message stringlengths 2 30.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 237 109k | cluster float64 10 10 | __index_level_0__ int64 474 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrew was very excited to participate in Olympiad of Metropolises. Days flew by quickly, and Andrew is already at the airport, ready to go home. He has n rubles left, and would like to exchange them to euro and dollar bills. Andrew can mix dollar bills and euro bills in whatever way he wants. The price of one dollar is d rubles, and one euro costs e rubles.
Recall that there exist the following dollar bills: 1, 2, 5, 10, 20, 50, 100, and the following euro bills — 5, 10, 20, 50, 100, 200 (note that, in this problem we do not consider the 500 euro bill, it is hard to find such bills in the currency exchange points). Andrew can buy any combination of bills, and his goal is to minimize the total number of rubles he will have after the exchange.
Help him — write a program that given integers n, e and d, finds the minimum number of rubles Andrew can get after buying dollar and euro bills.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^8) — the initial sum in rubles Andrew has.
The second line of the input contains one integer d (30 ≤ d ≤ 100) — the price of one dollar in rubles.
The third line of the input contains integer e (30 ≤ e ≤ 100) — the price of one euro in rubles.
Output
Output one integer — the minimum number of rubles Andrew can have after buying dollar and euro bills optimally.
Examples
Input
100
60
70
Output
40
Input
410
55
70
Output
5
Input
600
60
70
Output
0
Note
In the first example, we can buy just 1 dollar because there is no 1 euro bill.
In the second example, optimal exchange is to buy 5 euro and 1 dollar.
In the third example, optimal exchange is to buy 10 dollars in one bill. | instruction | 0 | 90,215 | 10 | 180,430 |
Tags: brute force, math
Correct Solution:
```
rubles = (int)(input())
dollar = (int)(input())
euro = 5*(int)(input())
remain_d = rubles%dollar
remain_e = rubles%euro
max_d = rubles//dollar
max_e = rubles//euro
remain = min(remain_d,remain_e)
for i in range(max_d):
if((rubles-(i*dollar))%euro<remain):
remain = (rubles-(i*dollar))%euro
for i in range(max_e):
if((rubles-(i*euro))%euro<remain):
remain = (rubles-(i*euro))%dollar
print(remain)
``` | output | 1 | 90,215 | 10 | 180,431 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrew was very excited to participate in Olympiad of Metropolises. Days flew by quickly, and Andrew is already at the airport, ready to go home. He has n rubles left, and would like to exchange them to euro and dollar bills. Andrew can mix dollar bills and euro bills in whatever way he wants. The price of one dollar is d rubles, and one euro costs e rubles.
Recall that there exist the following dollar bills: 1, 2, 5, 10, 20, 50, 100, and the following euro bills — 5, 10, 20, 50, 100, 200 (note that, in this problem we do not consider the 500 euro bill, it is hard to find such bills in the currency exchange points). Andrew can buy any combination of bills, and his goal is to minimize the total number of rubles he will have after the exchange.
Help him — write a program that given integers n, e and d, finds the minimum number of rubles Andrew can get after buying dollar and euro bills.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^8) — the initial sum in rubles Andrew has.
The second line of the input contains one integer d (30 ≤ d ≤ 100) — the price of one dollar in rubles.
The third line of the input contains integer e (30 ≤ e ≤ 100) — the price of one euro in rubles.
Output
Output one integer — the minimum number of rubles Andrew can have after buying dollar and euro bills optimally.
Examples
Input
100
60
70
Output
40
Input
410
55
70
Output
5
Input
600
60
70
Output
0
Note
In the first example, we can buy just 1 dollar because there is no 1 euro bill.
In the second example, optimal exchange is to buy 5 euro and 1 dollar.
In the third example, optimal exchange is to buy 10 dollars in one bill. | instruction | 0 | 90,216 | 10 | 180,432 |
Tags: brute force, math
Correct Solution:
```
def solve(n, d, e):
r = n
for f in range(n // (5 * e) + 1):
s = (n - f * 5 * e) % d
r = min(r,s)
return r
n = int(input())
d = int(input())
e = int(input())
print(solve(n,d,e))
``` | output | 1 | 90,216 | 10 | 180,433 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrew was very excited to participate in Olympiad of Metropolises. Days flew by quickly, and Andrew is already at the airport, ready to go home. He has n rubles left, and would like to exchange them to euro and dollar bills. Andrew can mix dollar bills and euro bills in whatever way he wants. The price of one dollar is d rubles, and one euro costs e rubles.
Recall that there exist the following dollar bills: 1, 2, 5, 10, 20, 50, 100, and the following euro bills — 5, 10, 20, 50, 100, 200 (note that, in this problem we do not consider the 500 euro bill, it is hard to find such bills in the currency exchange points). Andrew can buy any combination of bills, and his goal is to minimize the total number of rubles he will have after the exchange.
Help him — write a program that given integers n, e and d, finds the minimum number of rubles Andrew can get after buying dollar and euro bills.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^8) — the initial sum in rubles Andrew has.
The second line of the input contains one integer d (30 ≤ d ≤ 100) — the price of one dollar in rubles.
The third line of the input contains integer e (30 ≤ e ≤ 100) — the price of one euro in rubles.
Output
Output one integer — the minimum number of rubles Andrew can have after buying dollar and euro bills optimally.
Examples
Input
100
60
70
Output
40
Input
410
55
70
Output
5
Input
600
60
70
Output
0
Note
In the first example, we can buy just 1 dollar because there is no 1 euro bill.
In the second example, optimal exchange is to buy 5 euro and 1 dollar.
In the third example, optimal exchange is to buy 10 dollars in one bill. | instruction | 0 | 90,217 | 10 | 180,434 |
Tags: brute force, math
Correct Solution:
```
import sys
n = int(input())
d = int(input())
e = int(input())
e = e * 5
if e == d:
print(n%d)
sys.exit()
x = n // e
min = n
##print(x)
for i in range(0, x+1):
g = (n -(e*i))%d
if g < min:
min = g
print(min)
``` | output | 1 | 90,217 | 10 | 180,435 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrew was very excited to participate in Olympiad of Metropolises. Days flew by quickly, and Andrew is already at the airport, ready to go home. He has n rubles left, and would like to exchange them to euro and dollar bills. Andrew can mix dollar bills and euro bills in whatever way he wants. The price of one dollar is d rubles, and one euro costs e rubles.
Recall that there exist the following dollar bills: 1, 2, 5, 10, 20, 50, 100, and the following euro bills — 5, 10, 20, 50, 100, 200 (note that, in this problem we do not consider the 500 euro bill, it is hard to find such bills in the currency exchange points). Andrew can buy any combination of bills, and his goal is to minimize the total number of rubles he will have after the exchange.
Help him — write a program that given integers n, e and d, finds the minimum number of rubles Andrew can get after buying dollar and euro bills.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^8) — the initial sum in rubles Andrew has.
The second line of the input contains one integer d (30 ≤ d ≤ 100) — the price of one dollar in rubles.
The third line of the input contains integer e (30 ≤ e ≤ 100) — the price of one euro in rubles.
Output
Output one integer — the minimum number of rubles Andrew can have after buying dollar and euro bills optimally.
Examples
Input
100
60
70
Output
40
Input
410
55
70
Output
5
Input
600
60
70
Output
0
Note
In the first example, we can buy just 1 dollar because there is no 1 euro bill.
In the second example, optimal exchange is to buy 5 euro and 1 dollar.
In the third example, optimal exchange is to buy 10 dollars in one bill.
Submitted Solution:
```
n=int(input())
d=int(input())
e=int(input())
x=n//d+1
e*=5
ans=n
for i in range(x):
ans=min((n-i*d)%e,ans)
ans=min(ans,n%e)
print(ans)
``` | instruction | 0 | 90,218 | 10 | 180,436 |
Yes | output | 1 | 90,218 | 10 | 180,437 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrew was very excited to participate in Olympiad of Metropolises. Days flew by quickly, and Andrew is already at the airport, ready to go home. He has n rubles left, and would like to exchange them to euro and dollar bills. Andrew can mix dollar bills and euro bills in whatever way he wants. The price of one dollar is d rubles, and one euro costs e rubles.
Recall that there exist the following dollar bills: 1, 2, 5, 10, 20, 50, 100, and the following euro bills — 5, 10, 20, 50, 100, 200 (note that, in this problem we do not consider the 500 euro bill, it is hard to find such bills in the currency exchange points). Andrew can buy any combination of bills, and his goal is to minimize the total number of rubles he will have after the exchange.
Help him — write a program that given integers n, e and d, finds the minimum number of rubles Andrew can get after buying dollar and euro bills.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^8) — the initial sum in rubles Andrew has.
The second line of the input contains one integer d (30 ≤ d ≤ 100) — the price of one dollar in rubles.
The third line of the input contains integer e (30 ≤ e ≤ 100) — the price of one euro in rubles.
Output
Output one integer — the minimum number of rubles Andrew can have after buying dollar and euro bills optimally.
Examples
Input
100
60
70
Output
40
Input
410
55
70
Output
5
Input
600
60
70
Output
0
Note
In the first example, we can buy just 1 dollar because there is no 1 euro bill.
In the second example, optimal exchange is to buy 5 euro and 1 dollar.
In the third example, optimal exchange is to buy 10 dollars in one bill.
Submitted Solution:
```
a = int(input())
b = int(input())
c = int(input()) * 5
ans = a % b
k = 1
while k * c <= a:
ans = min(ans, (a - c * k) % b)
k += 1
print(ans)
``` | instruction | 0 | 90,219 | 10 | 180,438 |
Yes | output | 1 | 90,219 | 10 | 180,439 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrew was very excited to participate in Olympiad of Metropolises. Days flew by quickly, and Andrew is already at the airport, ready to go home. He has n rubles left, and would like to exchange them to euro and dollar bills. Andrew can mix dollar bills and euro bills in whatever way he wants. The price of one dollar is d rubles, and one euro costs e rubles.
Recall that there exist the following dollar bills: 1, 2, 5, 10, 20, 50, 100, and the following euro bills — 5, 10, 20, 50, 100, 200 (note that, in this problem we do not consider the 500 euro bill, it is hard to find such bills in the currency exchange points). Andrew can buy any combination of bills, and his goal is to minimize the total number of rubles he will have after the exchange.
Help him — write a program that given integers n, e and d, finds the minimum number of rubles Andrew can get after buying dollar and euro bills.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^8) — the initial sum in rubles Andrew has.
The second line of the input contains one integer d (30 ≤ d ≤ 100) — the price of one dollar in rubles.
The third line of the input contains integer e (30 ≤ e ≤ 100) — the price of one euro in rubles.
Output
Output one integer — the minimum number of rubles Andrew can have after buying dollar and euro bills optimally.
Examples
Input
100
60
70
Output
40
Input
410
55
70
Output
5
Input
600
60
70
Output
0
Note
In the first example, we can buy just 1 dollar because there is no 1 euro bill.
In the second example, optimal exchange is to buy 5 euro and 1 dollar.
In the third example, optimal exchange is to buy 10 dollars in one bill.
Submitted Solution:
```
total, usd, euro = int(input()), int(input()), int(input())
out = total % usd
euro_div = 5 * euro
while euro_div <= total:
out = min(out, (total - euro_div) % usd)
if out == 0:
break
euro_div += 5 * euro
print(out)
``` | instruction | 0 | 90,220 | 10 | 180,440 |
Yes | output | 1 | 90,220 | 10 | 180,441 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrew was very excited to participate in Olympiad of Metropolises. Days flew by quickly, and Andrew is already at the airport, ready to go home. He has n rubles left, and would like to exchange them to euro and dollar bills. Andrew can mix dollar bills and euro bills in whatever way he wants. The price of one dollar is d rubles, and one euro costs e rubles.
Recall that there exist the following dollar bills: 1, 2, 5, 10, 20, 50, 100, and the following euro bills — 5, 10, 20, 50, 100, 200 (note that, in this problem we do not consider the 500 euro bill, it is hard to find such bills in the currency exchange points). Andrew can buy any combination of bills, and his goal is to minimize the total number of rubles he will have after the exchange.
Help him — write a program that given integers n, e and d, finds the minimum number of rubles Andrew can get after buying dollar and euro bills.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^8) — the initial sum in rubles Andrew has.
The second line of the input contains one integer d (30 ≤ d ≤ 100) — the price of one dollar in rubles.
The third line of the input contains integer e (30 ≤ e ≤ 100) — the price of one euro in rubles.
Output
Output one integer — the minimum number of rubles Andrew can have after buying dollar and euro bills optimally.
Examples
Input
100
60
70
Output
40
Input
410
55
70
Output
5
Input
600
60
70
Output
0
Note
In the first example, we can buy just 1 dollar because there is no 1 euro bill.
In the second example, optimal exchange is to buy 5 euro and 1 dollar.
In the third example, optimal exchange is to buy 10 dollars in one bill.
Submitted Solution:
```
n = int(input())
d = int(input())
e = int(input())
e = e*5
edge = n//e
r = min(n%e, n%d)
while edge!=0:
x = n - e * edge
r = min(x%d, r)
edge -=1
print(r)
``` | instruction | 0 | 90,221 | 10 | 180,442 |
Yes | output | 1 | 90,221 | 10 | 180,443 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrew was very excited to participate in Olympiad of Metropolises. Days flew by quickly, and Andrew is already at the airport, ready to go home. He has n rubles left, and would like to exchange them to euro and dollar bills. Andrew can mix dollar bills and euro bills in whatever way he wants. The price of one dollar is d rubles, and one euro costs e rubles.
Recall that there exist the following dollar bills: 1, 2, 5, 10, 20, 50, 100, and the following euro bills — 5, 10, 20, 50, 100, 200 (note that, in this problem we do not consider the 500 euro bill, it is hard to find such bills in the currency exchange points). Andrew can buy any combination of bills, and his goal is to minimize the total number of rubles he will have after the exchange.
Help him — write a program that given integers n, e and d, finds the minimum number of rubles Andrew can get after buying dollar and euro bills.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^8) — the initial sum in rubles Andrew has.
The second line of the input contains one integer d (30 ≤ d ≤ 100) — the price of one dollar in rubles.
The third line of the input contains integer e (30 ≤ e ≤ 100) — the price of one euro in rubles.
Output
Output one integer — the minimum number of rubles Andrew can have after buying dollar and euro bills optimally.
Examples
Input
100
60
70
Output
40
Input
410
55
70
Output
5
Input
600
60
70
Output
0
Note
In the first example, we can buy just 1 dollar because there is no 1 euro bill.
In the second example, optimal exchange is to buy 5 euro and 1 dollar.
In the third example, optimal exchange is to buy 10 dollars in one bill.
Submitted Solution:
```
import math
import sys
from collections import defaultdict
# input = sys.stdin.readline
rt = lambda: map(int, input().split())
ri = lambda: int(input())
rl = lambda: list(map(int, input().split()))
def main():
n = ri()
e = ri()
d = ri()
print(min((n%(e))%(5*d), (n%(5*d))%e))
if __name__ == '__main__':
main()
``` | instruction | 0 | 90,222 | 10 | 180,444 |
No | output | 1 | 90,222 | 10 | 180,445 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrew was very excited to participate in Olympiad of Metropolises. Days flew by quickly, and Andrew is already at the airport, ready to go home. He has n rubles left, and would like to exchange them to euro and dollar bills. Andrew can mix dollar bills and euro bills in whatever way he wants. The price of one dollar is d rubles, and one euro costs e rubles.
Recall that there exist the following dollar bills: 1, 2, 5, 10, 20, 50, 100, and the following euro bills — 5, 10, 20, 50, 100, 200 (note that, in this problem we do not consider the 500 euro bill, it is hard to find such bills in the currency exchange points). Andrew can buy any combination of bills, and his goal is to minimize the total number of rubles he will have after the exchange.
Help him — write a program that given integers n, e and d, finds the minimum number of rubles Andrew can get after buying dollar and euro bills.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^8) — the initial sum in rubles Andrew has.
The second line of the input contains one integer d (30 ≤ d ≤ 100) — the price of one dollar in rubles.
The third line of the input contains integer e (30 ≤ e ≤ 100) — the price of one euro in rubles.
Output
Output one integer — the minimum number of rubles Andrew can have after buying dollar and euro bills optimally.
Examples
Input
100
60
70
Output
40
Input
410
55
70
Output
5
Input
600
60
70
Output
0
Note
In the first example, we can buy just 1 dollar because there is no 1 euro bill.
In the second example, optimal exchange is to buy 5 euro and 1 dollar.
In the third example, optimal exchange is to buy 10 dollars in one bill.
Submitted Solution:
```
n=int(input())
m=int(input())
k=int(input())*5
f=1
g=1
for i in range(k,m,k):
if i%m==0:
f=0
break
for j in range(m,n,m):
if j%k==0:
g=0
break
a=n%k
if a>=m:
a=a%m
b=n%m
if b>=k:
b=b%k
for i in range(m,n,m):
if i%k==0:
f=0
print(min(a,b,f,g))
``` | instruction | 0 | 90,223 | 10 | 180,446 |
No | output | 1 | 90,223 | 10 | 180,447 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrew was very excited to participate in Olympiad of Metropolises. Days flew by quickly, and Andrew is already at the airport, ready to go home. He has n rubles left, and would like to exchange them to euro and dollar bills. Andrew can mix dollar bills and euro bills in whatever way he wants. The price of one dollar is d rubles, and one euro costs e rubles.
Recall that there exist the following dollar bills: 1, 2, 5, 10, 20, 50, 100, and the following euro bills — 5, 10, 20, 50, 100, 200 (note that, in this problem we do not consider the 500 euro bill, it is hard to find such bills in the currency exchange points). Andrew can buy any combination of bills, and his goal is to minimize the total number of rubles he will have after the exchange.
Help him — write a program that given integers n, e and d, finds the minimum number of rubles Andrew can get after buying dollar and euro bills.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^8) — the initial sum in rubles Andrew has.
The second line of the input contains one integer d (30 ≤ d ≤ 100) — the price of one dollar in rubles.
The third line of the input contains integer e (30 ≤ e ≤ 100) — the price of one euro in rubles.
Output
Output one integer — the minimum number of rubles Andrew can have after buying dollar and euro bills optimally.
Examples
Input
100
60
70
Output
40
Input
410
55
70
Output
5
Input
600
60
70
Output
0
Note
In the first example, we can buy just 1 dollar because there is no 1 euro bill.
In the second example, optimal exchange is to buy 5 euro and 1 dollar.
In the third example, optimal exchange is to buy 10 dollars in one bill.
Submitted Solution:
```
n = int(input())
d = int(input())
e = int(input())
dollars = {1, 2, 5, 10, 20, 50, 100}
euros = {5, 10, 20, 50, 100, 200}
values = set()
for i in dollars:
values.add(i * d)
for i in euros:
values.add(i * e)
values = list(values)
values.sort()
len = len(values)
for i in range(len + 1):
if (i == len or values[i] > n):
break
if (i != 0):
i -= 1
n -= values[i]
while (i >= 0):
if (values[i] <= n):
n -= values[i]
else:
i -= 1
print(n)
``` | instruction | 0 | 90,224 | 10 | 180,448 |
No | output | 1 | 90,224 | 10 | 180,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrew was very excited to participate in Olympiad of Metropolises. Days flew by quickly, and Andrew is already at the airport, ready to go home. He has n rubles left, and would like to exchange them to euro and dollar bills. Andrew can mix dollar bills and euro bills in whatever way he wants. The price of one dollar is d rubles, and one euro costs e rubles.
Recall that there exist the following dollar bills: 1, 2, 5, 10, 20, 50, 100, and the following euro bills — 5, 10, 20, 50, 100, 200 (note that, in this problem we do not consider the 500 euro bill, it is hard to find such bills in the currency exchange points). Andrew can buy any combination of bills, and his goal is to minimize the total number of rubles he will have after the exchange.
Help him — write a program that given integers n, e and d, finds the minimum number of rubles Andrew can get after buying dollar and euro bills.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^8) — the initial sum in rubles Andrew has.
The second line of the input contains one integer d (30 ≤ d ≤ 100) — the price of one dollar in rubles.
The third line of the input contains integer e (30 ≤ e ≤ 100) — the price of one euro in rubles.
Output
Output one integer — the minimum number of rubles Andrew can have after buying dollar and euro bills optimally.
Examples
Input
100
60
70
Output
40
Input
410
55
70
Output
5
Input
600
60
70
Output
0
Note
In the first example, we can buy just 1 dollar because there is no 1 euro bill.
In the second example, optimal exchange is to buy 5 euro and 1 dollar.
In the third example, optimal exchange is to buy 10 dollars in one bill.
Submitted Solution:
```
n = int(input())
d = int(input())
e = int(input())
n -= ((n // (e * 200)) * (e * 200))
n -= ((n // (e * 100)) * (e * 100))
k1 = ((n // (e * 100)) * (e * 100))
k2 = ((n // (d * 100)) * (d * 100))
n -= max(k1, k2)
k1 = ((n // (e * 50)) * (e * 50))
k2 = ((n // (d * 50)) * (d * 50))
n -= max(k1, k2)
k1 = ((n // (e * 20)) * (e * 20))
k2 = ((n // (d * 20)) * (d * 20))
n -= max(k1, k2)
k1 = ((n // (e * 10)) * (e * 10))
k2 = ((n // (d * 10)) * (d * 10))
n -= max(k1, k2)
k1 = ((n // (e * 5)) * (e * 5))
k2 = ((n // (d * 5)) * (d * 5))
n -= max(k1, k2)
n -= ((n // (d * 2)) * (d * 2))
n -= ((n // d) * d)
print(n)
``` | instruction | 0 | 90,225 | 10 | 180,450 |
No | output | 1 | 90,225 | 10 | 180,451 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Bill is shopping at the store. There are some coins in his wallet (10-yen coins, 50-yen coins, 100-yen coins, 500-yen coins), but he is now trying to consume as much of this coin as possible. In other words, by paying for the goods with an appropriate number of coins, we are trying to minimize the total number of coins after receiving the change.
Fortunately, the clerk at this store is so disciplined and kind that change is always delivered in the best possible way. Therefore, for example, five 100-yen coins will not be given instead of one 500-yen coin. You can also take out 5 10-yen coins and receive 50-yen coins as change. However, you must not pay the coins of the same type as the coins you issued so that they will be returned as change. For example, if a 10-yen coin is paid out and another 10-yen coin is returned as change, a completely meaningless exchange will occur.
However, Mr. Bill was not good at calculating, so he could not find out how many coins he should actually use. So he asked you for help. Your job is to write a program to determine the type and number of coins to use based on the number of coins in his wallet and the payment price. The clerk does not use banknotes for change.
Input
The input contains several test cases.
Each test case consists of two lines. The first line contains a single integer representing Mr. Bill's payment in yen. The second line contains four integers, which in turn represent the number of 10-yen coins, 50-yen coins, 100-yen coins, and 500-yen coins in your wallet.
The payment amount is always in units of 10 yen. That is, the one-yen place of the payment amount is always 0. You can also assume that you can only have up to 20 coins of the same type in your wallet. No non-payable cases are given during input.
The end of input is represented by a line containing a single 0.
Output
For each test case, print out the type and number of coins that Mr. Bill should use.
Each line of output contains the two integers ci and ki. This means using ki coins for ci yen when paying. When using multiple types of coins, output as many lines as necessary in order from the smallest ci. Refer to the output example below.
The output must not include extra space. Separate consecutive test cases with a blank line.
Example
Input
160
1 1 2 0
160
1 0 2 10
0
Output
10 1
50 1
100 1
10 1
100 2 | instruction | 0 | 90,905 | 10 | 181,810 |
"Correct Solution:
```
def back_oturigation(fee, coin_values, coin_nums):
"""
1. ?????????????????????????????????£??????????????????
2. 1????????????????????????????????????????????????????????????
????????????§?????????????????????-> ??°?????????????????????¶???????????????????
3. 2?????????????????????°?????????????????????????????????????????????
"""
# 1.
oturi = coin_values[0] * coin_nums[0] \
+ coin_values[1] * coin_nums[1] \
+ coin_values[2] * coin_nums[2] \
+ coin_values[3] * coin_nums[3] \
- fee
use_coins = [0] * len(coin_values)
no_use_coins = [0] * len(coin_values)
# 2.
for i in range(len(coin_values) - 1, -1, -1):
no_use_coins[i] = int(oturi / coin_values[i])
oturi = oturi - (coin_values[i] * no_use_coins[i])
# 3.
for i in range(0, len(use_coins), 1):
use_coins[i] = coin_nums[i] - no_use_coins[i]
if use_coins[i] > 0:
print(str(coin_values[i]) + ' ' + str(use_coins[i]))
if __name__ == "__main__":
first = True
while (True):
total_fee = int(input(''))
if total_fee == 0:
break
if first:
first = False
else:
print('')
coins = input('')
coins = coins.split(' ') # 10, 50, 100, 500?????????§????????????
coins = [int(coin) for coin in coins]
coin_values = [10, 50, 100, 500]
back_oturigation(total_fee, coin_values, coins)
``` | output | 1 | 90,905 | 10 | 181,811 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Bill is shopping at the store. There are some coins in his wallet (10-yen coins, 50-yen coins, 100-yen coins, 500-yen coins), but he is now trying to consume as much of this coin as possible. In other words, by paying for the goods with an appropriate number of coins, we are trying to minimize the total number of coins after receiving the change.
Fortunately, the clerk at this store is so disciplined and kind that change is always delivered in the best possible way. Therefore, for example, five 100-yen coins will not be given instead of one 500-yen coin. You can also take out 5 10-yen coins and receive 50-yen coins as change. However, you must not pay the coins of the same type as the coins you issued so that they will be returned as change. For example, if a 10-yen coin is paid out and another 10-yen coin is returned as change, a completely meaningless exchange will occur.
However, Mr. Bill was not good at calculating, so he could not find out how many coins he should actually use. So he asked you for help. Your job is to write a program to determine the type and number of coins to use based on the number of coins in his wallet and the payment price. The clerk does not use banknotes for change.
Input
The input contains several test cases.
Each test case consists of two lines. The first line contains a single integer representing Mr. Bill's payment in yen. The second line contains four integers, which in turn represent the number of 10-yen coins, 50-yen coins, 100-yen coins, and 500-yen coins in your wallet.
The payment amount is always in units of 10 yen. That is, the one-yen place of the payment amount is always 0. You can also assume that you can only have up to 20 coins of the same type in your wallet. No non-payable cases are given during input.
The end of input is represented by a line containing a single 0.
Output
For each test case, print out the type and number of coins that Mr. Bill should use.
Each line of output contains the two integers ci and ki. This means using ki coins for ci yen when paying. When using multiple types of coins, output as many lines as necessary in order from the smallest ci. Refer to the output example below.
The output must not include extra space. Separate consecutive test cases with a blank line.
Example
Input
160
1 1 2 0
160
1 0 2 10
0
Output
10 1
50 1
100 1
10 1
100 2 | instruction | 0 | 90,906 | 10 | 181,812 |
"Correct Solution:
```
first = True
while True:
P = int(input())
if P== 0: break
if first:
first = False
else:
print('')
c1,c2,c3,c4 = map(int,input().split())
v = c1*10 + c2*50 + c3*100 + c4*500
n = c1 + c2 + c3 + c4
ans = {}
rem = v - P
ans[10] = c1 - (rem//10) % 5
ans[50] = c2 - (rem//50) % 2
ans[100] = c3 - (rem//100) % 5
ans[500] = c4 - rem//500
for k,v in sorted(ans.items()):
if v <= 0: continue
print('{0} {1}'.format(k,v))
``` | output | 1 | 90,906 | 10 | 181,813 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Bill is shopping at the store. There are some coins in his wallet (10-yen coins, 50-yen coins, 100-yen coins, 500-yen coins), but he is now trying to consume as much of this coin as possible. In other words, by paying for the goods with an appropriate number of coins, we are trying to minimize the total number of coins after receiving the change.
Fortunately, the clerk at this store is so disciplined and kind that change is always delivered in the best possible way. Therefore, for example, five 100-yen coins will not be given instead of one 500-yen coin. You can also take out 5 10-yen coins and receive 50-yen coins as change. However, you must not pay the coins of the same type as the coins you issued so that they will be returned as change. For example, if a 10-yen coin is paid out and another 10-yen coin is returned as change, a completely meaningless exchange will occur.
However, Mr. Bill was not good at calculating, so he could not find out how many coins he should actually use. So he asked you for help. Your job is to write a program to determine the type and number of coins to use based on the number of coins in his wallet and the payment price. The clerk does not use banknotes for change.
Input
The input contains several test cases.
Each test case consists of two lines. The first line contains a single integer representing Mr. Bill's payment in yen. The second line contains four integers, which in turn represent the number of 10-yen coins, 50-yen coins, 100-yen coins, and 500-yen coins in your wallet.
The payment amount is always in units of 10 yen. That is, the one-yen place of the payment amount is always 0. You can also assume that you can only have up to 20 coins of the same type in your wallet. No non-payable cases are given during input.
The end of input is represented by a line containing a single 0.
Output
For each test case, print out the type and number of coins that Mr. Bill should use.
Each line of output contains the two integers ci and ki. This means using ki coins for ci yen when paying. When using multiple types of coins, output as many lines as necessary in order from the smallest ci. Refer to the output example below.
The output must not include extra space. Separate consecutive test cases with a blank line.
Example
Input
160
1 1 2 0
160
1 0 2 10
0
Output
10 1
50 1
100 1
10 1
100 2 | instruction | 0 | 90,907 | 10 | 181,814 |
"Correct Solution:
```
def back_oturigation(fee, coin_values, coin_nums):
"""
1. ????????????????????????????????£??????????????????
2. 1????????????????????????????????????????????????????????????
???????????§?????????????????????-> ?°????????????????????¶???????????????????
3. 2????????????????????°?????????????????????????????????????????????
"""
# 1.
oturi = coin_values[0] * coin_nums[0] \
+ coin_values[1] * coin_nums[1] \
+ coin_values[2] * coin_nums[2] \
+ coin_values[3] * coin_nums[3] \
- fee
use_coins = [0] * len(coin_values)
no_use_coins = [0] * len(coin_values)
# 2.
for i in range(len(coin_values) - 1, -1, -1):
no_use_coins[i] = int(oturi / coin_values[i])
oturi = oturi - (coin_values[i] * no_use_coins[i])
# 3.
for i in range(0, len(use_coins), 1):
use_coins[i] = coin_nums[i] - no_use_coins[i]
if use_coins[i] > 0:
print(str(coin_values[i]) + ' ' + str(use_coins[i]))
if __name__ == "__main__":
first = True
while (True):
total_fee = int(input(''))
if total_fee == 0:
break
if first:
first = False
else:
print('')
coins = input('')
coins = coins.split(' ') # 10, 50, 100, 500????????§????????????
coins = [int(coin) for coin in coins]
coin_values = [10, 50, 100, 500]
back_oturigation(total_fee, coin_values, coins)
``` | output | 1 | 90,907 | 10 | 181,815 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Bill is shopping at the store. There are some coins in his wallet (10-yen coins, 50-yen coins, 100-yen coins, 500-yen coins), but he is now trying to consume as much of this coin as possible. In other words, by paying for the goods with an appropriate number of coins, we are trying to minimize the total number of coins after receiving the change.
Fortunately, the clerk at this store is so disciplined and kind that change is always delivered in the best possible way. Therefore, for example, five 100-yen coins will not be given instead of one 500-yen coin. You can also take out 5 10-yen coins and receive 50-yen coins as change. However, you must not pay the coins of the same type as the coins you issued so that they will be returned as change. For example, if a 10-yen coin is paid out and another 10-yen coin is returned as change, a completely meaningless exchange will occur.
However, Mr. Bill was not good at calculating, so he could not find out how many coins he should actually use. So he asked you for help. Your job is to write a program to determine the type and number of coins to use based on the number of coins in his wallet and the payment price. The clerk does not use banknotes for change.
Input
The input contains several test cases.
Each test case consists of two lines. The first line contains a single integer representing Mr. Bill's payment in yen. The second line contains four integers, which in turn represent the number of 10-yen coins, 50-yen coins, 100-yen coins, and 500-yen coins in your wallet.
The payment amount is always in units of 10 yen. That is, the one-yen place of the payment amount is always 0. You can also assume that you can only have up to 20 coins of the same type in your wallet. No non-payable cases are given during input.
The end of input is represented by a line containing a single 0.
Output
For each test case, print out the type and number of coins that Mr. Bill should use.
Each line of output contains the two integers ci and ki. This means using ki coins for ci yen when paying. When using multiple types of coins, output as many lines as necessary in order from the smallest ci. Refer to the output example below.
The output must not include extra space. Separate consecutive test cases with a blank line.
Example
Input
160
1 1 2 0
160
1 0 2 10
0
Output
10 1
50 1
100 1
10 1
100 2 | instruction | 0 | 90,908 | 10 | 181,816 |
"Correct Solution:
```
INF = 10 ** 15
MOD = 10 ** 9 + 7
money = (10,50,100,500)
def solve(N):
coins = list(map(int,input().split()))
tot = sum(coins[i]*money[i] for i in range(4))
ans = [0] * 4
tot -= N
for i in range(3,-1,-1):
X = tot//money[i]
tot %= money[i]
ans[i] = max(coins[i] - X,0)
for i in range(4):
if ans[i] == 0:
continue
print(money[i],ans[i])
def main():
i = 0
while True:
N = int(input())
if N == 0:
break
if i > 0:
print('')
i += 1
solve(N)
if __name__ == '__main__':
main()
``` | output | 1 | 90,908 | 10 | 181,817 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Bill is shopping at the store. There are some coins in his wallet (10-yen coins, 50-yen coins, 100-yen coins, 500-yen coins), but he is now trying to consume as much of this coin as possible. In other words, by paying for the goods with an appropriate number of coins, we are trying to minimize the total number of coins after receiving the change.
Fortunately, the clerk at this store is so disciplined and kind that change is always delivered in the best possible way. Therefore, for example, five 100-yen coins will not be given instead of one 500-yen coin. You can also take out 5 10-yen coins and receive 50-yen coins as change. However, you must not pay the coins of the same type as the coins you issued so that they will be returned as change. For example, if a 10-yen coin is paid out and another 10-yen coin is returned as change, a completely meaningless exchange will occur.
However, Mr. Bill was not good at calculating, so he could not find out how many coins he should actually use. So he asked you for help. Your job is to write a program to determine the type and number of coins to use based on the number of coins in his wallet and the payment price. The clerk does not use banknotes for change.
Input
The input contains several test cases.
Each test case consists of two lines. The first line contains a single integer representing Mr. Bill's payment in yen. The second line contains four integers, which in turn represent the number of 10-yen coins, 50-yen coins, 100-yen coins, and 500-yen coins in your wallet.
The payment amount is always in units of 10 yen. That is, the one-yen place of the payment amount is always 0. You can also assume that you can only have up to 20 coins of the same type in your wallet. No non-payable cases are given during input.
The end of input is represented by a line containing a single 0.
Output
For each test case, print out the type and number of coins that Mr. Bill should use.
Each line of output contains the two integers ci and ki. This means using ki coins for ci yen when paying. When using multiple types of coins, output as many lines as necessary in order from the smallest ci. Refer to the output example below.
The output must not include extra space. Separate consecutive test cases with a blank line.
Example
Input
160
1 1 2 0
160
1 0 2 10
0
Output
10 1
50 1
100 1
10 1
100 2 | instruction | 0 | 90,909 | 10 | 181,818 |
"Correct Solution:
```
ansOut = []
coin = [10, 50, 100, 500]
while True:
price = int(input())
if price == 0:
break
cash = list(map(int, input().split()))
sumCash = sum(c * n for c, n in zip(coin, cash))
change = sumCash - price
changeCoins = [(change % 50) // 10, (change % 100) // 50, (change % 500) // 100, change // 500]
out = []
for i in range(4):
if cash[i] > changeCoins[i]:
out.append('{} {}'.format(coin[i], cash[i] - changeCoins[i]))
ansOut.append('\n'.join(out))
print('\n\n'.join(ansOut))
``` | output | 1 | 90,909 | 10 | 181,819 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Bill is shopping at the store. There are some coins in his wallet (10-yen coins, 50-yen coins, 100-yen coins, 500-yen coins), but he is now trying to consume as much of this coin as possible. In other words, by paying for the goods with an appropriate number of coins, we are trying to minimize the total number of coins after receiving the change.
Fortunately, the clerk at this store is so disciplined and kind that change is always delivered in the best possible way. Therefore, for example, five 100-yen coins will not be given instead of one 500-yen coin. You can also take out 5 10-yen coins and receive 50-yen coins as change. However, you must not pay the coins of the same type as the coins you issued so that they will be returned as change. For example, if a 10-yen coin is paid out and another 10-yen coin is returned as change, a completely meaningless exchange will occur.
However, Mr. Bill was not good at calculating, so he could not find out how many coins he should actually use. So he asked you for help. Your job is to write a program to determine the type and number of coins to use based on the number of coins in his wallet and the payment price. The clerk does not use banknotes for change.
Input
The input contains several test cases.
Each test case consists of two lines. The first line contains a single integer representing Mr. Bill's payment in yen. The second line contains four integers, which in turn represent the number of 10-yen coins, 50-yen coins, 100-yen coins, and 500-yen coins in your wallet.
The payment amount is always in units of 10 yen. That is, the one-yen place of the payment amount is always 0. You can also assume that you can only have up to 20 coins of the same type in your wallet. No non-payable cases are given during input.
The end of input is represented by a line containing a single 0.
Output
For each test case, print out the type and number of coins that Mr. Bill should use.
Each line of output contains the two integers ci and ki. This means using ki coins for ci yen when paying. When using multiple types of coins, output as many lines as necessary in order from the smallest ci. Refer to the output example below.
The output must not include extra space. Separate consecutive test cases with a blank line.
Example
Input
160
1 1 2 0
160
1 0 2 10
0
Output
10 1
50 1
100 1
10 1
100 2 | instruction | 0 | 90,910 | 10 | 181,820 |
"Correct Solution:
```
coin = [10,50,100,500,999999]
J = True
while True:
n = int(input())
if n == 0:
break
else:
if not J:
print()
else:
J = False
L = list(map(int,input().split()))
S = 10*L[0]+50*L[1]+100*L[2]+500*L[3]
res = S-n #とりあえず硬貨を全部使ったものと仮定する
change = [res%coin[i+1]//coin[i] for i in range(4)] #ここでお釣りとしてどの硬貨が何枚戻ってくるかを計算している
for i in range(4):
if change[i] < L[i]: #お釣りとして戻ってくる枚数の方が少なければ採用
print(coin[i],L[i]-change[i])
``` | output | 1 | 90,910 | 10 | 181,821 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Bill is shopping at the store. There are some coins in his wallet (10-yen coins, 50-yen coins, 100-yen coins, 500-yen coins), but he is now trying to consume as much of this coin as possible. In other words, by paying for the goods with an appropriate number of coins, we are trying to minimize the total number of coins after receiving the change.
Fortunately, the clerk at this store is so disciplined and kind that change is always delivered in the best possible way. Therefore, for example, five 100-yen coins will not be given instead of one 500-yen coin. You can also take out 5 10-yen coins and receive 50-yen coins as change. However, you must not pay the coins of the same type as the coins you issued so that they will be returned as change. For example, if a 10-yen coin is paid out and another 10-yen coin is returned as change, a completely meaningless exchange will occur.
However, Mr. Bill was not good at calculating, so he could not find out how many coins he should actually use. So he asked you for help. Your job is to write a program to determine the type and number of coins to use based on the number of coins in his wallet and the payment price. The clerk does not use banknotes for change.
Input
The input contains several test cases.
Each test case consists of two lines. The first line contains a single integer representing Mr. Bill's payment in yen. The second line contains four integers, which in turn represent the number of 10-yen coins, 50-yen coins, 100-yen coins, and 500-yen coins in your wallet.
The payment amount is always in units of 10 yen. That is, the one-yen place of the payment amount is always 0. You can also assume that you can only have up to 20 coins of the same type in your wallet. No non-payable cases are given during input.
The end of input is represented by a line containing a single 0.
Output
For each test case, print out the type and number of coins that Mr. Bill should use.
Each line of output contains the two integers ci and ki. This means using ki coins for ci yen when paying. When using multiple types of coins, output as many lines as necessary in order from the smallest ci. Refer to the output example below.
The output must not include extra space. Separate consecutive test cases with a blank line.
Example
Input
160
1 1 2 0
160
1 0 2 10
0
Output
10 1
50 1
100 1
10 1
100 2 | instruction | 0 | 90,911 | 10 | 181,822 |
"Correct Solution:
```
clst = [10,50,100,500]
def sgn(t):
for I in range(0,4):
if clst[I] == t:
return I
def cnt(l1,l2,N):
s = sgn(l1[N])
l2[s] = l2[s] + 1
def shiharai(bill,purse):
B = bill
lst1 = []
lst2 = [0]
for i in range(0,4):
for j in range(0,purse[i]):
lst1.append(clst[i])
lst2[0] = lst1[0]
for i in range(1,len(lst1)):
k = lst1[i] + lst2[i-1]
lst2.append(k)
pay = [0,0,0,0]
while B > 0:
i = 0
while lst2[i] < B:
i = i + 1
if lst2[i] == B:
for j in range(0,i+1):
cnt(lst1,pay,j)
B = B - lst1[j]
else:
cnt(lst1,pay,i)
B = B - lst1[i]
for i in range(4):
purse[i] = purse[i] - pay[i]
purse[0] = purse[0] - B//10
return purse
def exchng(purse,n,r):
while purse[n] >= r:
purse[n] = purse[n] - r
purse[n+1] = purse[n+1] + 1
def ryogae(purse):
exchng(purse,0,5)
exchng(purse,1,2)
exchng(purse,2,5)
return purse
L = 0
while True:
bill = int(input().strip())
if bill == 0:
break
if L != 0:
print()
lst = list(map(int,input().strip().split(" ")))
purse = [0,0,0,0]
for A in range(0,4):
purse[A] = lst[A]
purse = shiharai(bill,purse)
purse = ryogae(purse)
shouldpay = [0,0,0,0]
for B in range(0,4):
if lst[B] > purse[B]:
shouldpay[B] = lst[B] - purse[B]
for i in range(0,4):
if shouldpay[i] != 0:
print(clst[i],shouldpay[i])
L = L + 1
``` | output | 1 | 90,911 | 10 | 181,823 |
Provide a correct Python 3 solution for this coding contest problem.
Mr. Bill is shopping at the store. There are some coins in his wallet (10-yen coins, 50-yen coins, 100-yen coins, 500-yen coins), but he is now trying to consume as much of this coin as possible. In other words, by paying for the goods with an appropriate number of coins, we are trying to minimize the total number of coins after receiving the change.
Fortunately, the clerk at this store is so disciplined and kind that change is always delivered in the best possible way. Therefore, for example, five 100-yen coins will not be given instead of one 500-yen coin. You can also take out 5 10-yen coins and receive 50-yen coins as change. However, you must not pay the coins of the same type as the coins you issued so that they will be returned as change. For example, if a 10-yen coin is paid out and another 10-yen coin is returned as change, a completely meaningless exchange will occur.
However, Mr. Bill was not good at calculating, so he could not find out how many coins he should actually use. So he asked you for help. Your job is to write a program to determine the type and number of coins to use based on the number of coins in his wallet and the payment price. The clerk does not use banknotes for change.
Input
The input contains several test cases.
Each test case consists of two lines. The first line contains a single integer representing Mr. Bill's payment in yen. The second line contains four integers, which in turn represent the number of 10-yen coins, 50-yen coins, 100-yen coins, and 500-yen coins in your wallet.
The payment amount is always in units of 10 yen. That is, the one-yen place of the payment amount is always 0. You can also assume that you can only have up to 20 coins of the same type in your wallet. No non-payable cases are given during input.
The end of input is represented by a line containing a single 0.
Output
For each test case, print out the type and number of coins that Mr. Bill should use.
Each line of output contains the two integers ci and ki. This means using ki coins for ci yen when paying. When using multiple types of coins, output as many lines as necessary in order from the smallest ci. Refer to the output example below.
The output must not include extra space. Separate consecutive test cases with a blank line.
Example
Input
160
1 1 2 0
160
1 0 2 10
0
Output
10 1
50 1
100 1
10 1
100 2 | instruction | 0 | 90,912 | 10 | 181,824 |
"Correct Solution:
```
coins = [10, 50, 100, 500]
def s(value):
a = list(map(int, input().split()))
pay = [0, 0, 0, 0]
ms = sum([a[i] * coins[i] for i in range(4)]) - value
m = [ms % 50 // 10, ms % 100 // 50, ms % 500 // 100, ms // 500]
for cc, aa, mm in zip(coins, a, m):
if aa > mm:
print(cc, aa - mm)
s(int(input()))
while True:
value = int(input())
if not value:
break
print()
s(value)
``` | output | 1 | 90,912 | 10 | 181,825 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Bill is shopping at the store. There are some coins in his wallet (10-yen coins, 50-yen coins, 100-yen coins, 500-yen coins), but he is now trying to consume as much of this coin as possible. In other words, by paying for the goods with an appropriate number of coins, we are trying to minimize the total number of coins after receiving the change.
Fortunately, the clerk at this store is so disciplined and kind that change is always delivered in the best possible way. Therefore, for example, five 100-yen coins will not be given instead of one 500-yen coin. You can also take out 5 10-yen coins and receive 50-yen coins as change. However, you must not pay the coins of the same type as the coins you issued so that they will be returned as change. For example, if a 10-yen coin is paid out and another 10-yen coin is returned as change, a completely meaningless exchange will occur.
However, Mr. Bill was not good at calculating, so he could not find out how many coins he should actually use. So he asked you for help. Your job is to write a program to determine the type and number of coins to use based on the number of coins in his wallet and the payment price. The clerk does not use banknotes for change.
Input
The input contains several test cases.
Each test case consists of two lines. The first line contains a single integer representing Mr. Bill's payment in yen. The second line contains four integers, which in turn represent the number of 10-yen coins, 50-yen coins, 100-yen coins, and 500-yen coins in your wallet.
The payment amount is always in units of 10 yen. That is, the one-yen place of the payment amount is always 0. You can also assume that you can only have up to 20 coins of the same type in your wallet. No non-payable cases are given during input.
The end of input is represented by a line containing a single 0.
Output
For each test case, print out the type and number of coins that Mr. Bill should use.
Each line of output contains the two integers ci and ki. This means using ki coins for ci yen when paying. When using multiple types of coins, output as many lines as necessary in order from the smallest ci. Refer to the output example below.
The output must not include extra space. Separate consecutive test cases with a blank line.
Example
Input
160
1 1 2 0
160
1 0 2 10
0
Output
10 1
50 1
100 1
10 1
100 2
Submitted Solution:
```
coin=[10, 50, 100, 500, 500000]
first = True
while True:
bill = int(input())
if bill == 0:
break;
if not first:
print()
else:
first = False
posses = list(map(int, input().split()))
Sumcoin= sum(coin[i] * posses[i] for i in range(4))
change = Sumcoin - bill
ChangeCoin = [change % coin[i+1] // coin[i] for i in range(4)]
hand = [posses[i] - ChangeCoin[i] for i in range(4)]
for i in range(4):
if posses[i] > ChangeCoin[i]:
print(coin[i], hand[i])
``` | instruction | 0 | 90,913 | 10 | 181,826 |
Yes | output | 1 | 90,913 | 10 | 181,827 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Bill is shopping at the store. There are some coins in his wallet (10-yen coins, 50-yen coins, 100-yen coins, 500-yen coins), but he is now trying to consume as much of this coin as possible. In other words, by paying for the goods with an appropriate number of coins, we are trying to minimize the total number of coins after receiving the change.
Fortunately, the clerk at this store is so disciplined and kind that change is always delivered in the best possible way. Therefore, for example, five 100-yen coins will not be given instead of one 500-yen coin. You can also take out 5 10-yen coins and receive 50-yen coins as change. However, you must not pay the coins of the same type as the coins you issued so that they will be returned as change. For example, if a 10-yen coin is paid out and another 10-yen coin is returned as change, a completely meaningless exchange will occur.
However, Mr. Bill was not good at calculating, so he could not find out how many coins he should actually use. So he asked you for help. Your job is to write a program to determine the type and number of coins to use based on the number of coins in his wallet and the payment price. The clerk does not use banknotes for change.
Input
The input contains several test cases.
Each test case consists of two lines. The first line contains a single integer representing Mr. Bill's payment in yen. The second line contains four integers, which in turn represent the number of 10-yen coins, 50-yen coins, 100-yen coins, and 500-yen coins in your wallet.
The payment amount is always in units of 10 yen. That is, the one-yen place of the payment amount is always 0. You can also assume that you can only have up to 20 coins of the same type in your wallet. No non-payable cases are given during input.
The end of input is represented by a line containing a single 0.
Output
For each test case, print out the type and number of coins that Mr. Bill should use.
Each line of output contains the two integers ci and ki. This means using ki coins for ci yen when paying. When using multiple types of coins, output as many lines as necessary in order from the smallest ci. Refer to the output example below.
The output must not include extra space. Separate consecutive test cases with a blank line.
Example
Input
160
1 1 2 0
160
1 0 2 10
0
Output
10 1
50 1
100 1
10 1
100 2
Submitted Solution:
```
cnt = 0
while True:
N = int(input())
if N == 0:
break
if cnt != 0:
print ("")
cnt = 1
A,B,C,D = map(int,input().split())
mon = A * 10 + B * 50 + C * 100 + D * 500
mon -= N
d = mon // 500
mon %= 500
c = mon // 100
mon %= 100
b = mon // 50
mon %= 50
a = mon // 10
if A-a > 0:
print (10,A-a)
if B-b > 0:
print (50,B-b)
if C-c > 0:
print (100,C-c)
if D-d > 0:
print (500,D-d)
``` | instruction | 0 | 90,914 | 10 | 181,828 |
Yes | output | 1 | 90,914 | 10 | 181,829 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Bill is shopping at the store. There are some coins in his wallet (10-yen coins, 50-yen coins, 100-yen coins, 500-yen coins), but he is now trying to consume as much of this coin as possible. In other words, by paying for the goods with an appropriate number of coins, we are trying to minimize the total number of coins after receiving the change.
Fortunately, the clerk at this store is so disciplined and kind that change is always delivered in the best possible way. Therefore, for example, five 100-yen coins will not be given instead of one 500-yen coin. You can also take out 5 10-yen coins and receive 50-yen coins as change. However, you must not pay the coins of the same type as the coins you issued so that they will be returned as change. For example, if a 10-yen coin is paid out and another 10-yen coin is returned as change, a completely meaningless exchange will occur.
However, Mr. Bill was not good at calculating, so he could not find out how many coins he should actually use. So he asked you for help. Your job is to write a program to determine the type and number of coins to use based on the number of coins in his wallet and the payment price. The clerk does not use banknotes for change.
Input
The input contains several test cases.
Each test case consists of two lines. The first line contains a single integer representing Mr. Bill's payment in yen. The second line contains four integers, which in turn represent the number of 10-yen coins, 50-yen coins, 100-yen coins, and 500-yen coins in your wallet.
The payment amount is always in units of 10 yen. That is, the one-yen place of the payment amount is always 0. You can also assume that you can only have up to 20 coins of the same type in your wallet. No non-payable cases are given during input.
The end of input is represented by a line containing a single 0.
Output
For each test case, print out the type and number of coins that Mr. Bill should use.
Each line of output contains the two integers ci and ki. This means using ki coins for ci yen when paying. When using multiple types of coins, output as many lines as necessary in order from the smallest ci. Refer to the output example below.
The output must not include extra space. Separate consecutive test cases with a blank line.
Example
Input
160
1 1 2 0
160
1 0 2 10
0
Output
10 1
50 1
100 1
10 1
100 2
Submitted Solution:
```
coins = [10, 50, 100, 500]
flg = 0
while 1:
subtotal = int(input())
if subtotal == 0:
break
if flg == 1:
print()
s_purse = list(map(int, input().split()))
ans = [0, 0, 0, 0]
for id in range(len(coins)):
if id == 0:
continue
payable = 0
for i in range(len(coins) - id):
payable += coins[i] * s_purse[i]
while 1:
if payable < subtotal:
ans[-id] += 1
s_purse[-id] -= 1
subtotal -= coins[-id]
else:
break
if 0 < subtotal:
ans[0] += subtotal // 10
s_purse[0] -= subtotal // 10
subtotal -= coins[0] * ans[0]
charge = -subtotal
cha_coins = [charge//10, 0, 0, 0]
hitsuyou = [5, 2, 5]
for i in range(len(coins) - 1):
if hitsuyou[i] <= cha_coins[i]:
cha_coins[i + 1] = cha_coins[i] // hitsuyou[i]
cha_coins[i] = cha_coins[i] % hitsuyou[i]
if hitsuyou[i] <= cha_coins[i] + s_purse[i]:
ans[i] += (cha_coins[i] + s_purse[i]) // hitsuyou[i] * hitsuyou[i] - cha_coins[i]
cha_coins[i + 1] += (cha_coins[i] + s_purse[i]) // hitsuyou[i]
for i in range(len(coins)):
if ans[i]:
print(coins[i], int(ans[i]))
flg = 1
``` | instruction | 0 | 90,915 | 10 | 181,830 |
Yes | output | 1 | 90,915 | 10 | 181,831 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Bill is shopping at the store. There are some coins in his wallet (10-yen coins, 50-yen coins, 100-yen coins, 500-yen coins), but he is now trying to consume as much of this coin as possible. In other words, by paying for the goods with an appropriate number of coins, we are trying to minimize the total number of coins after receiving the change.
Fortunately, the clerk at this store is so disciplined and kind that change is always delivered in the best possible way. Therefore, for example, five 100-yen coins will not be given instead of one 500-yen coin. You can also take out 5 10-yen coins and receive 50-yen coins as change. However, you must not pay the coins of the same type as the coins you issued so that they will be returned as change. For example, if a 10-yen coin is paid out and another 10-yen coin is returned as change, a completely meaningless exchange will occur.
However, Mr. Bill was not good at calculating, so he could not find out how many coins he should actually use. So he asked you for help. Your job is to write a program to determine the type and number of coins to use based on the number of coins in his wallet and the payment price. The clerk does not use banknotes for change.
Input
The input contains several test cases.
Each test case consists of two lines. The first line contains a single integer representing Mr. Bill's payment in yen. The second line contains four integers, which in turn represent the number of 10-yen coins, 50-yen coins, 100-yen coins, and 500-yen coins in your wallet.
The payment amount is always in units of 10 yen. That is, the one-yen place of the payment amount is always 0. You can also assume that you can only have up to 20 coins of the same type in your wallet. No non-payable cases are given during input.
The end of input is represented by a line containing a single 0.
Output
For each test case, print out the type and number of coins that Mr. Bill should use.
Each line of output contains the two integers ci and ki. This means using ki coins for ci yen when paying. When using multiple types of coins, output as many lines as necessary in order from the smallest ci. Refer to the output example below.
The output must not include extra space. Separate consecutive test cases with a blank line.
Example
Input
160
1 1 2 0
160
1 0 2 10
0
Output
10 1
50 1
100 1
10 1
100 2
Submitted Solution:
```
def best_coins(r, i=3, bc=[0] * 4):
global coins
bc[i], r = divmod(r, coins[i])
if i:
best_coins(r, i - 1, bc)
return bc
coins = (10, 50, 100, 500)
output = []
while True:
price = int(input())
if not price:
break
pocket = tuple(map(int, input().split()))
remains = best_coins(sum(c * p for c, p in zip(coins, pocket)) - price)
output.append('\n'.join('{} {}'.format(c, p - r) for c, p, r in zip(coins, pocket, remains) if p - r > 0))
print('\n\n'.join(output))
``` | instruction | 0 | 90,916 | 10 | 181,832 |
Yes | output | 1 | 90,916 | 10 | 181,833 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Bill is shopping at the store. There are some coins in his wallet (10-yen coins, 50-yen coins, 100-yen coins, 500-yen coins), but he is now trying to consume as much of this coin as possible. In other words, by paying for the goods with an appropriate number of coins, we are trying to minimize the total number of coins after receiving the change.
Fortunately, the clerk at this store is so disciplined and kind that change is always delivered in the best possible way. Therefore, for example, five 100-yen coins will not be given instead of one 500-yen coin. You can also take out 5 10-yen coins and receive 50-yen coins as change. However, you must not pay the coins of the same type as the coins you issued so that they will be returned as change. For example, if a 10-yen coin is paid out and another 10-yen coin is returned as change, a completely meaningless exchange will occur.
However, Mr. Bill was not good at calculating, so he could not find out how many coins he should actually use. So he asked you for help. Your job is to write a program to determine the type and number of coins to use based on the number of coins in his wallet and the payment price. The clerk does not use banknotes for change.
Input
The input contains several test cases.
Each test case consists of two lines. The first line contains a single integer representing Mr. Bill's payment in yen. The second line contains four integers, which in turn represent the number of 10-yen coins, 50-yen coins, 100-yen coins, and 500-yen coins in your wallet.
The payment amount is always in units of 10 yen. That is, the one-yen place of the payment amount is always 0. You can also assume that you can only have up to 20 coins of the same type in your wallet. No non-payable cases are given during input.
The end of input is represented by a line containing a single 0.
Output
For each test case, print out the type and number of coins that Mr. Bill should use.
Each line of output contains the two integers ci and ki. This means using ki coins for ci yen when paying. When using multiple types of coins, output as many lines as necessary in order from the smallest ci. Refer to the output example below.
The output must not include extra space. Separate consecutive test cases with a blank line.
Example
Input
160
1 1 2 0
160
1 0 2 10
0
Output
10 1
50 1
100 1
10 1
100 2
Submitted Solution:
```
while 1:
n=int(input())
if not n:break
a,b,c,d=map(int,input().split())
y=(10,50,100,500)
l=[[i,j,k,l] for i in range(a+1) for j in range(b+1) for k in range(c+1) for l in range(d+1) if i*10+j*50+k*100+l*500==n]
t=sorted(l,key=lambda x:sum(x))[-1]
[print(i,j) for (i,j) in zip(y,t) if j]
``` | instruction | 0 | 90,917 | 10 | 181,834 |
No | output | 1 | 90,917 | 10 | 181,835 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Bill is shopping at the store. There are some coins in his wallet (10-yen coins, 50-yen coins, 100-yen coins, 500-yen coins), but he is now trying to consume as much of this coin as possible. In other words, by paying for the goods with an appropriate number of coins, we are trying to minimize the total number of coins after receiving the change.
Fortunately, the clerk at this store is so disciplined and kind that change is always delivered in the best possible way. Therefore, for example, five 100-yen coins will not be given instead of one 500-yen coin. You can also take out 5 10-yen coins and receive 50-yen coins as change. However, you must not pay the coins of the same type as the coins you issued so that they will be returned as change. For example, if a 10-yen coin is paid out and another 10-yen coin is returned as change, a completely meaningless exchange will occur.
However, Mr. Bill was not good at calculating, so he could not find out how many coins he should actually use. So he asked you for help. Your job is to write a program to determine the type and number of coins to use based on the number of coins in his wallet and the payment price. The clerk does not use banknotes for change.
Input
The input contains several test cases.
Each test case consists of two lines. The first line contains a single integer representing Mr. Bill's payment in yen. The second line contains four integers, which in turn represent the number of 10-yen coins, 50-yen coins, 100-yen coins, and 500-yen coins in your wallet.
The payment amount is always in units of 10 yen. That is, the one-yen place of the payment amount is always 0. You can also assume that you can only have up to 20 coins of the same type in your wallet. No non-payable cases are given during input.
The end of input is represented by a line containing a single 0.
Output
For each test case, print out the type and number of coins that Mr. Bill should use.
Each line of output contains the two integers ci and ki. This means using ki coins for ci yen when paying. When using multiple types of coins, output as many lines as necessary in order from the smallest ci. Refer to the output example below.
The output must not include extra space. Separate consecutive test cases with a blank line.
Example
Input
160
1 1 2 0
160
1 0 2 10
0
Output
10 1
50 1
100 1
10 1
100 2
Submitted Solution:
```
def back_oturigation(fee, coin_values, coin_nums):
"""
1. ????????????????????????????????£??????????????????
2. 1????????????????????????????????????????????????????????????
???????????§?????????????????????-> ?°????????????????????¶???????????????????
3. 2????????????????????°?????????????????????????????????????????????
"""
# 1.
oturi = coin_values[0] * coin_nums[0] \
+ coin_values[1] * coin_nums[1] \
+ coin_values[2] * coin_nums[2] \
+ coin_values[3] * coin_nums[3] \
- fee
use_coins = [0] * len(coin_values)
no_use_coins = [0] * len(coin_values)
# 2.
for i in range(len(coin_values) - 1, -1, -1):
no_use_coins[i] = int(oturi / coin_values[i])
oturi = oturi - (coin_values[i] * no_use_coins[i])
# 3.
for i in range(0, len(use_coins), 1):
use_coins[i] = coin_nums[i] - no_use_coins[i]
if use_coins[i] > 0:
print(str(coin_values[i]) + ' ' + str(use_coins[i]))
if __name__ == "__main__":
while (True):
total_fee = int(input('total_fee:'))
if total_fee == 0:
break
coins = input('total_coins:')
coins = coins.split(' ') # 10, 50, 100, 500????????§????????????
coins = [int(coin) for coin in coins]
coin_values = [10, 50, 100, 500]
back_oturigation(total_fee, coin_values, coins)
``` | instruction | 0 | 90,918 | 10 | 181,836 |
No | output | 1 | 90,918 | 10 | 181,837 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Bill is shopping at the store. There are some coins in his wallet (10-yen coins, 50-yen coins, 100-yen coins, 500-yen coins), but he is now trying to consume as much of this coin as possible. In other words, by paying for the goods with an appropriate number of coins, we are trying to minimize the total number of coins after receiving the change.
Fortunately, the clerk at this store is so disciplined and kind that change is always delivered in the best possible way. Therefore, for example, five 100-yen coins will not be given instead of one 500-yen coin. You can also take out 5 10-yen coins and receive 50-yen coins as change. However, you must not pay the coins of the same type as the coins you issued so that they will be returned as change. For example, if a 10-yen coin is paid out and another 10-yen coin is returned as change, a completely meaningless exchange will occur.
However, Mr. Bill was not good at calculating, so he could not find out how many coins he should actually use. So he asked you for help. Your job is to write a program to determine the type and number of coins to use based on the number of coins in his wallet and the payment price. The clerk does not use banknotes for change.
Input
The input contains several test cases.
Each test case consists of two lines. The first line contains a single integer representing Mr. Bill's payment in yen. The second line contains four integers, which in turn represent the number of 10-yen coins, 50-yen coins, 100-yen coins, and 500-yen coins in your wallet.
The payment amount is always in units of 10 yen. That is, the one-yen place of the payment amount is always 0. You can also assume that you can only have up to 20 coins of the same type in your wallet. No non-payable cases are given during input.
The end of input is represented by a line containing a single 0.
Output
For each test case, print out the type and number of coins that Mr. Bill should use.
Each line of output contains the two integers ci and ki. This means using ki coins for ci yen when paying. When using multiple types of coins, output as many lines as necessary in order from the smallest ci. Refer to the output example below.
The output must not include extra space. Separate consecutive test cases with a blank line.
Example
Input
160
1 1 2 0
160
1 0 2 10
0
Output
10 1
50 1
100 1
10 1
100 2
Submitted Solution:
```
coins = [10, 50, 100, 500]
def oturi(a, b):
s = b - a
l = [0, 0, 0, 0]
for i in reversed(range(4)):
n = s // coins[i]
s -= coins[i] * n
l[i] = n
return l
first = True
while True:
value = int(input())
if not value:
break
cs = list(map(int, input().split()))
if first:
first = False
else:
print()
pay = [0, 0, 0, 0]
pay_sum = 0
for i, c in enumerate(cs):
pay[i] += c
pay_sum += coins[i] * c
if pay_sum >= value:
break
o = oturi(value, pay_sum)
for i, v in enumerate(o):
pay[i] -= v
if pay[i] < 0:
pay[i] = 0
for i, v in enumerate(pay):
if v:
print(coins[i], v)
``` | instruction | 0 | 90,919 | 10 | 181,838 |
No | output | 1 | 90,919 | 10 | 181,839 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Bill is shopping at the store. There are some coins in his wallet (10-yen coins, 50-yen coins, 100-yen coins, 500-yen coins), but he is now trying to consume as much of this coin as possible. In other words, by paying for the goods with an appropriate number of coins, we are trying to minimize the total number of coins after receiving the change.
Fortunately, the clerk at this store is so disciplined and kind that change is always delivered in the best possible way. Therefore, for example, five 100-yen coins will not be given instead of one 500-yen coin. You can also take out 5 10-yen coins and receive 50-yen coins as change. However, you must not pay the coins of the same type as the coins you issued so that they will be returned as change. For example, if a 10-yen coin is paid out and another 10-yen coin is returned as change, a completely meaningless exchange will occur.
However, Mr. Bill was not good at calculating, so he could not find out how many coins he should actually use. So he asked you for help. Your job is to write a program to determine the type and number of coins to use based on the number of coins in his wallet and the payment price. The clerk does not use banknotes for change.
Input
The input contains several test cases.
Each test case consists of two lines. The first line contains a single integer representing Mr. Bill's payment in yen. The second line contains four integers, which in turn represent the number of 10-yen coins, 50-yen coins, 100-yen coins, and 500-yen coins in your wallet.
The payment amount is always in units of 10 yen. That is, the one-yen place of the payment amount is always 0. You can also assume that you can only have up to 20 coins of the same type in your wallet. No non-payable cases are given during input.
The end of input is represented by a line containing a single 0.
Output
For each test case, print out the type and number of coins that Mr. Bill should use.
Each line of output contains the two integers ci and ki. This means using ki coins for ci yen when paying. When using multiple types of coins, output as many lines as necessary in order from the smallest ci. Refer to the output example below.
The output must not include extra space. Separate consecutive test cases with a blank line.
Example
Input
160
1 1 2 0
160
1 0 2 10
0
Output
10 1
50 1
100 1
10 1
100 2
Submitted Solution:
```
c = [10, 50, 100, 500]
while True:
n = int(input())
if n == 0:
break
m = list(map(int, input().split()))
r = [0]*4
n = sum(x[0]*x[1] for x in zip(m, c)) - n
for i in range(4)[::-1]:
r[i] = n // c[i]
n -= r[i] * c[i]
for i in range(4):
if m[i] > r[i]:
print('{} {}'.format(c[i], m[i]-r[i]))
print()
``` | instruction | 0 | 90,920 | 10 | 181,840 |
No | output | 1 | 90,920 | 10 | 181,841 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Serge came to the school dining room and discovered that there is a big queue here. There are m pupils in the queue. He's not sure now if he wants to wait until the queue will clear, so he wants to know which dish he will receive if he does. As Serge is very tired, he asks you to compute it instead of him.
Initially there are n dishes with costs a_1, a_2, …, a_n. As you already know, there are the queue of m pupils who have b_1, …, b_m togrogs respectively (pupils are enumerated by queue order, i.e the first pupil in the queue has b_1 togrogs and the last one has b_m togrogs)
Pupils think that the most expensive dish is the most delicious one, so every pupil just buys the most expensive dish for which he has money (every dish has a single copy, so when a pupil has bought it nobody can buy it later), and if a pupil doesn't have money for any dish, he just leaves the queue (so brutal capitalism...)
But money isn't a problem at all for Serge, so Serge is buying the most expensive dish if there is at least one remaining.
Moreover, Serge's school has a very unstable economic situation and the costs of some dishes or number of togrogs of some pupils can change. More formally, you must process q queries:
* change a_i to x. It means that the price of the i-th dish becomes x togrogs.
* change b_i to x. It means that the i-th pupil in the queue has x togrogs now.
Nobody leaves the queue during those queries because a saleswoman is late.
After every query, you must tell Serge price of the dish which he will buy if he has waited until the queue is clear, or -1 if there are no dishes at this point, according to rules described above.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 300\ 000) — number of dishes and pupils respectively. The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^{6}) — elements of array a. The third line contains m integers b_1, b_2, …, b_{m} (1 ≤ b_i ≤ 10^{6}) — elements of array b. The fourth line conatins integer q (1 ≤ q ≤ 300\ 000) — number of queries.
Each of the following q lines contains as follows:
* if a query changes price of some dish, it contains 1, and two integers i and x (1 ≤ i ≤ n, 1 ≤ x ≤ 10^{6}), what means a_i becomes x.
* if a query changes number of togrogs of some pupil, it contains 2, and two integers i and x (1 ≤ i ≤ m, 1 ≤ x ≤ 10^{6}), what means b_i becomes x.
Output
For each of q queries prints the answer as the statement describes, the answer of the i-th query in the i-th line (the price of the dish which Serge will buy or -1 if nothing remains)
Examples
Input
1 1
1
1
1
1 1 100
Output
100
Input
1 1
1
1
1
2 1 100
Output
-1
Input
4 6
1 8 2 4
3 3 6 1 5 2
3
1 1 1
2 5 10
1 1 6
Output
8
-1
4
Note
In the first sample after the first query, there is one dish with price 100 togrogs and one pupil with one togrog, so Serge will buy the dish with price 100 togrogs.
In the second sample after the first query, there is one dish with price one togrog and one pupil with 100 togrogs, so Serge will get nothing.
In the third sample after the first query, nobody can buy the dish with price 8, so Serge will take it. After the second query, all dishes will be bought, after the third one the third and fifth pupils will by the first and the second dishes respectively and nobody will by the fourth one.
Submitted Solution:
```
print ("Down Journey :)\n" * 2)
``` | instruction | 0 | 91,000 | 10 | 182,000 |
No | output | 1 | 91,000 | 10 | 182,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Serge came to the school dining room and discovered that there is a big queue here. There are m pupils in the queue. He's not sure now if he wants to wait until the queue will clear, so he wants to know which dish he will receive if he does. As Serge is very tired, he asks you to compute it instead of him.
Initially there are n dishes with costs a_1, a_2, …, a_n. As you already know, there are the queue of m pupils who have b_1, …, b_m togrogs respectively (pupils are enumerated by queue order, i.e the first pupil in the queue has b_1 togrogs and the last one has b_m togrogs)
Pupils think that the most expensive dish is the most delicious one, so every pupil just buys the most expensive dish for which he has money (every dish has a single copy, so when a pupil has bought it nobody can buy it later), and if a pupil doesn't have money for any dish, he just leaves the queue (so brutal capitalism...)
But money isn't a problem at all for Serge, so Serge is buying the most expensive dish if there is at least one remaining.
Moreover, Serge's school has a very unstable economic situation and the costs of some dishes or number of togrogs of some pupils can change. More formally, you must process q queries:
* change a_i to x. It means that the price of the i-th dish becomes x togrogs.
* change b_i to x. It means that the i-th pupil in the queue has x togrogs now.
Nobody leaves the queue during those queries because a saleswoman is late.
After every query, you must tell Serge price of the dish which he will buy if he has waited until the queue is clear, or -1 if there are no dishes at this point, according to rules described above.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 300\ 000) — number of dishes and pupils respectively. The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^{6}) — elements of array a. The third line contains m integers b_1, b_2, …, b_{m} (1 ≤ b_i ≤ 10^{6}) — elements of array b. The fourth line conatins integer q (1 ≤ q ≤ 300\ 000) — number of queries.
Each of the following q lines contains as follows:
* if a query changes price of some dish, it contains 1, and two integers i and x (1 ≤ i ≤ n, 1 ≤ x ≤ 10^{6}), what means a_i becomes x.
* if a query changes number of togrogs of some pupil, it contains 2, and two integers i and x (1 ≤ i ≤ m, 1 ≤ x ≤ 10^{6}), what means b_i becomes x.
Output
For each of q queries prints the answer as the statement describes, the answer of the i-th query in the i-th line (the price of the dish which Serge will buy or -1 if nothing remains)
Examples
Input
1 1
1
1
1
1 1 100
Output
100
Input
1 1
1
1
1
2 1 100
Output
-1
Input
4 6
1 8 2 4
3 3 6 1 5 2
3
1 1 1
2 5 10
1 1 6
Output
8
-1
4
Note
In the first sample after the first query, there is one dish with price 100 togrogs and one pupil with one togrog, so Serge will buy the dish with price 100 togrogs.
In the second sample after the first query, there is one dish with price one togrog and one pupil with 100 togrogs, so Serge will get nothing.
In the third sample after the first query, nobody can buy the dish with price 8, so Serge will take it. After the second query, all dishes will be bought, after the third one the third and fifth pupils will by the first and the second dishes respectively and nobody will by the fourth one.
Submitted Solution:
```
print("Para que no se le olvide hacerlo despues perro")
``` | instruction | 0 | 91,001 | 10 | 182,002 |
No | output | 1 | 91,001 | 10 | 182,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Serge came to the school dining room and discovered that there is a big queue here. There are m pupils in the queue. He's not sure now if he wants to wait until the queue will clear, so he wants to know which dish he will receive if he does. As Serge is very tired, he asks you to compute it instead of him.
Initially there are n dishes with costs a_1, a_2, …, a_n. As you already know, there are the queue of m pupils who have b_1, …, b_m togrogs respectively (pupils are enumerated by queue order, i.e the first pupil in the queue has b_1 togrogs and the last one has b_m togrogs)
Pupils think that the most expensive dish is the most delicious one, so every pupil just buys the most expensive dish for which he has money (every dish has a single copy, so when a pupil has bought it nobody can buy it later), and if a pupil doesn't have money for any dish, he just leaves the queue (so brutal capitalism...)
But money isn't a problem at all for Serge, so Serge is buying the most expensive dish if there is at least one remaining.
Moreover, Serge's school has a very unstable economic situation and the costs of some dishes or number of togrogs of some pupils can change. More formally, you must process q queries:
* change a_i to x. It means that the price of the i-th dish becomes x togrogs.
* change b_i to x. It means that the i-th pupil in the queue has x togrogs now.
Nobody leaves the queue during those queries because a saleswoman is late.
After every query, you must tell Serge price of the dish which he will buy if he has waited until the queue is clear, or -1 if there are no dishes at this point, according to rules described above.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 300\ 000) — number of dishes and pupils respectively. The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^{6}) — elements of array a. The third line contains m integers b_1, b_2, …, b_{m} (1 ≤ b_i ≤ 10^{6}) — elements of array b. The fourth line conatins integer q (1 ≤ q ≤ 300\ 000) — number of queries.
Each of the following q lines contains as follows:
* if a query changes price of some dish, it contains 1, and two integers i and x (1 ≤ i ≤ n, 1 ≤ x ≤ 10^{6}), what means a_i becomes x.
* if a query changes number of togrogs of some pupil, it contains 2, and two integers i and x (1 ≤ i ≤ m, 1 ≤ x ≤ 10^{6}), what means b_i becomes x.
Output
For each of q queries prints the answer as the statement describes, the answer of the i-th query in the i-th line (the price of the dish which Serge will buy or -1 if nothing remains)
Examples
Input
1 1
1
1
1
1 1 100
Output
100
Input
1 1
1
1
1
2 1 100
Output
-1
Input
4 6
1 8 2 4
3 3 6 1 5 2
3
1 1 1
2 5 10
1 1 6
Output
8
-1
4
Note
In the first sample after the first query, there is one dish with price 100 togrogs and one pupil with one togrog, so Serge will buy the dish with price 100 togrogs.
In the second sample after the first query, there is one dish with price one togrog and one pupil with 100 togrogs, so Serge will get nothing.
In the third sample after the first query, nobody can buy the dish with price 8, so Serge will take it. After the second query, all dishes will be bought, after the third one the third and fifth pupils will by the first and the second dishes respectively and nobody will by the fourth one.
Submitted Solution:
```
nm = list(map(int,input().split()))
AA = list(map(int,input().split()))
BB = list(map(int,input().split()))
Q = eval(input())
for i in range(Q):
A=AA.copy()
B=BB.copy()
aix=list(map(int,input().split()))
if(aix[0]==1):
A[aix[1]-1]=aix[2]
else:
B[aix[1]-1]=aix[2]
for j in range(len(B)):
T=B[i]
TT=list(filter(lambda x: x<=T, A))
if(len(A)>0 and len(TT)>0):
A.pop(A.index(max(TT)))
if(len(A)==0):
print(-1)
else:
print(max(A))
``` | instruction | 0 | 91,002 | 10 | 182,004 |
No | output | 1 | 91,002 | 10 | 182,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Serge came to the school dining room and discovered that there is a big queue here. There are m pupils in the queue. He's not sure now if he wants to wait until the queue will clear, so he wants to know which dish he will receive if he does. As Serge is very tired, he asks you to compute it instead of him.
Initially there are n dishes with costs a_1, a_2, …, a_n. As you already know, there are the queue of m pupils who have b_1, …, b_m togrogs respectively (pupils are enumerated by queue order, i.e the first pupil in the queue has b_1 togrogs and the last one has b_m togrogs)
Pupils think that the most expensive dish is the most delicious one, so every pupil just buys the most expensive dish for which he has money (every dish has a single copy, so when a pupil has bought it nobody can buy it later), and if a pupil doesn't have money for any dish, he just leaves the queue (so brutal capitalism...)
But money isn't a problem at all for Serge, so Serge is buying the most expensive dish if there is at least one remaining.
Moreover, Serge's school has a very unstable economic situation and the costs of some dishes or number of togrogs of some pupils can change. More formally, you must process q queries:
* change a_i to x. It means that the price of the i-th dish becomes x togrogs.
* change b_i to x. It means that the i-th pupil in the queue has x togrogs now.
Nobody leaves the queue during those queries because a saleswoman is late.
After every query, you must tell Serge price of the dish which he will buy if he has waited until the queue is clear, or -1 if there are no dishes at this point, according to rules described above.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 300\ 000) — number of dishes and pupils respectively. The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^{6}) — elements of array a. The third line contains m integers b_1, b_2, …, b_{m} (1 ≤ b_i ≤ 10^{6}) — elements of array b. The fourth line conatins integer q (1 ≤ q ≤ 300\ 000) — number of queries.
Each of the following q lines contains as follows:
* if a query changes price of some dish, it contains 1, and two integers i and x (1 ≤ i ≤ n, 1 ≤ x ≤ 10^{6}), what means a_i becomes x.
* if a query changes number of togrogs of some pupil, it contains 2, and two integers i and x (1 ≤ i ≤ m, 1 ≤ x ≤ 10^{6}), what means b_i becomes x.
Output
For each of q queries prints the answer as the statement describes, the answer of the i-th query in the i-th line (the price of the dish which Serge will buy or -1 if nothing remains)
Examples
Input
1 1
1
1
1
1 1 100
Output
100
Input
1 1
1
1
1
2 1 100
Output
-1
Input
4 6
1 8 2 4
3 3 6 1 5 2
3
1 1 1
2 5 10
1 1 6
Output
8
-1
4
Note
In the first sample after the first query, there is one dish with price 100 togrogs and one pupil with one togrog, so Serge will buy the dish with price 100 togrogs.
In the second sample after the first query, there is one dish with price one togrog and one pupil with 100 togrogs, so Serge will get nothing.
In the third sample after the first query, nobody can buy the dish with price 8, so Serge will take it. After the second query, all dishes will be bought, after the third one the third and fifth pupils will by the first and the second dishes respectively and nobody will by the fourth one.
Submitted Solution:
```
from typing import Any, Callable, List
class SegmentTree():
T = Any
E = Any
F = Callable[[T, T], T]
G = Callable[[T, E], T]
H = Callable[[E, E], E]
C = Any
def __init__(self, f: F, g: G, h: H,
ti: T, ei: E) -> None:
self.f = f
self.g = g
self.h = h
self.ti = ti
self.ei = ei
def init(self, n_: int) -> int:
self.n = 1
self.height = 0
while self.n < n_:
self.n <<= 1
self.height += 1
self.dat = [self.ti for _ in range(2*self.n)]
self.laz = [self.ei for _ in range(2*self.n)]
def build(self, v: List) -> None:
n_ = len(v)
self.init(n_)
for i in range(n_):
self.dat[self.n+i] = v[i]
for i in range(self.n-1, 0, -1):
self.dat[i] = self.f(self.dat[(i << 1) | 0],
self.dat[(i << 1) | 1])
def reflect(self, k: int) -> T:
if self.laz[k] == self.ei:
return self.dat[k]
else:
return self.g(self.dat[k], self.laz[k])
def propagate(self, k: int) -> None:
if self.laz[k] == self.ei:
return
self.laz[(k << 1) | 0] = \
self.h(self.laz[(k << 1) | 0], self.laz[k])
self.laz[(k << 1) | 1] = \
self.h(self.laz[(k << 1) | 1], self.laz[k])
self.dat[k] = self.reflect(k)
self.laz[k] = self.ei
def thrust(self, k: int) -> None:
for i in range(self.height, 0, -1):
self.propagate(k >> i)
def recalc(self, k: int) -> None:
while True:
k >>= 1
if k == 0:
break
self.dat[k] = self.f(self.reflect((k << 1) | 0),
self.reflect((k << 1) | 1))
def update(self, a: int, b: int, x: E) -> None:
if a >= b:
return
a += self.n
self.thrust(a)
b += self.n-1
self.thrust(b)
l, r = a, b
while l < r:
if l & 1:
self.laz[l] = self.h(self.laz[l], x)
l += 1
if r & 1:
r -= 1
self.laz[r] = self.h(self.laz[r], x)
l, r = l >> 1, r >> 1
self.recalc(a)
self.recalc(b)
def set_val(self, a: int, x: T) -> None:
a += self.n
self.thrust(a)
self.dat[a] = x
self.laz[a] = self.ei
self.recalc(a)
def query(self, a: int, b: int) -> T:
if a >= b:
return self.ti
a += self.n
self.thrust(a)
b += self.n-1
self.thrust(b)
vl, vr = self.ti, self.ti
l, r = a, b
while l < r:
if l & 1:
vl = self.f(vl, self.reflect(l))
l += 1
if r & 1:
r -= 1
vr = self.f(self.reflect(r), vr)
return self.f(vl, vr)
def find_(self, st: int, check: C, acc: T, k: int, l: int, r: int) -> int:
if l+1 == r:
acc = self.f(acc, self.reflect(k))
if check(acc):
return k - self.n
else:
return -1
self.propagate(k)
m = (l+r) >> 1
if (m <= st):
return self.find_(st, check, acc, (k << 1) | 1, m, r)
if st <= l and not check(self.f(acc, self.dat[k])):
acc = self.f(acc, self.dat[k])
return -1
vl = self.find_(st, check, acc, (k << 1) | 0, l, m)
if vl != -1:
return vl
return self.find_(st, check, acc, (k << 1) | 1, m, r)
def find(self, st: int, check: C) -> int:
acc = self.ti
return self.find_(st, check, acc, 1, 0, self.n)
def CFR569_C():
from typing import Callable
import sys
input = sys.stdin.buffer.readline
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
f: Callable[[int, int], int] = lambda a, b: max(a, b)
g: Callable[[int, int], int] = lambda a, b: a+b
ti = ei = 0
seg = SegmentTree(f, g, g, ti, ei)
sz = 1
while sz < 3e5:
sz <<= 1
seg.build([0 for _ in range(sz)])
for i in range(n):
seg.update(sz-a[i], sz, 1)
for i in range(m):
seg.update(sz-b[i], sz, -1)
q = int(input())
check: Callable[[int], bool] = lambda d: d > 0
for i in range(q):
t, k, v = map(int, input().split())
k -= 1
if t == 1:
seg.update(sz-a[k], sz, -1)
a[k] = v
seg.update(sz-a[k], sz, 1)
if t == 2:
seg.update(sz-b[k], sz, 1)
b[k] = v
seg.update(sz-b[k], sz, -1)
pos = seg.find(0, check)
print(pos if pos < 0 else sz - pos)
def main():
CFR569_C()
if __name__ == '__main__':
main()
``` | instruction | 0 | 91,003 | 10 | 182,006 |
No | output | 1 | 91,003 | 10 | 182,007 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES | instruction | 0 | 91,018 | 10 | 182,036 |
Tags: math
Correct Solution:
```
t = int(input())
for _ in range(t):
a,b,n,s = list(map(int, input().split()))
coins = min(s//n,a)
s = s - coins*n
if(s<=b):
print("YES")
else:
print("NO")
``` | output | 1 | 91,018 | 10 | 182,037 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES | instruction | 0 | 91,019 | 10 | 182,038 |
Tags: math
Correct Solution:
```
""" should avoid multiplaction"""
import math
q=int(input())
for _ in range(q):
a , b, n, s=map(float,input().split())
t = s//n
t2 = s%n
if t <= a :
if t2 <= b:
print("YES")
else:
print("NO")
elif (abs(t-a) * n + t2) <= b:
print("YES")
else:
print("NO")
``` | output | 1 | 91,019 | 10 | 182,039 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES | instruction | 0 | 91,020 | 10 | 182,040 |
Tags: math
Correct Solution:
```
for _ in range(int(input())):
a , b , n , s = map(int,input().split())
au=min(a,s//n)
if s-(au)*n <=b:
print("YES")
else:
print("NO")
``` | output | 1 | 91,020 | 10 | 182,041 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES | instruction | 0 | 91,021 | 10 | 182,042 |
Tags: math
Correct Solution:
```
# cook your dish here
for _ in range(int(input())):
a,b,n,s = map(int ,input().split())
ans = s//n
su = 0
flag = 0
if(ans<=a):
su = su + ans*n
if(s-su<=b):
flag = 1
elif(ans> a):
su = su + a*n
if(s-su<=b):
flag = 1
print("YES" if flag==1 else "NO")
``` | output | 1 | 91,021 | 10 | 182,043 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES | instruction | 0 | 91,022 | 10 | 182,044 |
Tags: math
Correct Solution:
```
T = int(input())
for O in range(T):
a, b, n, s = map(int, input().split(' '))
c = s % n
if (c <= b and a*n+b >= s):
print("YES")
else:
print("NO")
``` | output | 1 | 91,022 | 10 | 182,045 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES | instruction | 0 | 91,023 | 10 | 182,046 |
Tags: math
Correct Solution:
```
t=int(input())
while t>0:
a,b,n,s=map(int,input().split())
if s//n <=a and s-((s//n)*n) <=b:
print("YES")
elif s//n > a and s-(a*n)<=b:
print('YES')
else:
print('NO')
t-=1
``` | output | 1 | 91,023 | 10 | 182,047 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES | instruction | 0 | 91,024 | 10 | 182,048 |
Tags: math
Correct Solution:
```
t=int(input())
while(t):
t-=1
q=list(map(int,input().split()))
a=q[0]
b=q[1]
n=q[2]
s=q[3]
if(b+(a*n)<s):
print("NO")
else:
if(s//n <= a):
if(s%n==0):
print("YES")
else:
s=s%n
if(b>=s):
print("YES")
else:
print("NO")
elif(s//n > a):
s=s%n
if(b>=s):
print("YES")
else:
print("NO")
``` | output | 1 | 91,024 | 10 | 182,049 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES | instruction | 0 | 91,025 | 10 | 182,050 |
Tags: math
Correct Solution:
```
from collections import defaultdict as dd
from collections import deque
import bisect
import heapq
def ri():
return int(input())
def rl():
return list(map(int, input().split()))
def solve():
a, b, n, s = rl()
if a * n + b < s:
print ("NO")
return
d = n * (s // n)
if s - d <= b:
print ("YES")
else:
print ("NO")
mode = 'T'
if mode == 'T':
t = ri()
for i in range(t):
solve()
else:
solve()
``` | output | 1 | 91,025 | 10 | 182,051 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Submitted Solution:
```
##for debug comment out
import sys,atexit
from io import BytesIO
inp = BytesIO(sys.stdin.buffer.read())
input = lambda:inp.readline().decode('ascii')
buf = BytesIO()
#sys.stdout.write = lambda s: buf.write(s.encode('ascii'))
#print = lambda s: buf.write(s.encode('ascii'))
atexit.register(lambda:sys.__stdout__.buffer.write(buf.getvalue()))
for i in range(int(input())):
a,b,n,S=map(int,input().split())
a=min(S//n,a)
S-=n*a
#print(S)
if b>=S:
print("YES")
else:
print("NO")
``` | instruction | 0 | 91,026 | 10 | 182,052 |
Yes | output | 1 | 91,026 | 10 | 182,053 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Submitted Solution:
```
T = int(input())
for _ in range(T):
a, b, n, S = map(int, input().split())
if min(S//n, a) * n + b >= S:
print('YES')
else:
print('NO')
``` | instruction | 0 | 91,027 | 10 | 182,054 |
Yes | output | 1 | 91,027 | 10 | 182,055 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Submitted Solution:
```
q = int(input())
for _ in range(q):
a, b, n, s = list(map(int, input().split()))
if a * n == s or b == s or (a*n < s and b >= (s-(a*n))):
print('YES')
elif b >= (s - (n * min((s // n), a))):
print('YES')
else:
print('NO')
``` | instruction | 0 | 91,028 | 10 | 182,056 |
Yes | output | 1 | 91,028 | 10 | 182,057 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Submitted Solution:
```
#RAVENS
#TEAM_2
#ESSI-DAYI_MOHSEN-LORENZO
#STARTED IN 5:15 :)
check=False
for i in range(int(input())):
check=False
a,b,n,s=map(int,input().split())
if (a * n >= s):
if(s%n<=b):
print("YES")
else:
print("NO")
elif (((a * n) + b) >= s):
tmp=s%(a*n)
if (tmp <= b):
print("YES")
else:
print("NO")
else:
print("NO")
'''if(a*n>=s):
print("s : ",s)
print(((s//n)*a))
s=s-((s//n)*a)
if(b>=s):
check=True
else:
check=False
elif(((a*n)+b)>=s):
print("s : ",s)
s-=a*n
print("s : ",s)
if (b >= s):
check = True
else:
check = False
print("YES" if check else "NO")'''
``` | instruction | 0 | 91,029 | 10 | 182,058 |
Yes | output | 1 | 91,029 | 10 | 182,059 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Submitted Solution:
```
q = int(input())
for tests in range(q):
inputs = input()
inputs = inputs.split(" ")
a = int(inputs[0])
b = int(inputs[1])
n = int(inputs[2])
S = int(inputs[3])
if(b > S):
print("yes")
continue
a = a + b//n
b = b % n
if(b != 0):
if(S > a*n + b):
print("NO")
continue
quotient = S // n
remainder = S - n * quotient
if(quotient > a):
print("NO")
continue
elif(remainder > b):
print("NO")
continue
print("YES")
``` | instruction | 0 | 91,030 | 10 | 182,060 |
No | output | 1 | 91,030 | 10 | 182,061 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Submitted Solution:
```
for _ in range(int(input())):
a,b,n,s=map(int,input().split())
value=int(s/n)
if value<=a:
rem=value%n
if rem<=b:
print("YES")
else:
print("NO")
else:
s=s-(n*a)
if s>b:
print("NO")
else:
print("YES")
``` | instruction | 0 | 91,031 | 10 | 182,062 |
No | output | 1 | 91,031 | 10 | 182,063 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Submitted Solution:
```
from __future__ import print_function
q=int(input())
while q > 0:
q-=1
lst=input()
#taking intgers as input
#a,b,n,s
#print("yes")
arr=list(map(int,lst.split()))
s=arr[3]
if arr[0]*arr[2] > s:
if s%arr[0] <= arr[1]:
print("YES")
else:
print("NO")
else :
if arr[0]*arr[2]+arr[1] < s:
print("NO")
else:
print("YES")
``` | instruction | 0 | 91,032 | 10 | 182,064 |
No | output | 1 | 91,032 | 10 | 182,065 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Submitted Solution:
```
t=int(input())
for _ in range(t):
a,b,n,s=map(int,input().split())
f='NO'
for i in range(1,b+1):
if (s-i)%n==0 and (s-i)//n<=a and (s-i)//n>=0:
f='YES'
break;
print(f)
``` | instruction | 0 | 91,033 | 10 | 182,066 |
No | output | 1 | 91,033 | 10 | 182,067 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money. | instruction | 0 | 91,066 | 10 | 182,132 |
Tags: greedy, math
Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
ans=[]
for _ in range(int(input())):
x,y=map(int,input().split())
a,b=map(int,input().split())
c1=(max(x,y)-min(x,y))*a+min(x,y)*b
c2=(x+y)*a
ans.append(min(c1,c2))
print("\n".join(map(str,ans)))
``` | output | 1 | 91,066 | 10 | 182,133 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money. | instruction | 0 | 91,067 | 10 | 182,134 |
Tags: greedy, math
Correct Solution:
```
t = int(input())
for i in range(t):
x,y = map(int, input().split())
a,b = map(int, input().split())
print( min(abs(x-y)*a + min(x,y)*b, x*a+y*a) )
``` | output | 1 | 91,067 | 10 | 182,135 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money. | instruction | 0 | 91,068 | 10 | 182,136 |
Tags: greedy, math
Correct Solution:
```
for _ in range(int(input())):
x,y=map(int,input().split())
a,b=map(int,input().split())
t=[]
if x==y==0:
print(0)
else:
z=abs(x-y)
z=z*a
l=min(x,y)
l=l*b
t.append(l+z)
t.append(a*(x+y))
print(min(t))
``` | output | 1 | 91,068 | 10 | 182,137 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.