0 for j in range(m+1)] for i in range(n+1)]
for i in range(n+1):
dp[i][0] = i
for j in range(m+1):
dp[0][j] = j
for i in range(1,n+1):
for j in range(1,m+1):
cost = 0 if s1[i-1] == s2[j-1] else 1
dp[i][j] = min(dp[i-1][j]+1,dp[i][j-1]+1,dp[i-1][j-1]+cost)
print(dp[n][m])
"""
#1_F
"""
import bisect
n,W = map(int,input().split())
b = [list(map(int, input().split())) for i in range(n)]
dp = [float("INF") for i in range(10001)]
dp[0] = 0
for v,w in b:
for i in range(10001):
if 10000-i+v < 10001:
dp[10000-i+v] = min(dp[10000-i+v], dp[10000-i]+w)
for i in range(1,10000):
dp[10000-i] = min(dp[10000-i], dp[10001-i])
print(bisect.bisect_right(dp,W)-1)
"""
#1_G
"""
n,W = LI()
g = LIR(n)
"""
#1_H
n,W = LI()
v = [None for i in range(n)]
w = [None for i in range(n)]
for i in range(n):
v[i],w[i] = LI()
a = [[0,0]]
b = [[0,0]]
m = n//2
for i in range(m):
for j in range(1<<i):
a.append([a[j][0]+v[i],a[j][1]+w[i]])
for i in range(n-m):
for j in range(1<<i):
b.append([b[j][0]+v[i+m],b[j][1]+w[i+m]])
a.sort(key = lambda x:x[1])
b.sort(key = lambda x:x[1])
ma = b[0][0]
for i in range(1,len(b)):
if ma > b[i][0]:
b[i][0] = ma
else:
ma = b[i][0]
ans = 0
i = len(b)-1
for v,w in a:
rem = W-w
while b[i][1] > rem:
if i == 0:break
i -= 1
if b[i][1] <= rem:
ans = max(ans,v+b[i][0])
print(ans)
#2:permutaion/path
#2_A
"""
n,e = LI()
v = [[] for i in range(n)]
f = [-1 for i in range(n)]
for i in range(e):
s,t,d = LI()
v[s].append([t,d])
if t == 0:
f[s] = d
po = [1 for i in range(n+1)]
for i in range(n):
po[i+1] = po[i]*2
dp = [[float("inf") for i in range(po[n])] for j in range(n)]
dp[0][1] = 0
for j in range(po[n]):
for x in range(n):
if po[x]&j:
for y,d in v[x]:
if not po[y]&j:
dp[y][j|po[y]] = min(dp[y][j|po[y]],dp[x][j]+d)
ans = float("inf")
for i in range(n):
if f[i] != -1:
ans = min(ans,dp[i][-1]+f[i])
if ans == float("inf"):
print(-1)
else:
print(ans)
"""
#3:pattern
#3_A
"""
h,w = map(int, input().split())
c = [list(map(int, input().split())) for i in range(h)]
dp = [[0 for i in range(w)] for j in range(h)]
for i in range(h):
dp[i][0] = 1 if c[i][0] == 0 else 0
for i in range(w):
dp[0][i] = 1 if c[0][i] == 0 else 0
for i in range(1,h):
for j in range(1,w):
if c[i][j] == 0:
dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1])+1
ans = 0
for i in dp:
for j in i:
ans = max(ans, j)
print(ans**2)
"""
#3_B
#4:counting
#4_A
"""
n,v = LI()
a = LI()
b = LI()
c = LI()
d = LI()
p = defaultdict(int)
q = defaultdict(int)
for i in a:
for j in b:
if i+j <= v-2:
p[i+j] += 1
for i in c:
for j in d:
if i+j <= v-2:
q[i+j] += 1
ans = 0
for i,j in p.items():
ans += j*q[v-i]
print(ans)
"""
#5:Twelvedfold_Way
#5_A
"""
n,k = map(int, input().split())
ans = k
for i in range(n-1):
ans *= k
ans %= 1000000007
print(ans)
"""
#5_B
"""
import sys
sys.setrecursionlimit(10000)
memo = [1,1]
def fact(a):
if len(memo) > a:
return memo[a]
b = a*fact(a-1)
memo.append(b)
return b
n,k = map(int, input().split())
if n > k:
ans = 0
else:
ans = fact(k)//fact(k-n)
print(ans%1000000007)
"""
#5_C
"""
def comb(a,b):
return fac[a]*inv[b]*inv[a-b]%mod
n,k = LI()
if n < k:
print(0)
quit()
fac = [1 for i in range(k+1)]
for i in range(k):
fac[i+1] = fac[i]*(i+1)%mod
inv = [None for i in range(k+1)]
inv[k] = pow(fac[k],mod-2,mod)
for i in range(k)[::-1]:
inv[i] = inv[i+1]*(i+1)%mod
ans = 0
for i in range(k+1):
if i%2:
ans -= comb(k,i)*pow(k-i,n,mod)%mod
else:
ans += comb(k,i)*pow(k-i,n,mod)%mod
ans %= mod
print(ans)
"""
#5_D
"""
n,k = map(int, input().split())
import sys
sys.setrecursionlimit(10000)
memo = [1,1]
def fact(a):
if len(memo) > a:
return memo[a]
b = a*fact(a-1)
memo.append(b)
return b
ans = fact(n+k-1)//(fact(n)*fact(k-1))%1000000007
print(ans)
"""
#5_E
"""
def comb(a,b):
return fac[a]*inv[b]*inv[a-b]%mod
n,k = LI()
if n > k:
print(0)
quit()
fac = [1 for i in range(k+1)]
for i in range(k):
fac[i+1] = fac[i]*(i+1)%mod
inv = [None for i in range(k+1)]
inv[k] = pow(fac[k],mod-2,mod)
for i in range(k)[::-1]:
inv[i] = inv[i+1]*(i+1)%mod
print(comb(k,n))
"""
#5_F
"""
def comb(a,b):
return fac[a]*inv[b]*inv[a-b]%mod
n,k = LI()
if n < k:
print(0)
quit()
fac = [1 for i in range(n+1)]
for i in range(n):
fac[i+1] = fac[i]*(i+1)%mod
inv = [None for i in range(n+1)]
inv[n] = pow(fac[n],mod-2,mod)
for i in range(n)[::-1]:
inv[i] = inv[i+1]*(i+1)%mod
print(comb(n-1,n-k))
"""
#5_G
#5_H
#5_I
#5_J
#5_K
#5_L
``` | instruction | 0 | 15,039 | 10 | 30,078 |
Yes | output | 1 | 15,039 | 10 | 30,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 40
* 1 ≤ vi ≤ 1015
* 1 ≤ wi ≤ 1015
* 1 ≤ W ≤ 1015
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9
Submitted Solution:
```
#!/usr/bin/env python3
# DPL_1_H: Huge Knapsack Problem
# 0/1 Knapsack using Branch and Bound
# https://www.youtube.com/watch?v=yV1d-b_NeK8
from heapq import heappush, heappop
def max_value(t, xs):
def _cost(weight, value, i):
for j in range(i, n):
v, w = xs[j]
if weight + w > t:
return (-value, -value - (v/w)*(t-weight))
weight += w
value += v
return (-value, -value*1.0)
n = len(xs)
xs.sort(key=lambda x: x[0]/x[1], reverse=True)
u, c = _cost(0, 0, 0)
heap = [(c, u, 0, 0, 0)]
maxcost = u
while heap:
cost, upper, weight, value, i = heappop(heap)
if cost > maxcost:
break
if i < n:
v, w = xs[i]
if weight+w <= t:
heappush(heap, (cost, upper, weight+w, value+v, i+1))
u, c = _cost(weight, value, i+1)
if c < maxcost:
if u < maxcost:
maxcost = u
heappush(heap, (c, u, weight, value, i+1))
return -maxcost
def run():
n, t = [int(i) for i in input().split()]
xs = []
for _ in range(n):
v, w = [int(i) for i in input().split()]
xs.append((v, w))
print(max_value(t, xs))
if __name__ == '__main__':
run()
``` | instruction | 0 | 15,040 | 10 | 30,080 |
Yes | output | 1 | 15,040 | 10 | 30,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 40
* 1 ≤ vi ≤ 1015
* 1 ≤ wi ≤ 1015
* 1 ≤ W ≤ 1015
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9
Submitted Solution:
```
def make_combination(fro, to, value, weight, W):
"""ビット演算が重いのでグレイコードを使った"""
combination = []
w = 0
v = 0
combination.append([w, v])
for i in range(1, 1 << (to - fro)):
code = i ^ (i >> 1)
bitpos = 0
while not (1 << bitpos & i):
bitpos += 1
if (1 << bitpos & code):
w += weight[bitpos + fro]
v += value[bitpos + fro]
else:
w -= weight[bitpos + fro]
v -= value[bitpos + fro]
combination.append([w, v])
# 必要のない組み合わせを弾く
combination.sort()
for i in range(len(combination) - 1):
if combination[i][1] > combination[i + 1][1]:
combination[i + 1][1] = combination[i][1]
return combination
def cal_ans(from_0_to_half_combination, from_half_to_N_combination, W):
ans = 0
""" 尺取り法 """
left = 0
right = len(from_half_to_N_combination) - 1
for left in range(len(from_0_to_half_combination)):
now_weight = from_0_to_half_combination[left][0]
now_value = from_0_to_half_combination[left][1]
while now_weight + from_half_to_N_combination[right][0] > W and right != 0:
right -= 1
if now_weight + from_half_to_N_combination[right][0] <= W:
ans = max(ans, now_value + from_half_to_N_combination[right][1])
""" 二分探索 """
# for now in range(len(from_0_to_half_combination)):
# now_weight = from_0_to_half_combination[now][0]
# now_value = from_0_to_half_combination[now][1]
# count = 0
# left = 0
# right = len(from_0_to_half_combination) - 1
# while count < 45:
# mid = (left + right) // 2
# if now_weight + from_half_to_N_combination[mid][0] > W:
# right = mid - 1
# else:
# ans = max(ans, now_value + from_half_to_N_combination[mid][1])
# left = mid + 1
# count += 1
return ans
def main():
N, W = map(int, input().split())
half = N // 2
value = []
weight = []
for i in range(N):
v, w = map(int, input().split())
value.append(v)
weight.append(w)
# 半分全列挙の組み合わせを前半と後半で作る
from_0_to_half_combination = make_combination(0, half, value, weight, W)
from_half_to_N_combination = make_combination(half, N, value, weight, W)
# 尺取り法か二分探索で計算
ans = cal_ans(from_0_to_half_combination, from_half_to_N_combination, W)
print(ans)
if __name__ == "__main__":
main()
``` | instruction | 0 | 15,041 | 10 | 30,082 |
Yes | output | 1 | 15,041 | 10 | 30,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 40
* 1 ≤ vi ≤ 1015
* 1 ≤ wi ≤ 1015
* 1 ≤ W ≤ 1015
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9
Submitted Solution:
```
N,W = map(int,input().split())
v = [0]*N;w = [0]*N
for i in range(N):
v[i],w[i] = map(int,input().split())
value = [0 for i in range(W+1)]
for i in range(N):
for j in range(W,-1,-1):
if j<w[i]:
value[j] = value[j]
else:
value[j] = max(value[j],value[j-w[i]]+v[i])
print(value[W])
``` | instruction | 0 | 15,042 | 10 | 30,084 |
No | output | 1 | 15,042 | 10 | 30,085 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 40
* 1 ≤ vi ≤ 1015
* 1 ≤ wi ≤ 1015
* 1 ≤ W ≤ 1015
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9
Submitted Solution:
```
def solve():
from bisect import bisect_left
N, W = map(int, input().split())
n1 = N//2
n2 = N-n1
a1 = [tuple(map(int, input().split())) for _ in [0]*n1]
a2 = [tuple(map(int, input().split())) for _ in [0]*n2]
dp1, dp2 = [(0, 0) for _ in [0]*(2**n1)], [(0, 0) for _ in [0]*(2**n2)]
for a, dp in ((a1, dp1), (a2, dp2)):
n = len(a)
append = dp.append
for bit in range(1, 2**n):
v, w = 0, 0
for i in range(n):
if 2**i & bit:
v += a[i][0]
w += a[i][1]
if w <= W:
dp[bit] = (w, v)
dp = sorted(dp)
dp = [pair for w in [[0]] for sa in [w.__setitem__] for pair in dp if pair[1] > w[0] and not sa(0, pair[1])]
result = 0
inf = float("inf")
l = len(dp2)
for w, v in dp1:
i = bisect_left(dp2, (W-w, inf))
newv = v + dp2[i-1][1] if i > 0 else 0
if newv > result:
result = newv
print(result)
if __name__ == "__main__":
solve()
``` | instruction | 0 | 15,043 | 10 | 30,086 |
No | output | 1 | 15,043 | 10 | 30,087 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most W, that is capacity of the knapsack.
Find the maximum total value of items in the knapsack.
Constraints
* 1 ≤ N ≤ 40
* 1 ≤ vi ≤ 1015
* 1 ≤ wi ≤ 1015
* 1 ≤ W ≤ 1015
Input
N W
v1 w1
v2 w2
:
vN wN
The first line consists of the integers N and W. In the following N lines, the value and weight of the i-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 5
4 2
5 2
2 1
8 3
Output
13
Input
2 20
5 9
4 10
Output
9
Submitted Solution:
```
def solve():
from bisect import bisect_left
N, W = map(int, input().split())
n1 = N//2
n2 = N-n1
a = [[tuple(map(int, input().split())) for _ in [0]*n1], [tuple(map(int, input().split())) for _ in [0]*n2]]
dp = [[(0, 0) for _ in [0]*(2**n1)], [(0, 0) for _ in [0]*(2**n2)]]
for i in range(2):
n = len(a[i])
cdp = dp[i]
ca = a[i]
for j in range(n):
_v, _w = ca[j]
if _w > W:
continue
for bitset in range(2**j, 2**(j+1)):
v, w = _v, _w
j = 1
for k in range(j-1):
if 2**k & bitset:
v += ca[k][0]
w += ca[k][1]
if w <= W:
cdp[bitset] = (w, v)
cdp = sorted(cdp)
dp[i] = [pair for w in [[-1]] for sa in [w.__setitem__] for pair in cdp if pair[1] > w[0] and not sa(0, pair[1])]
result = 0
inf = float("inf")
l = len(dp[1])
for w, v in dp[0]:
i = bisect_left(dp[1], (W-w, inf))
newv = v + dp[1][i-1][1] if i > 0 else 0
#print(w, v, i, newv)
if newv > result:
result = newv
print(result)
if __name__ == "__main__":
solve()
``` | instruction | 0 | 15,044 | 10 | 30,088 |
No | output | 1 | 15,044 | 10 | 30,089 |
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