message stringlengths 2 30.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 237 109k | cluster float64 10 10 | __index_level_0__ int64 474 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Summer holidays! Someone is going on trips, someone is visiting grandparents, but someone is trying to get a part-time job. This summer Noora decided that she wants to earn some money, and took a job in a shop as an assistant.
Shop, where Noora is working, has a plan on the following n days. For each day sales manager knows exactly, that in i-th day ki products will be put up for sale and exactly li clients will come to the shop that day. Also, the manager is sure, that everyone, who comes to the shop, buys exactly one product or, if there aren't any left, leaves the shop without buying anything. Moreover, due to the short shelf-life of the products, manager established the following rule: if some part of the products left on the shelves at the end of the day, that products aren't kept on the next day and are sent to the dump.
For advertising purposes manager offered to start a sell-out in the shop. He asked Noora to choose any f days from n next for sell-outs. On each of f chosen days the number of products were put up for sale would be doubled. Thus, if on i-th day shop planned to put up for sale ki products and Noora has chosen this day for sell-out, shelves of the shop would keep 2·ki products. Consequently, there is an opportunity to sell two times more products on days of sell-out.
Noora's task is to choose f days to maximize total number of sold products. She asks you to help her with such a difficult problem.
Input
The first line contains two integers n and f (1 ≤ n ≤ 105, 0 ≤ f ≤ n) denoting the number of days in shop's plan and the number of days that Noora has to choose for sell-out.
Each line of the following n subsequent lines contains two integers ki, li (0 ≤ ki, li ≤ 109) denoting the number of products on the shelves of the shop on the i-th day and the number of clients that will come to the shop on i-th day.
Output
Print a single integer denoting the maximal number of products that shop can sell.
Examples
Input
4 2
2 1
3 5
2 3
1 5
Output
10
Input
4 1
0 2
0 3
3 5
0 6
Output
5
Note
In the first example we can choose days with numbers 2 and 4 for sell-out. In this case new numbers of products for sale would be equal to [2, 6, 2, 2] respectively. So on the first day shop will sell 1 product, on the second — 5, on the third — 2, on the fourth — 2. In total 1 + 5 + 2 + 2 = 10 product units.
In the second example it is possible to sell 5 products, if you choose third day for sell-out.
Submitted Solution:
```
n, f = [int(i) for i in input().split()]
a = [0] * n
a2 = [0] * n
for i in range(n):
k, l = [int(i) for i in input().split()]
a[i] = (k, l)
a.sort(key=lambda t: min(t[0], t[1] - t[0]), reverse=True)
ans = 0
for i in range(len(a)):
t = a[i]
ans += max(0, min(t[0] + t[0], t[1])) if i < f else max(0, min(t[0], t[1]))
print(ans)
``` | instruction | 0 | 17,343 | 10 | 34,686 |
Yes | output | 1 | 17,343 | 10 | 34,687 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Summer holidays! Someone is going on trips, someone is visiting grandparents, but someone is trying to get a part-time job. This summer Noora decided that she wants to earn some money, and took a job in a shop as an assistant.
Shop, where Noora is working, has a plan on the following n days. For each day sales manager knows exactly, that in i-th day ki products will be put up for sale and exactly li clients will come to the shop that day. Also, the manager is sure, that everyone, who comes to the shop, buys exactly one product or, if there aren't any left, leaves the shop without buying anything. Moreover, due to the short shelf-life of the products, manager established the following rule: if some part of the products left on the shelves at the end of the day, that products aren't kept on the next day and are sent to the dump.
For advertising purposes manager offered to start a sell-out in the shop. He asked Noora to choose any f days from n next for sell-outs. On each of f chosen days the number of products were put up for sale would be doubled. Thus, if on i-th day shop planned to put up for sale ki products and Noora has chosen this day for sell-out, shelves of the shop would keep 2·ki products. Consequently, there is an opportunity to sell two times more products on days of sell-out.
Noora's task is to choose f days to maximize total number of sold products. She asks you to help her with such a difficult problem.
Input
The first line contains two integers n and f (1 ≤ n ≤ 105, 0 ≤ f ≤ n) denoting the number of days in shop's plan and the number of days that Noora has to choose for sell-out.
Each line of the following n subsequent lines contains two integers ki, li (0 ≤ ki, li ≤ 109) denoting the number of products on the shelves of the shop on the i-th day and the number of clients that will come to the shop on i-th day.
Output
Print a single integer denoting the maximal number of products that shop can sell.
Examples
Input
4 2
2 1
3 5
2 3
1 5
Output
10
Input
4 1
0 2
0 3
3 5
0 6
Output
5
Note
In the first example we can choose days with numbers 2 and 4 for sell-out. In this case new numbers of products for sale would be equal to [2, 6, 2, 2] respectively. So on the first day shop will sell 1 product, on the second — 5, on the third — 2, on the fourth — 2. In total 1 + 5 + 2 + 2 = 10 product units.
In the second example it is possible to sell 5 products, if you choose third day for sell-out.
Submitted Solution:
```
# Problem: B. Summer sell-off
# Contest: Codeforces - Codeforces Round #415 (Div. 2)
# URL: https://codeforces.com/contest/810/problem/B
# Memory Limit: 256 MB
# Time Limit: 1000 ms
# Powered by CP Editor (https://github.com/cpeditor/cpeditor)
from sys import stdin
def get_ints(): return list(map(int, stdin.readline().strip().split()))
n,f = get_ints()
a = []
b = []
for _ in range(n):
x,y = get_ints()
a.append([x,y])
sum =0
a = sorted(a,key= lambda day:day[1] if day[0] >0 else day[0],reverse = True )
for i in range(n):
if f >0:
a[i][0] *=2
f-=1
# print(a[i][1] , a[i][0])
sum+=min(a[i][1] , a[i][0])
print(sum)
``` | instruction | 0 | 17,344 | 10 | 34,688 |
No | output | 1 | 17,344 | 10 | 34,689 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Summer holidays! Someone is going on trips, someone is visiting grandparents, but someone is trying to get a part-time job. This summer Noora decided that she wants to earn some money, and took a job in a shop as an assistant.
Shop, where Noora is working, has a plan on the following n days. For each day sales manager knows exactly, that in i-th day ki products will be put up for sale and exactly li clients will come to the shop that day. Also, the manager is sure, that everyone, who comes to the shop, buys exactly one product or, if there aren't any left, leaves the shop without buying anything. Moreover, due to the short shelf-life of the products, manager established the following rule: if some part of the products left on the shelves at the end of the day, that products aren't kept on the next day and are sent to the dump.
For advertising purposes manager offered to start a sell-out in the shop. He asked Noora to choose any f days from n next for sell-outs. On each of f chosen days the number of products were put up for sale would be doubled. Thus, if on i-th day shop planned to put up for sale ki products and Noora has chosen this day for sell-out, shelves of the shop would keep 2·ki products. Consequently, there is an opportunity to sell two times more products on days of sell-out.
Noora's task is to choose f days to maximize total number of sold products. She asks you to help her with such a difficult problem.
Input
The first line contains two integers n and f (1 ≤ n ≤ 105, 0 ≤ f ≤ n) denoting the number of days in shop's plan and the number of days that Noora has to choose for sell-out.
Each line of the following n subsequent lines contains two integers ki, li (0 ≤ ki, li ≤ 109) denoting the number of products on the shelves of the shop on the i-th day and the number of clients that will come to the shop on i-th day.
Output
Print a single integer denoting the maximal number of products that shop can sell.
Examples
Input
4 2
2 1
3 5
2 3
1 5
Output
10
Input
4 1
0 2
0 3
3 5
0 6
Output
5
Note
In the first example we can choose days with numbers 2 and 4 for sell-out. In this case new numbers of products for sale would be equal to [2, 6, 2, 2] respectively. So on the first day shop will sell 1 product, on the second — 5, on the third — 2, on the fourth — 2. In total 1 + 5 + 2 + 2 = 10 product units.
In the second example it is possible to sell 5 products, if you choose third day for sell-out.
Submitted Solution:
```
n, f = map(int, input().split())
final = []
for i in range(n):
products, people = map(int, input().split())
final.append([products, people])
final.sort(key = lambda x: x[1] - x[0], reverse = True)
pair = count = 0
while count < f and pair < len(final):
if final[pair][0] != 0 and final[pair][1] != 0:
final[pair][0] *= 2
count += 1
pair += 1
print(sum(min(i[0], i[1]) for i in final))
``` | instruction | 0 | 17,345 | 10 | 34,690 |
No | output | 1 | 17,345 | 10 | 34,691 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Summer holidays! Someone is going on trips, someone is visiting grandparents, but someone is trying to get a part-time job. This summer Noora decided that she wants to earn some money, and took a job in a shop as an assistant.
Shop, where Noora is working, has a plan on the following n days. For each day sales manager knows exactly, that in i-th day ki products will be put up for sale and exactly li clients will come to the shop that day. Also, the manager is sure, that everyone, who comes to the shop, buys exactly one product or, if there aren't any left, leaves the shop without buying anything. Moreover, due to the short shelf-life of the products, manager established the following rule: if some part of the products left on the shelves at the end of the day, that products aren't kept on the next day and are sent to the dump.
For advertising purposes manager offered to start a sell-out in the shop. He asked Noora to choose any f days from n next for sell-outs. On each of f chosen days the number of products were put up for sale would be doubled. Thus, if on i-th day shop planned to put up for sale ki products and Noora has chosen this day for sell-out, shelves of the shop would keep 2·ki products. Consequently, there is an opportunity to sell two times more products on days of sell-out.
Noora's task is to choose f days to maximize total number of sold products. She asks you to help her with such a difficult problem.
Input
The first line contains two integers n and f (1 ≤ n ≤ 105, 0 ≤ f ≤ n) denoting the number of days in shop's plan and the number of days that Noora has to choose for sell-out.
Each line of the following n subsequent lines contains two integers ki, li (0 ≤ ki, li ≤ 109) denoting the number of products on the shelves of the shop on the i-th day and the number of clients that will come to the shop on i-th day.
Output
Print a single integer denoting the maximal number of products that shop can sell.
Examples
Input
4 2
2 1
3 5
2 3
1 5
Output
10
Input
4 1
0 2
0 3
3 5
0 6
Output
5
Note
In the first example we can choose days with numbers 2 and 4 for sell-out. In this case new numbers of products for sale would be equal to [2, 6, 2, 2] respectively. So on the first day shop will sell 1 product, on the second — 5, on the third — 2, on the fourth — 2. In total 1 + 5 + 2 + 2 = 10 product units.
In the second example it is possible to sell 5 products, if you choose third day for sell-out.
Submitted Solution:
```
n, f = map(int, input().split())
s = [[int(j) for j in input().split()] for i in range(n)]
a = 0
c = 0
x2 = dict()
x1 = list()
for i in range(n):
x2[min(s[i][0]*2, s[i][1])] = i
x2 = sorted(x2.items(), reverse=True)
for i in range(f):
c += x2[i][0]
x1.append(x2[i][1])
for i in range(n):
if i in x1:
continue
else:
c += min(s[i][0], s[i][1])
print(c)
# for i in range(n):
# if (min(s[i][0], s[i][1]) < min((s[i][0]*2), s[i][1])) & (c != f):
# s[i][0] *= 2
# c += 1
# a += min(s[i][0], s[i][1])
# print(a)
``` | instruction | 0 | 17,346 | 10 | 34,692 |
No | output | 1 | 17,346 | 10 | 34,693 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Summer holidays! Someone is going on trips, someone is visiting grandparents, but someone is trying to get a part-time job. This summer Noora decided that she wants to earn some money, and took a job in a shop as an assistant.
Shop, where Noora is working, has a plan on the following n days. For each day sales manager knows exactly, that in i-th day ki products will be put up for sale and exactly li clients will come to the shop that day. Also, the manager is sure, that everyone, who comes to the shop, buys exactly one product or, if there aren't any left, leaves the shop without buying anything. Moreover, due to the short shelf-life of the products, manager established the following rule: if some part of the products left on the shelves at the end of the day, that products aren't kept on the next day and are sent to the dump.
For advertising purposes manager offered to start a sell-out in the shop. He asked Noora to choose any f days from n next for sell-outs. On each of f chosen days the number of products were put up for sale would be doubled. Thus, if on i-th day shop planned to put up for sale ki products and Noora has chosen this day for sell-out, shelves of the shop would keep 2·ki products. Consequently, there is an opportunity to sell two times more products on days of sell-out.
Noora's task is to choose f days to maximize total number of sold products. She asks you to help her with such a difficult problem.
Input
The first line contains two integers n and f (1 ≤ n ≤ 105, 0 ≤ f ≤ n) denoting the number of days in shop's plan and the number of days that Noora has to choose for sell-out.
Each line of the following n subsequent lines contains two integers ki, li (0 ≤ ki, li ≤ 109) denoting the number of products on the shelves of the shop on the i-th day and the number of clients that will come to the shop on i-th day.
Output
Print a single integer denoting the maximal number of products that shop can sell.
Examples
Input
4 2
2 1
3 5
2 3
1 5
Output
10
Input
4 1
0 2
0 3
3 5
0 6
Output
5
Note
In the first example we can choose days with numbers 2 and 4 for sell-out. In this case new numbers of products for sale would be equal to [2, 6, 2, 2] respectively. So on the first day shop will sell 1 product, on the second — 5, on the third — 2, on the fourth — 2. In total 1 + 5 + 2 + 2 = 10 product units.
In the second example it is possible to sell 5 products, if you choose third day for sell-out.
Submitted Solution:
```
#python3
import sys, threading, os.path
import collections, heapq, math,bisect
import string
from platform import python_version
import itertools
sys.setrecursionlimit(10**6)
threading.stack_size(2**27)
def main():
if os.path.exists('input.txt'):
input = open('input.txt', 'r')
else:
input = sys.stdin
#--------------------------------INPUT---------------------------------
n, f = list(map(int, input.readline().split()))
lis=[]
for i in range(n):
lis.append(list(map(int, input.readline().split())))
results=[]
for x,y in lis:
if x>=y:
results.append([y,y])
elif x<y and x*2>=y:
results.append([y,x])
else:
results.append([x*2,x])
#print(results)
results = sorted(results,reverse = True)
#print(results)
sum = 0
for i in range(n):
if i+1>f:
sum+=results[i][1]
else:
sum+=results[i][0]
output=sum
#-------------------------------OUTPUT----------------------------------
if os.path.exists('output.txt'):
open('output.txt', 'w').writelines(str(output))
else:
sys.stdout.write(str(output))
if __name__ == '__main__':
main()
#threading.Thread(target=main).start()
``` | instruction | 0 | 17,347 | 10 | 34,694 |
No | output | 1 | 17,347 | 10 | 34,695 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After waking up at hh:mm, Andrew realised that he had forgotten to feed his only cat for yet another time (guess why there's only one cat). The cat's current hunger level is H points, moreover each minute without food increases his hunger by D points.
At any time Andrew can visit the store where tasty buns are sold (you can assume that is doesn't take time to get to the store and back). One such bun costs C roubles and decreases hunger by N points. Since the demand for bakery drops heavily in the evening, there is a special 20% discount for buns starting from 20:00 (note that the cost might become rational). Of course, buns cannot be sold by parts.
Determine the minimum amount of money Andrew has to spend in order to feed his cat. The cat is considered fed if its hunger level is less than or equal to zero.
Input
The first line contains two integers hh and mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59) — the time of Andrew's awakening.
The second line contains four integers H, D, C and N (1 ≤ H ≤ 105, 1 ≤ D, C, N ≤ 102).
Output
Output the minimum amount of money to within three decimal digits. You answer is considered correct, if its absolute or relative error does not exceed 10 - 4.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.
Examples
Input
19 00
255 1 100 1
Output
25200.0000
Input
17 41
1000 6 15 11
Output
1365.0000
Note
In the first sample Andrew can visit the store at exactly 20:00. The cat's hunger will be equal to 315, hence it will be necessary to purchase 315 buns. The discount makes the final answer 25200 roubles.
In the second sample it's optimal to visit the store right after he wakes up. Then he'll have to buy 91 bins per 15 roubles each and spend a total of 1365 roubles. | instruction | 0 | 17,408 | 10 | 34,816 |
Tags: greedy, math
Correct Solution:
```
h,m = map(int, input().split())
k,d,c,n =map(int, input().split())
from math import ceil
if h >= 20:
print(ceil(k/n)*(c*0.8))
else:
a1 = ceil(k/n)*c
r = 60*(20-h) - m
a2 = (ceil((k+r*d)/n))*(c*0.8)
print(min(a1,a2))
``` | output | 1 | 17,408 | 10 | 34,817 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After waking up at hh:mm, Andrew realised that he had forgotten to feed his only cat for yet another time (guess why there's only one cat). The cat's current hunger level is H points, moreover each minute without food increases his hunger by D points.
At any time Andrew can visit the store where tasty buns are sold (you can assume that is doesn't take time to get to the store and back). One such bun costs C roubles and decreases hunger by N points. Since the demand for bakery drops heavily in the evening, there is a special 20% discount for buns starting from 20:00 (note that the cost might become rational). Of course, buns cannot be sold by parts.
Determine the minimum amount of money Andrew has to spend in order to feed his cat. The cat is considered fed if its hunger level is less than or equal to zero.
Input
The first line contains two integers hh and mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59) — the time of Andrew's awakening.
The second line contains four integers H, D, C and N (1 ≤ H ≤ 105, 1 ≤ D, C, N ≤ 102).
Output
Output the minimum amount of money to within three decimal digits. You answer is considered correct, if its absolute or relative error does not exceed 10 - 4.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.
Examples
Input
19 00
255 1 100 1
Output
25200.0000
Input
17 41
1000 6 15 11
Output
1365.0000
Note
In the first sample Andrew can visit the store at exactly 20:00. The cat's hunger will be equal to 315, hence it will be necessary to purchase 315 buns. The discount makes the final answer 25200 roubles.
In the second sample it's optimal to visit the store right after he wakes up. Then he'll have to buy 91 bins per 15 roubles each and spend a total of 1365 roubles. | instruction | 0 | 17,409 | 10 | 34,818 |
Tags: greedy, math
Correct Solution:
```
I = lambda: map(int, input().split())
hh, mm = I()
H, D, C, N = I()
print(min((H+N-1)//N * C, (H+D*max(0,1200-60*hh-mm)+N-1)//N * 0.8*C))
``` | output | 1 | 17,409 | 10 | 34,819 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After waking up at hh:mm, Andrew realised that he had forgotten to feed his only cat for yet another time (guess why there's only one cat). The cat's current hunger level is H points, moreover each minute without food increases his hunger by D points.
At any time Andrew can visit the store where tasty buns are sold (you can assume that is doesn't take time to get to the store and back). One such bun costs C roubles and decreases hunger by N points. Since the demand for bakery drops heavily in the evening, there is a special 20% discount for buns starting from 20:00 (note that the cost might become rational). Of course, buns cannot be sold by parts.
Determine the minimum amount of money Andrew has to spend in order to feed his cat. The cat is considered fed if its hunger level is less than or equal to zero.
Input
The first line contains two integers hh and mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59) — the time of Andrew's awakening.
The second line contains four integers H, D, C and N (1 ≤ H ≤ 105, 1 ≤ D, C, N ≤ 102).
Output
Output the minimum amount of money to within three decimal digits. You answer is considered correct, if its absolute or relative error does not exceed 10 - 4.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.
Examples
Input
19 00
255 1 100 1
Output
25200.0000
Input
17 41
1000 6 15 11
Output
1365.0000
Note
In the first sample Andrew can visit the store at exactly 20:00. The cat's hunger will be equal to 315, hence it will be necessary to purchase 315 buns. The discount makes the final answer 25200 roubles.
In the second sample it's optimal to visit the store right after he wakes up. Then he'll have to buy 91 bins per 15 roubles each and spend a total of 1365 roubles. | instruction | 0 | 17,410 | 10 | 34,820 |
Tags: greedy, math
Correct Solution:
```
import math
h, m = [int(x) for x in input().split()]
H, D, C, N = [int(x) for x in input().split()]
start = math.ceil(H/N)*C
if h>=20:
stop = math.ceil(H/N)*C*0.8
else:
stop = math.ceil((H+(20*60 - (h*60+m))*D)/N)*C*0.8
print(min(start, stop))
``` | output | 1 | 17,410 | 10 | 34,821 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After waking up at hh:mm, Andrew realised that he had forgotten to feed his only cat for yet another time (guess why there's only one cat). The cat's current hunger level is H points, moreover each minute without food increases his hunger by D points.
At any time Andrew can visit the store where tasty buns are sold (you can assume that is doesn't take time to get to the store and back). One such bun costs C roubles and decreases hunger by N points. Since the demand for bakery drops heavily in the evening, there is a special 20% discount for buns starting from 20:00 (note that the cost might become rational). Of course, buns cannot be sold by parts.
Determine the minimum amount of money Andrew has to spend in order to feed his cat. The cat is considered fed if its hunger level is less than or equal to zero.
Input
The first line contains two integers hh and mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59) — the time of Andrew's awakening.
The second line contains four integers H, D, C and N (1 ≤ H ≤ 105, 1 ≤ D, C, N ≤ 102).
Output
Output the minimum amount of money to within three decimal digits. You answer is considered correct, if its absolute or relative error does not exceed 10 - 4.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.
Examples
Input
19 00
255 1 100 1
Output
25200.0000
Input
17 41
1000 6 15 11
Output
1365.0000
Note
In the first sample Andrew can visit the store at exactly 20:00. The cat's hunger will be equal to 315, hence it will be necessary to purchase 315 buns. The discount makes the final answer 25200 roubles.
In the second sample it's optimal to visit the store right after he wakes up. Then he'll have to buy 91 bins per 15 roubles each and spend a total of 1365 roubles. | instruction | 0 | 17,411 | 10 | 34,822 |
Tags: greedy, math
Correct Solution:
```
from math import ceil
hh, mm = [int(x) for x in input().split()]
h, d, c, n = [int(x) for x in input().split()]
cost = 0.8 * c if hh >= 20 else c
res = int(ceil(h / n)) * cost
if hh < 20:
diff = (20 - hh) * 60 - mm
diff *= d
h += diff
res = min(res, int(ceil(h / n)) * 0.8 * c)
print(res)
``` | output | 1 | 17,411 | 10 | 34,823 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After waking up at hh:mm, Andrew realised that he had forgotten to feed his only cat for yet another time (guess why there's only one cat). The cat's current hunger level is H points, moreover each minute without food increases his hunger by D points.
At any time Andrew can visit the store where tasty buns are sold (you can assume that is doesn't take time to get to the store and back). One such bun costs C roubles and decreases hunger by N points. Since the demand for bakery drops heavily in the evening, there is a special 20% discount for buns starting from 20:00 (note that the cost might become rational). Of course, buns cannot be sold by parts.
Determine the minimum amount of money Andrew has to spend in order to feed his cat. The cat is considered fed if its hunger level is less than or equal to zero.
Input
The first line contains two integers hh and mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59) — the time of Andrew's awakening.
The second line contains four integers H, D, C and N (1 ≤ H ≤ 105, 1 ≤ D, C, N ≤ 102).
Output
Output the minimum amount of money to within three decimal digits. You answer is considered correct, if its absolute or relative error does not exceed 10 - 4.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.
Examples
Input
19 00
255 1 100 1
Output
25200.0000
Input
17 41
1000 6 15 11
Output
1365.0000
Note
In the first sample Andrew can visit the store at exactly 20:00. The cat's hunger will be equal to 315, hence it will be necessary to purchase 315 buns. The discount makes the final answer 25200 roubles.
In the second sample it's optimal to visit the store right after he wakes up. Then he'll have to buy 91 bins per 15 roubles each and spend a total of 1365 roubles. | instruction | 0 | 17,412 | 10 | 34,824 |
Tags: greedy, math
Correct Solution:
```
hh, mm = map(int, input().split())
H,D,C,N = map(int, input().split())
def ceildiv(a, b):
return -(-a // b)
left = max(20*60 - hh*60-mm,0)
B = ceildiv(H + D*left,N)*C*0.8
C = ceildiv(H,N)*C*1.0
print(min(B,C))
``` | output | 1 | 17,412 | 10 | 34,825 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After waking up at hh:mm, Andrew realised that he had forgotten to feed his only cat for yet another time (guess why there's only one cat). The cat's current hunger level is H points, moreover each minute without food increases his hunger by D points.
At any time Andrew can visit the store where tasty buns are sold (you can assume that is doesn't take time to get to the store and back). One such bun costs C roubles and decreases hunger by N points. Since the demand for bakery drops heavily in the evening, there is a special 20% discount for buns starting from 20:00 (note that the cost might become rational). Of course, buns cannot be sold by parts.
Determine the minimum amount of money Andrew has to spend in order to feed his cat. The cat is considered fed if its hunger level is less than or equal to zero.
Input
The first line contains two integers hh and mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59) — the time of Andrew's awakening.
The second line contains four integers H, D, C and N (1 ≤ H ≤ 105, 1 ≤ D, C, N ≤ 102).
Output
Output the minimum amount of money to within three decimal digits. You answer is considered correct, if its absolute or relative error does not exceed 10 - 4.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.
Examples
Input
19 00
255 1 100 1
Output
25200.0000
Input
17 41
1000 6 15 11
Output
1365.0000
Note
In the first sample Andrew can visit the store at exactly 20:00. The cat's hunger will be equal to 315, hence it will be necessary to purchase 315 buns. The discount makes the final answer 25200 roubles.
In the second sample it's optimal to visit the store right after he wakes up. Then he'll have to buy 91 bins per 15 roubles each and spend a total of 1365 roubles. | instruction | 0 | 17,413 | 10 | 34,826 |
Tags: greedy, math
Correct Solution:
```
hr, m = [int(x) for x in input().strip().split(" ")]
h, d, c, n = [int(x) for x in input().strip().split(" ")]
buns = h//n
if h % n > 0:
buns += 1
rightAway = buns * c
later = rightAway
if hr >= 20 and hr <= 23:
rightAway -= (rightAway * .2)
else:
dMin = 60 - m
dHr = 20 - (hr + 1)
dMin += 60 * dHr
h += d * dMin
buns = h // n
if h%n > 0:
buns += 1
later = (buns * c) - ((buns*c) * .2)
print(min(rightAway, later))
``` | output | 1 | 17,413 | 10 | 34,827 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After waking up at hh:mm, Andrew realised that he had forgotten to feed his only cat for yet another time (guess why there's only one cat). The cat's current hunger level is H points, moreover each minute without food increases his hunger by D points.
At any time Andrew can visit the store where tasty buns are sold (you can assume that is doesn't take time to get to the store and back). One such bun costs C roubles and decreases hunger by N points. Since the demand for bakery drops heavily in the evening, there is a special 20% discount for buns starting from 20:00 (note that the cost might become rational). Of course, buns cannot be sold by parts.
Determine the minimum amount of money Andrew has to spend in order to feed his cat. The cat is considered fed if its hunger level is less than or equal to zero.
Input
The first line contains two integers hh and mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59) — the time of Andrew's awakening.
The second line contains four integers H, D, C and N (1 ≤ H ≤ 105, 1 ≤ D, C, N ≤ 102).
Output
Output the minimum amount of money to within three decimal digits. You answer is considered correct, if its absolute or relative error does not exceed 10 - 4.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.
Examples
Input
19 00
255 1 100 1
Output
25200.0000
Input
17 41
1000 6 15 11
Output
1365.0000
Note
In the first sample Andrew can visit the store at exactly 20:00. The cat's hunger will be equal to 315, hence it will be necessary to purchase 315 buns. The discount makes the final answer 25200 roubles.
In the second sample it's optimal to visit the store right after he wakes up. Then he'll have to buy 91 bins per 15 roubles each and spend a total of 1365 roubles. | instruction | 0 | 17,414 | 10 | 34,828 |
Tags: greedy, math
Correct Solution:
```
import collections
import os
import sys
from io import BytesIO, IOBase
import math
ins = lambda: [int(x) for x in input()]
inp = lambda: int(input())
inps = lambda: [int(x) for x in input().split()]
def main():
#t=int(input())
#for _ in range(t):
hh,mm=inps()
#print(hh,mm)
#n=int(input())
h,d,c,n=list(map(int,input().split()))
#p=max(max(a,b),2*min(a,b))
# region fastio
tm=0
cost=0
if hh<20:
tm+=60-mm
tm+=(20-hh-1)*60
#print("min",tm)
tm*=d
# print("min",tm)
#print("c1",c*math.ceil(h/n))
#print("c2",0.8*c*math.ceil((tm+h)/n))
cost=min(c*math.ceil(h/n),0.8*c*math.ceil((tm+h)/n))
else:
cost=0.8*c*math.ceil(h/n)
#print("{0:.15f}".format(cost))
print (round(cost,4)) #print ('%.2f'%a)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | output | 1 | 17,414 | 10 | 34,829 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After waking up at hh:mm, Andrew realised that he had forgotten to feed his only cat for yet another time (guess why there's only one cat). The cat's current hunger level is H points, moreover each minute without food increases his hunger by D points.
At any time Andrew can visit the store where tasty buns are sold (you can assume that is doesn't take time to get to the store and back). One such bun costs C roubles and decreases hunger by N points. Since the demand for bakery drops heavily in the evening, there is a special 20% discount for buns starting from 20:00 (note that the cost might become rational). Of course, buns cannot be sold by parts.
Determine the minimum amount of money Andrew has to spend in order to feed his cat. The cat is considered fed if its hunger level is less than or equal to zero.
Input
The first line contains two integers hh and mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59) — the time of Andrew's awakening.
The second line contains four integers H, D, C and N (1 ≤ H ≤ 105, 1 ≤ D, C, N ≤ 102).
Output
Output the minimum amount of money to within three decimal digits. You answer is considered correct, if its absolute or relative error does not exceed 10 - 4.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.
Examples
Input
19 00
255 1 100 1
Output
25200.0000
Input
17 41
1000 6 15 11
Output
1365.0000
Note
In the first sample Andrew can visit the store at exactly 20:00. The cat's hunger will be equal to 315, hence it will be necessary to purchase 315 buns. The discount makes the final answer 25200 roubles.
In the second sample it's optimal to visit the store right after he wakes up. Then he'll have to buy 91 bins per 15 roubles each and spend a total of 1365 roubles. | instruction | 0 | 17,415 | 10 | 34,830 |
Tags: greedy, math
Correct Solution:
```
from math import ceil
[hh, mm] = list(map(int, input().split()))
[H, D, C, N] = list(map(int, input().split()))
S = 0
if 20 <= hh <= 23:
S = 0.8 * C * ceil(H / N)
elif 0 <= hh <= 19:
S = min(C * ceil(H / N), 0.8 * C * ceil((H + D * (1200 - (hh * 60 + mm))) / N))
print (round(S, 3))
``` | output | 1 | 17,415 | 10 | 34,831 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After waking up at hh:mm, Andrew realised that he had forgotten to feed his only cat for yet another time (guess why there's only one cat). The cat's current hunger level is H points, moreover each minute without food increases his hunger by D points.
At any time Andrew can visit the store where tasty buns are sold (you can assume that is doesn't take time to get to the store and back). One such bun costs C roubles and decreases hunger by N points. Since the demand for bakery drops heavily in the evening, there is a special 20% discount for buns starting from 20:00 (note that the cost might become rational). Of course, buns cannot be sold by parts.
Determine the minimum amount of money Andrew has to spend in order to feed his cat. The cat is considered fed if its hunger level is less than or equal to zero.
Input
The first line contains two integers hh and mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59) — the time of Andrew's awakening.
The second line contains four integers H, D, C and N (1 ≤ H ≤ 105, 1 ≤ D, C, N ≤ 102).
Output
Output the minimum amount of money to within three decimal digits. You answer is considered correct, if its absolute or relative error does not exceed 10 - 4.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.
Examples
Input
19 00
255 1 100 1
Output
25200.0000
Input
17 41
1000 6 15 11
Output
1365.0000
Note
In the first sample Andrew can visit the store at exactly 20:00. The cat's hunger will be equal to 315, hence it will be necessary to purchase 315 buns. The discount makes the final answer 25200 roubles.
In the second sample it's optimal to visit the store right after he wakes up. Then he'll have to buy 91 bins per 15 roubles each and spend a total of 1365 roubles.
Submitted Solution:
```
t=input()
t=t.split()
t1=int(t[0])
t2=int(t[1])
a=input()
a=a.split()
h=int(a[0])
d=int(a[1])
c=int(a[2])
n=int(a[3])
if t1>=20:
c=c*(8/10)
q=h//n
if h//n-h/n!=0:
q=q+1
print((q)*c)
else:
td=(20-t1)*60-t2
h2=h+td*d
q2=h2//n
q=h//n
if h2//n-h2/n!=0:
q2=q2+1
if h//n-h/n!=0:
q=q+1
print(min((q2)*c*(8/10),(q)*c))
``` | instruction | 0 | 17,416 | 10 | 34,832 |
Yes | output | 1 | 17,416 | 10 | 34,833 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After waking up at hh:mm, Andrew realised that he had forgotten to feed his only cat for yet another time (guess why there's only one cat). The cat's current hunger level is H points, moreover each minute without food increases his hunger by D points.
At any time Andrew can visit the store where tasty buns are sold (you can assume that is doesn't take time to get to the store and back). One such bun costs C roubles and decreases hunger by N points. Since the demand for bakery drops heavily in the evening, there is a special 20% discount for buns starting from 20:00 (note that the cost might become rational). Of course, buns cannot be sold by parts.
Determine the minimum amount of money Andrew has to spend in order to feed his cat. The cat is considered fed if its hunger level is less than or equal to zero.
Input
The first line contains two integers hh and mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59) — the time of Andrew's awakening.
The second line contains four integers H, D, C and N (1 ≤ H ≤ 105, 1 ≤ D, C, N ≤ 102).
Output
Output the minimum amount of money to within three decimal digits. You answer is considered correct, if its absolute or relative error does not exceed 10 - 4.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.
Examples
Input
19 00
255 1 100 1
Output
25200.0000
Input
17 41
1000 6 15 11
Output
1365.0000
Note
In the first sample Andrew can visit the store at exactly 20:00. The cat's hunger will be equal to 315, hence it will be necessary to purchase 315 buns. The discount makes the final answer 25200 roubles.
In the second sample it's optimal to visit the store right after he wakes up. Then he'll have to buy 91 bins per 15 roubles each and spend a total of 1365 roubles.
Submitted Solution:
```
hh, mm = map(int, input().split())
H, D, C, N = map(int, input().split())
time = hh * 60 + mm
if time >= 20 * 60:
print((H + N - 1) // N * C * 0.8)
else:
print(min((H + N - 1) // N * C, (N - 1 + (20 * 60 - time) * D + H) // N * C * 0.8))
``` | instruction | 0 | 17,417 | 10 | 34,834 |
Yes | output | 1 | 17,417 | 10 | 34,835 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After waking up at hh:mm, Andrew realised that he had forgotten to feed his only cat for yet another time (guess why there's only one cat). The cat's current hunger level is H points, moreover each minute without food increases his hunger by D points.
At any time Andrew can visit the store where tasty buns are sold (you can assume that is doesn't take time to get to the store and back). One such bun costs C roubles and decreases hunger by N points. Since the demand for bakery drops heavily in the evening, there is a special 20% discount for buns starting from 20:00 (note that the cost might become rational). Of course, buns cannot be sold by parts.
Determine the minimum amount of money Andrew has to spend in order to feed his cat. The cat is considered fed if its hunger level is less than or equal to zero.
Input
The first line contains two integers hh and mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59) — the time of Andrew's awakening.
The second line contains four integers H, D, C and N (1 ≤ H ≤ 105, 1 ≤ D, C, N ≤ 102).
Output
Output the minimum amount of money to within three decimal digits. You answer is considered correct, if its absolute or relative error does not exceed 10 - 4.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.
Examples
Input
19 00
255 1 100 1
Output
25200.0000
Input
17 41
1000 6 15 11
Output
1365.0000
Note
In the first sample Andrew can visit the store at exactly 20:00. The cat's hunger will be equal to 315, hence it will be necessary to purchase 315 buns. The discount makes the final answer 25200 roubles.
In the second sample it's optimal to visit the store right after he wakes up. Then he'll have to buy 91 bins per 15 roubles each and spend a total of 1365 roubles.
Submitted Solution:
```
stroka1=input().split()
stroka2=input().split()
hh, mm = int(stroka1[0]), int(stroka1[1])
H, D, C, N = int(stroka2[0]), int(stroka2[1]), float(stroka2[2]), int(stroka2[3])
price1=float(0)
price2=float(0)
eda=0
while eda<H:
eda+=N
n=eda/N
eda=0
cena1=n*C
if hh<20:
while eda<((20-hh-1)*60+60-mm)*D:
eda+=D
n1=(eda/D)+n
cena2=n1*(C-C/5)
if cena1<cena2:
print(round(cena1, 3))
elif cena1>round(cena2, 3):
print(round(cena2, 3))
elif cena1==cena2:
print(round(cena2, 3))
else:
print(round(n*(C-C/5), 3))
``` | instruction | 0 | 17,420 | 10 | 34,840 |
No | output | 1 | 17,420 | 10 | 34,841 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After waking up at hh:mm, Andrew realised that he had forgotten to feed his only cat for yet another time (guess why there's only one cat). The cat's current hunger level is H points, moreover each minute without food increases his hunger by D points.
At any time Andrew can visit the store where tasty buns are sold (you can assume that is doesn't take time to get to the store and back). One such bun costs C roubles and decreases hunger by N points. Since the demand for bakery drops heavily in the evening, there is a special 20% discount for buns starting from 20:00 (note that the cost might become rational). Of course, buns cannot be sold by parts.
Determine the minimum amount of money Andrew has to spend in order to feed his cat. The cat is considered fed if its hunger level is less than or equal to zero.
Input
The first line contains two integers hh and mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59) — the time of Andrew's awakening.
The second line contains four integers H, D, C and N (1 ≤ H ≤ 105, 1 ≤ D, C, N ≤ 102).
Output
Output the minimum amount of money to within three decimal digits. You answer is considered correct, if its absolute or relative error does not exceed 10 - 4.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.
Examples
Input
19 00
255 1 100 1
Output
25200.0000
Input
17 41
1000 6 15 11
Output
1365.0000
Note
In the first sample Andrew can visit the store at exactly 20:00. The cat's hunger will be equal to 315, hence it will be necessary to purchase 315 buns. The discount makes the final answer 25200 roubles.
In the second sample it's optimal to visit the store right after he wakes up. Then he'll have to buy 91 bins per 15 roubles each and spend a total of 1365 roubles.
Submitted Solution:
```
from datetime import timedelta
a ,b = map(int , input().split())
x = timedelta(hours=a , minutes=b)
y= timedelta(hours=20)
if x < y:
diff = y - x
diff = diff.total_seconds() // 60
else:
diff = y - x
diff = diff.total_seconds() // 60 + (24 * 60)
h , d , c , n = map(int , input().split())
need = (h + n - 1) // n;
cost1 = need * c
c = 0.8 * c
h += diff * d
need = (h + n - 1) // n;
cost2 = need * c
ans = (min(cost2 , cost1))
print(f"{ans:.4f}")
``` | instruction | 0 | 17,421 | 10 | 34,842 |
No | output | 1 | 17,421 | 10 | 34,843 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After waking up at hh:mm, Andrew realised that he had forgotten to feed his only cat for yet another time (guess why there's only one cat). The cat's current hunger level is H points, moreover each minute without food increases his hunger by D points.
At any time Andrew can visit the store where tasty buns are sold (you can assume that is doesn't take time to get to the store and back). One such bun costs C roubles and decreases hunger by N points. Since the demand for bakery drops heavily in the evening, there is a special 20% discount for buns starting from 20:00 (note that the cost might become rational). Of course, buns cannot be sold by parts.
Determine the minimum amount of money Andrew has to spend in order to feed his cat. The cat is considered fed if its hunger level is less than or equal to zero.
Input
The first line contains two integers hh and mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59) — the time of Andrew's awakening.
The second line contains four integers H, D, C and N (1 ≤ H ≤ 105, 1 ≤ D, C, N ≤ 102).
Output
Output the minimum amount of money to within three decimal digits. You answer is considered correct, if its absolute or relative error does not exceed 10 - 4.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.
Examples
Input
19 00
255 1 100 1
Output
25200.0000
Input
17 41
1000 6 15 11
Output
1365.0000
Note
In the first sample Andrew can visit the store at exactly 20:00. The cat's hunger will be equal to 315, hence it will be necessary to purchase 315 buns. The discount makes the final answer 25200 roubles.
In the second sample it's optimal to visit the store right after he wakes up. Then he'll have to buy 91 bins per 15 roubles each and spend a total of 1365 roubles.
Submitted Solution:
```
from datetime import datetime, timedelta
def main():
HH, MM = input().split()
H, D, C, N = map(int, input().split())
d1 = datetime.strptime('20:00', '%H:%M')
t = HH + ':' + MM
d2 = datetime.strptime(t, '%H:%M')
s = abs(d1 - d2).total_seconds() / 60
a = -(-H // N) * C
b = -(-(D * s + H) // N) * C * 0.8
print(min(a, b))
main()
``` | instruction | 0 | 17,422 | 10 | 34,844 |
No | output | 1 | 17,422 | 10 | 34,845 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After waking up at hh:mm, Andrew realised that he had forgotten to feed his only cat for yet another time (guess why there's only one cat). The cat's current hunger level is H points, moreover each minute without food increases his hunger by D points.
At any time Andrew can visit the store where tasty buns are sold (you can assume that is doesn't take time to get to the store and back). One such bun costs C roubles and decreases hunger by N points. Since the demand for bakery drops heavily in the evening, there is a special 20% discount for buns starting from 20:00 (note that the cost might become rational). Of course, buns cannot be sold by parts.
Determine the minimum amount of money Andrew has to spend in order to feed his cat. The cat is considered fed if its hunger level is less than or equal to zero.
Input
The first line contains two integers hh and mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59) — the time of Andrew's awakening.
The second line contains four integers H, D, C and N (1 ≤ H ≤ 105, 1 ≤ D, C, N ≤ 102).
Output
Output the minimum amount of money to within three decimal digits. You answer is considered correct, if its absolute or relative error does not exceed 10 - 4.
Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.
Examples
Input
19 00
255 1 100 1
Output
25200.0000
Input
17 41
1000 6 15 11
Output
1365.0000
Note
In the first sample Andrew can visit the store at exactly 20:00. The cat's hunger will be equal to 315, hence it will be necessary to purchase 315 buns. The discount makes the final answer 25200 roubles.
In the second sample it's optimal to visit the store right after he wakes up. Then he'll have to buy 91 bins per 15 roubles each and spend a total of 1365 roubles.
Submitted Solution:
```
import math
hh, mm = [int(x) for x in input().split()]
h,d,c,n = [int(x) for x in input().split()]
dift = 20*60-(hh*60+mm)
z=(dift*d+h)/n*c*0.8
s=(h)/n*c
if hh>=20:
print((math.ceil(h/n)*c*0.8))
elif s>z:
print((math.ceil((dift*d+h)/n)*c*0.8))
else:
print((math.ceil((h)/n)*c))
``` | instruction | 0 | 17,423 | 10 | 34,846 |
No | output | 1 | 17,423 | 10 | 34,847 |
Provide a correct Python 3 solution for this coding contest problem.
Constraints
* All values in input are integers.
* 1\leq N, M\leq 12
* 1\leq X\leq 10^5
* 1\leq C_i \leq 10^5
* 0\leq A_{i, j} \leq 10^5
Input
Input is given from Standard Input in the following format:
N M X
C_1 A_{1,1} A_{1,2} \cdots A_{1,M}
C_2 A_{2,1} A_{2,2} \cdots A_{2,M}
\vdots
C_N A_{N,1} A_{N,2} \cdots A_{N,M}
Output
If the objective is not achievable, print `-1`; otherwise, print the minimum amount of money needed to achieve it.
Examples
Input
3 3 10
60 2 2 4
70 8 7 9
50 2 3 9
Output
120
Input
3 3 10
100 3 1 4
100 1 5 9
100 2 6 5
Output
-1
Input
8 5 22
100 3 7 5 3 1
164 4 5 2 7 8
334 7 2 7 2 9
234 4 7 2 8 2
541 5 4 3 3 6
235 4 8 6 9 7
394 3 6 1 6 2
872 8 4 3 7 2
Output
1067 | instruction | 0 | 17,461 | 10 | 34,922 |
"Correct Solution:
```
n, m, x = map(int, input().split())
ca = [list(map(int, input().split())) for i in range(n)]
ans = 10 ** 15
for i in range(2 ** n):
rikai = [0] * m
money = 0
for j in range(n):
if (i >> j) & 1:
for k in range(m):
rikai[k] += ca[j][k+1]
money += ca[j][0]
if min(rikai) >= x:
ans = min(ans, money)
if ans == 10 ** 15:
print(-1)
else:
print(ans)
``` | output | 1 | 17,461 | 10 | 34,923 |
Provide a correct Python 3 solution for this coding contest problem.
Constraints
* All values in input are integers.
* 1\leq N, M\leq 12
* 1\leq X\leq 10^5
* 1\leq C_i \leq 10^5
* 0\leq A_{i, j} \leq 10^5
Input
Input is given from Standard Input in the following format:
N M X
C_1 A_{1,1} A_{1,2} \cdots A_{1,M}
C_2 A_{2,1} A_{2,2} \cdots A_{2,M}
\vdots
C_N A_{N,1} A_{N,2} \cdots A_{N,M}
Output
If the objective is not achievable, print `-1`; otherwise, print the minimum amount of money needed to achieve it.
Examples
Input
3 3 10
60 2 2 4
70 8 7 9
50 2 3 9
Output
120
Input
3 3 10
100 3 1 4
100 1 5 9
100 2 6 5
Output
-1
Input
8 5 22
100 3 7 5 3 1
164 4 5 2 7 8
334 7 2 7 2 9
234 4 7 2 8 2
541 5 4 3 3 6
235 4 8 6 9 7
394 3 6 1 6 2
872 8 4 3 7 2
Output
1067 | instruction | 0 | 17,462 | 10 | 34,924 |
"Correct Solution:
```
N, M, X = map(int, input().split())
books = [list(map(int, input().split())) for _ in range(N)]
ans = -1
for bits in range(1 << N):
cost = 0
a = [0] * M
for i in range(N):
if (bits >> i) & 1 == 1:
cost += books[i][0]
for j in range(M):
a[j] += books[i][j + 1]
if min(a) >= X:
ans = min(cost, ans) if ans >= 0 else cost
print(ans)
``` | output | 1 | 17,462 | 10 | 34,925 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has K 500-yen coins. (Yen is the currency of Japan.) If these coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Constraints
* 1 \leq K \leq 100
* 1 \leq X \leq 10^5
Input
Input is given from Standard Input in the following format:
K X
Output
If the coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Examples
Input
2 900
Output
Yes
Input
1 501
Output
No
Input
4 2000
Output
Yes | instruction | 0 | 17,472 | 10 | 34,944 |
"Correct Solution:
```
K, X = map(int, input().split())
print(("Yes" if 500*K >= X else "No"))
``` | output | 1 | 17,472 | 10 | 34,945 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has K 500-yen coins. (Yen is the currency of Japan.) If these coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Constraints
* 1 \leq K \leq 100
* 1 \leq X \leq 10^5
Input
Input is given from Standard Input in the following format:
K X
Output
If the coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Examples
Input
2 900
Output
Yes
Input
1 501
Output
No
Input
4 2000
Output
Yes | instruction | 0 | 17,473 | 10 | 34,946 |
"Correct Solution:
```
print("YNeos"[eval(input().replace(" ", "*500<"))::2])
``` | output | 1 | 17,473 | 10 | 34,947 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has K 500-yen coins. (Yen is the currency of Japan.) If these coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Constraints
* 1 \leq K \leq 100
* 1 \leq X \leq 10^5
Input
Input is given from Standard Input in the following format:
K X
Output
If the coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Examples
Input
2 900
Output
Yes
Input
1 501
Output
No
Input
4 2000
Output
Yes | instruction | 0 | 17,474 | 10 | 34,948 |
"Correct Solution:
```
a = list(map(int, input().split()))
print("Yes" if a[0]*500 >= a[1] else "No")
``` | output | 1 | 17,474 | 10 | 34,949 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has K 500-yen coins. (Yen is the currency of Japan.) If these coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Constraints
* 1 \leq K \leq 100
* 1 \leq X \leq 10^5
Input
Input is given from Standard Input in the following format:
K X
Output
If the coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Examples
Input
2 900
Output
Yes
Input
1 501
Output
No
Input
4 2000
Output
Yes | instruction | 0 | 17,475 | 10 | 34,950 |
"Correct Solution:
```
X,Y=map(int,input().split())
print('Yes' if X*500 >= Y else 'No')
``` | output | 1 | 17,475 | 10 | 34,951 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has K 500-yen coins. (Yen is the currency of Japan.) If these coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Constraints
* 1 \leq K \leq 100
* 1 \leq X \leq 10^5
Input
Input is given from Standard Input in the following format:
K X
Output
If the coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Examples
Input
2 900
Output
Yes
Input
1 501
Output
No
Input
4 2000
Output
Yes | instruction | 0 | 17,476 | 10 | 34,952 |
"Correct Solution:
```
k,x=map(int,input().split())
print('Yes' if (k*500) >= x else 'No')
``` | output | 1 | 17,476 | 10 | 34,953 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has K 500-yen coins. (Yen is the currency of Japan.) If these coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Constraints
* 1 \leq K \leq 100
* 1 \leq X \leq 10^5
Input
Input is given from Standard Input in the following format:
K X
Output
If the coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Examples
Input
2 900
Output
Yes
Input
1 501
Output
No
Input
4 2000
Output
Yes | instruction | 0 | 17,477 | 10 | 34,954 |
"Correct Solution:
```
a,k=map(int,input().split())
if a*500 >=k:
print("Yes")
else:
print("No")
``` | output | 1 | 17,477 | 10 | 34,955 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has K 500-yen coins. (Yen is the currency of Japan.) If these coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Constraints
* 1 \leq K \leq 100
* 1 \leq X \leq 10^5
Input
Input is given from Standard Input in the following format:
K X
Output
If the coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Examples
Input
2 900
Output
Yes
Input
1 501
Output
No
Input
4 2000
Output
Yes | instruction | 0 | 17,478 | 10 | 34,956 |
"Correct Solution:
```
K,X=map(int,input().split())
f=500*K>=X
print('Yes' if f else 'No')
``` | output | 1 | 17,478 | 10 | 34,957 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi has K 500-yen coins. (Yen is the currency of Japan.) If these coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Constraints
* 1 \leq K \leq 100
* 1 \leq X \leq 10^5
Input
Input is given from Standard Input in the following format:
K X
Output
If the coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Examples
Input
2 900
Output
Yes
Input
1 501
Output
No
Input
4 2000
Output
Yes | instruction | 0 | 17,479 | 10 | 34,958 |
"Correct Solution:
```
print("Yes" if eval(input().replace(" ","*500>=")) else "No")
``` | output | 1 | 17,479 | 10 | 34,959 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has K 500-yen coins. (Yen is the currency of Japan.) If these coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Constraints
* 1 \leq K \leq 100
* 1 \leq X \leq 10^5
Input
Input is given from Standard Input in the following format:
K X
Output
If the coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Examples
Input
2 900
Output
Yes
Input
1 501
Output
No
Input
4 2000
Output
Yes
Submitted Solution:
```
K, X = map(int, input().split());print('Yes') if K * 500 >= X else print('No')
``` | instruction | 0 | 17,480 | 10 | 34,960 |
Yes | output | 1 | 17,480 | 10 | 34,961 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has K 500-yen coins. (Yen is the currency of Japan.) If these coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Constraints
* 1 \leq K \leq 100
* 1 \leq X \leq 10^5
Input
Input is given from Standard Input in the following format:
K X
Output
If the coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Examples
Input
2 900
Output
Yes
Input
1 501
Output
No
Input
4 2000
Output
Yes
Submitted Solution:
```
k,x=map(int,input().split())
print("YNeos"[k*500<x:5:2])
``` | instruction | 0 | 17,481 | 10 | 34,962 |
Yes | output | 1 | 17,481 | 10 | 34,963 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has K 500-yen coins. (Yen is the currency of Japan.) If these coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Constraints
* 1 \leq K \leq 100
* 1 \leq X \leq 10^5
Input
Input is given from Standard Input in the following format:
K X
Output
If the coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Examples
Input
2 900
Output
Yes
Input
1 501
Output
No
Input
4 2000
Output
Yes
Submitted Solution:
```
K, X = list(map(int, input().split()))
print('Yes' if X <= K*500 else 'No')
``` | instruction | 0 | 17,482 | 10 | 34,964 |
Yes | output | 1 | 17,482 | 10 | 34,965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has K 500-yen coins. (Yen is the currency of Japan.) If these coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Constraints
* 1 \leq K \leq 100
* 1 \leq X \leq 10^5
Input
Input is given from Standard Input in the following format:
K X
Output
If the coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Examples
Input
2 900
Output
Yes
Input
1 501
Output
No
Input
4 2000
Output
Yes
Submitted Solution:
```
k,x = map(int,input().split())
print('Yes'if 500*k >= x else "No")
``` | instruction | 0 | 17,483 | 10 | 34,966 |
Yes | output | 1 | 17,483 | 10 | 34,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has K 500-yen coins. (Yen is the currency of Japan.) If these coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Constraints
* 1 \leq K \leq 100
* 1 \leq X \leq 10^5
Input
Input is given from Standard Input in the following format:
K X
Output
If the coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Examples
Input
2 900
Output
Yes
Input
1 501
Output
No
Input
4 2000
Output
Yes
Submitted Solution:
```
a,b = map(int, input().split())
if 500*a > b:
print("YES")
else:
print("NO")
``` | instruction | 0 | 17,484 | 10 | 34,968 |
No | output | 1 | 17,484 | 10 | 34,969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has K 500-yen coins. (Yen is the currency of Japan.) If these coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Constraints
* 1 \leq K \leq 100
* 1 \leq X \leq 10^5
Input
Input is given from Standard Input in the following format:
K X
Output
If the coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Examples
Input
2 900
Output
Yes
Input
1 501
Output
No
Input
4 2000
Output
Yes
Submitted Solution:
```
a,b=map(int,input().split())
if a*500>=b :
print('YES')
else:
print('NO')
``` | instruction | 0 | 17,485 | 10 | 34,970 |
No | output | 1 | 17,485 | 10 | 34,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has K 500-yen coins. (Yen is the currency of Japan.) If these coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Constraints
* 1 \leq K \leq 100
* 1 \leq X \leq 10^5
Input
Input is given from Standard Input in the following format:
K X
Output
If the coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Examples
Input
2 900
Output
Yes
Input
1 501
Output
No
Input
4 2000
Output
Yes
Submitted Solution:
```
K, X = input().split()
if K * 500 > X:
print("Yes")
else:
print("No")
``` | instruction | 0 | 17,486 | 10 | 34,972 |
No | output | 1 | 17,486 | 10 | 34,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi has K 500-yen coins. (Yen is the currency of Japan.) If these coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Constraints
* 1 \leq K \leq 100
* 1 \leq X \leq 10^5
Input
Input is given from Standard Input in the following format:
K X
Output
If the coins add up to X yen or more, print `Yes`; otherwise, print `No`.
Examples
Input
2 900
Output
Yes
Input
1 501
Output
No
Input
4 2000
Output
Yes
Submitted Solution:
```
a,b = map(int,input().split(" "))
if a * 500 >= b:
print("yes")
else:
print("no")
``` | instruction | 0 | 17,487 | 10 | 34,974 |
No | output | 1 | 17,487 | 10 | 34,975 |
Provide a correct Python 3 solution for this coding contest problem.
There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of B_i after A_i days from the day you do it.
You can take and complete at most one of these jobs in a day.
However, you cannot retake a job that you have already done.
Find the maximum total reward that you can earn no later than M days from today.
You can already start working today.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^5
* 1 \leq A_i \leq 10^5
* 1 \leq B_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N
Output
Print the maximum total reward that you can earn no later than M days from today.
Examples
Input
3 4
4 3
4 1
2 2
Output
5
Input
5 3
1 2
1 3
1 4
2 1
2 3
Output
10
Input
1 1
2 1
Output
0 | instruction | 0 | 17,488 | 10 | 34,976 |
"Correct Solution:
```
import heapq as hq
n,m=map(int,input().split())
a=[list(map(int,input().split())) for i in range(n)]
a.sort()
for i in range(n):
a[i][1]*=-1
b=[]
hq.heapify(b)
ans,i=0,0
for k in range(m-1,-1,-1):
while i<n and a[i][0]+k<=m:
hq.heappush(b,a[i][1])
i+=1
if b:
ans-=hq.heappop(b)
print(ans)
``` | output | 1 | 17,488 | 10 | 34,977 |
Provide a correct Python 3 solution for this coding contest problem.
There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of B_i after A_i days from the day you do it.
You can take and complete at most one of these jobs in a day.
However, you cannot retake a job that you have already done.
Find the maximum total reward that you can earn no later than M days from today.
You can already start working today.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^5
* 1 \leq A_i \leq 10^5
* 1 \leq B_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N
Output
Print the maximum total reward that you can earn no later than M days from today.
Examples
Input
3 4
4 3
4 1
2 2
Output
5
Input
5 3
1 2
1 3
1 4
2 1
2 3
Output
10
Input
1 1
2 1
Output
0 | instruction | 0 | 17,489 | 10 | 34,978 |
"Correct Solution:
```
import heapq
n,m=map(int,input().split())
L=[[] for _ in range(m)]
for _ in range(n):
a,b=map(int,input().split())
if a<m+1:
L[a-1].append(-b)
h=[]
heapq.heapify(h)
ans=0
for i in range(m):
for j in L[i]:
heapq.heappush(h,j)
if len(h)==0:
continue
ans+=heapq.heappop(h)*(-1)
print(ans)
``` | output | 1 | 17,489 | 10 | 34,979 |
Provide a correct Python 3 solution for this coding contest problem.
There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of B_i after A_i days from the day you do it.
You can take and complete at most one of these jobs in a day.
However, you cannot retake a job that you have already done.
Find the maximum total reward that you can earn no later than M days from today.
You can already start working today.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^5
* 1 \leq A_i \leq 10^5
* 1 \leq B_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N
Output
Print the maximum total reward that you can earn no later than M days from today.
Examples
Input
3 4
4 3
4 1
2 2
Output
5
Input
5 3
1 2
1 3
1 4
2 1
2 3
Output
10
Input
1 1
2 1
Output
0 | instruction | 0 | 17,490 | 10 | 34,980 |
"Correct Solution:
```
from heapq import heappop, heappush
n, m = map(int, input().split())
d = []
for _ in range(n):
a, b = map(int, input().split())
d.append((m - a, b))
d.sort()
ans = 0
q = []
for i in range(m - 1, -1, -1):
while d and d[-1][0] == i:
heappush(q, -d.pop()[1])
if q:
ans -= heappop(q)
print(ans)
``` | output | 1 | 17,490 | 10 | 34,981 |
Provide a correct Python 3 solution for this coding contest problem.
There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of B_i after A_i days from the day you do it.
You can take and complete at most one of these jobs in a day.
However, you cannot retake a job that you have already done.
Find the maximum total reward that you can earn no later than M days from today.
You can already start working today.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^5
* 1 \leq A_i \leq 10^5
* 1 \leq B_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N
Output
Print the maximum total reward that you can earn no later than M days from today.
Examples
Input
3 4
4 3
4 1
2 2
Output
5
Input
5 3
1 2
1 3
1 4
2 1
2 3
Output
10
Input
1 1
2 1
Output
0 | instruction | 0 | 17,491 | 10 | 34,982 |
"Correct Solution:
```
import heapq
N, M = map(int, input().split())
W = [[] for _ in range(100005)]
for i in range(N):
A, B = map(int, input().split())
W[A].append(B)
hq = []
ans = 0
for A in range(1, M+1):
for B in W[A]:
heapq.heappush(hq, -B)
if hq:
B = heapq.heappop(hq)
ans -= B
print(ans)
``` | output | 1 | 17,491 | 10 | 34,983 |
Provide a correct Python 3 solution for this coding contest problem.
There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of B_i after A_i days from the day you do it.
You can take and complete at most one of these jobs in a day.
However, you cannot retake a job that you have already done.
Find the maximum total reward that you can earn no later than M days from today.
You can already start working today.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^5
* 1 \leq A_i \leq 10^5
* 1 \leq B_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N
Output
Print the maximum total reward that you can earn no later than M days from today.
Examples
Input
3 4
4 3
4 1
2 2
Output
5
Input
5 3
1 2
1 3
1 4
2 1
2 3
Output
10
Input
1 1
2 1
Output
0 | instruction | 0 | 17,492 | 10 | 34,984 |
"Correct Solution:
```
from collections import defaultdict
import heapq
n, m = map(int, input().split())
d = defaultdict(list)
for i in range(n):
a, b = map(int, input().split())
d[a-1].append(-b)
cnt = 0
a = []
heapq.heapify(a)
for i in range(m):
for j in d[i]:
heapq.heappush(a, j)
if a:
cnt += -heapq.heappop(a)
print(cnt)
``` | output | 1 | 17,492 | 10 | 34,985 |
Provide a correct Python 3 solution for this coding contest problem.
There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of B_i after A_i days from the day you do it.
You can take and complete at most one of these jobs in a day.
However, you cannot retake a job that you have already done.
Find the maximum total reward that you can earn no later than M days from today.
You can already start working today.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^5
* 1 \leq A_i \leq 10^5
* 1 \leq B_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N
Output
Print the maximum total reward that you can earn no later than M days from today.
Examples
Input
3 4
4 3
4 1
2 2
Output
5
Input
5 3
1 2
1 3
1 4
2 1
2 3
Output
10
Input
1 1
2 1
Output
0 | instruction | 0 | 17,493 | 10 | 34,986 |
"Correct Solution:
```
from heapq import *
N,M=map(int,input().split())
X=[[] for i in range(M)]
A,B=0,0
for i in range(N):
A,B=map(int,input().split())
if A>M:
continue
X[-A].append(B)
P=0
Q=[]
heapify(Q)
for i in range(M-1,-1,-1):
for j in range(len(X[i])):
heappush(Q,-X[i][j])
if len(Q):
P-=heappop(Q)
print(P)
``` | output | 1 | 17,493 | 10 | 34,987 |
Provide a correct Python 3 solution for this coding contest problem.
There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of B_i after A_i days from the day you do it.
You can take and complete at most one of these jobs in a day.
However, you cannot retake a job that you have already done.
Find the maximum total reward that you can earn no later than M days from today.
You can already start working today.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^5
* 1 \leq A_i \leq 10^5
* 1 \leq B_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N
Output
Print the maximum total reward that you can earn no later than M days from today.
Examples
Input
3 4
4 3
4 1
2 2
Output
5
Input
5 3
1 2
1 3
1 4
2 1
2 3
Output
10
Input
1 1
2 1
Output
0 | instruction | 0 | 17,494 | 10 | 34,988 |
"Correct Solution:
```
import heapq
N,M = map(int,input().split())
AB = [list(map(int,input().split())) for _ in range (N)]
AB.sort()
ans = 0
q = []
heapq.heapify(q)
idx = 0
for i in range(1,M+1):
while idx<N and AB[idx][0]<=i:
heapq.heappush(q, -AB[idx][1])
idx += 1
if q:
p = -heapq.heappop(q)
ans += p
print(ans)
``` | output | 1 | 17,494 | 10 | 34,989 |
Provide a correct Python 3 solution for this coding contest problem.
There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of B_i after A_i days from the day you do it.
You can take and complete at most one of these jobs in a day.
However, you cannot retake a job that you have already done.
Find the maximum total reward that you can earn no later than M days from today.
You can already start working today.
Constraints
* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq M \leq 10^5
* 1 \leq A_i \leq 10^5
* 1 \leq B_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N
Output
Print the maximum total reward that you can earn no later than M days from today.
Examples
Input
3 4
4 3
4 1
2 2
Output
5
Input
5 3
1 2
1 3
1 4
2 1
2 3
Output
10
Input
1 1
2 1
Output
0 | instruction | 0 | 17,495 | 10 | 34,990 |
"Correct Solution:
```
from heapq import*
(N,M),*t=[map(int,s.split())for s in open(0)]
q=[0]*99*M
v=[[]for _ in q]
z=0
for a,b in t:v[a-1]+=b,
for i in v[:M]:
for j in i:heappush(q,-j)
z-=heappop(q)
print(z)
``` | output | 1 | 17,495 | 10 | 34,991 |
Provide a correct Python 3 solution for this coding contest problem.
Taro, a boy who hates any inefficiencies, pays coins so that the number of coins to be returned as change is minimized in order to do smoothly when he buys something.
One day, however, he doubt if this way is really efficient. When he pays more number of coins, a clerk consumes longer time to find the total value. Maybe he should pay with least possible number of coins.
Thinking for a while, he has decided to take the middle course. So he tries to minimize total number of paid coins and returned coins as change.
Now he is going to buy a product of P yen having several coins. Since he is not good at calculation, please write a program that computes the minimal number of coins.
You may assume following things:
* There are 6 kinds of coins, 1 yen, 5 yen, 10 yen, 50 yen, 100 yen and 500 yen.
* The total value of coins he has is at least P yen.
* A clerk will return the change with least number of coins.
Constraints
* Judge data contains at most 100 data sets.
* 0 ≤ Ni ≤ 1000
Input
Input file contains several data sets. One data set has following format:
P N1 N5 N10 N50 N100 N500
Ni is an integer and is the number of coins of i yen that he have.
The end of input is denoted by a case where P = 0. You should output nothing for this data set.
Output
Output total number of coins that are paid and are returned.
Example
Input
123 3 0 2 0 1 1
999 9 9 9 9 9 9
0 0 0 0 0 0 0
Output
6
3 | instruction | 0 | 17,635 | 10 | 35,270 |
"Correct Solution:
```
price = [1, 5, 10, 50, 100, 500]
while True:
values = list(map(int, input().split()))
p, n = values[0], values[1:]
ans = 1e100
if p == 0:
break
p_sum = sum(map(lambda s: s[0] * s[1], zip(price, n)))
for change in range(1000):
total = p + change
pay = [0] * 6
for i in reversed(range(6)):
if total >= price[i]:
pay[i] = min(n[i], int(total / price[i]))
total -= pay[i] * price[i]
if total > 0:
break
coins = sum(pay)
_change = change
for i in reversed(range(6)):
if _change >= price[i]:
coins += int(_change / price[i])
_change %= price[i]
ans = min(ans, coins)
print(ans)
``` | output | 1 | 17,635 | 10 | 35,271 |
Provide a correct Python 3 solution for this coding contest problem.
Taro, a boy who hates any inefficiencies, pays coins so that the number of coins to be returned as change is minimized in order to do smoothly when he buys something.
One day, however, he doubt if this way is really efficient. When he pays more number of coins, a clerk consumes longer time to find the total value. Maybe he should pay with least possible number of coins.
Thinking for a while, he has decided to take the middle course. So he tries to minimize total number of paid coins and returned coins as change.
Now he is going to buy a product of P yen having several coins. Since he is not good at calculation, please write a program that computes the minimal number of coins.
You may assume following things:
* There are 6 kinds of coins, 1 yen, 5 yen, 10 yen, 50 yen, 100 yen and 500 yen.
* The total value of coins he has is at least P yen.
* A clerk will return the change with least number of coins.
Constraints
* Judge data contains at most 100 data sets.
* 0 ≤ Ni ≤ 1000
Input
Input file contains several data sets. One data set has following format:
P N1 N5 N10 N50 N100 N500
Ni is an integer and is the number of coins of i yen that he have.
The end of input is denoted by a case where P = 0. You should output nothing for this data set.
Output
Output total number of coins that are paid and are returned.
Example
Input
123 3 0 2 0 1 1
999 9 9 9 9 9 9
0 0 0 0 0 0 0
Output
6
3 | instruction | 0 | 17,636 | 10 | 35,272 |
"Correct Solution:
```
while True:
p, n1, n5, n10, n50, n100, n500 = map(int, input().split())
if p == 0:break
values = (500, 100, 50, 10, 5, 1)
values_cnt = (n500, n100, n50, n10, n5, n1)
INF = 10 ** 20
def return_cnt(x):
ret = 0
for value in values:
ret += x // value
x %= value
return ret
def make_price(x):
ret = 0
for value, cnt in zip(values, values_cnt):
available_cnt = x // value
ret += min(available_cnt, cnt)
x -= value * min(available_cnt, cnt)
if x == 0:return ret
return INF
print(min([make_price(i) + return_cnt(i - p) for i in range(p, p + 500)]))
``` | output | 1 | 17,636 | 10 | 35,273 |
Provide a correct Python 3 solution for this coding contest problem.
Taro, a boy who hates any inefficiencies, pays coins so that the number of coins to be returned as change is minimized in order to do smoothly when he buys something.
One day, however, he doubt if this way is really efficient. When he pays more number of coins, a clerk consumes longer time to find the total value. Maybe he should pay with least possible number of coins.
Thinking for a while, he has decided to take the middle course. So he tries to minimize total number of paid coins and returned coins as change.
Now he is going to buy a product of P yen having several coins. Since he is not good at calculation, please write a program that computes the minimal number of coins.
You may assume following things:
* There are 6 kinds of coins, 1 yen, 5 yen, 10 yen, 50 yen, 100 yen and 500 yen.
* The total value of coins he has is at least P yen.
* A clerk will return the change with least number of coins.
Constraints
* Judge data contains at most 100 data sets.
* 0 ≤ Ni ≤ 1000
Input
Input file contains several data sets. One data set has following format:
P N1 N5 N10 N50 N100 N500
Ni is an integer and is the number of coins of i yen that he have.
The end of input is denoted by a case where P = 0. You should output nothing for this data set.
Output
Output total number of coins that are paid and are returned.
Example
Input
123 3 0 2 0 1 1
999 9 9 9 9 9 9
0 0 0 0 0 0 0
Output
6
3 | instruction | 0 | 17,637 | 10 | 35,274 |
"Correct Solution:
```
# AOJ 1028: ICPC: Ideal Coin Payment and Change
# Python3 2018.7.5 bal4u
c = (500, 100, 50, 10, 5, 1)
def pay(p):
cnt = 0
for i in range(6):
if c[i] <= p:
k = p//c[i]
if k > n[i]: k = n[i]
cnt += k
p -= c[i]*k
return -1 if p > 0 else cnt
k = [0]*1001
for p in range(1, 1001):
cnt, q = 0, p
for j in range(6):
if c[j] <= q:
cnt += q//c[j]
q %= c[j]
k[p] = cnt;
while True:
a = list(map(int, input().split()))
p = a[0]
if p == 0: break
del a[0]
n = a[::-1]
ans = 0x7fffffff
for i in range(p, p+1001):
j = pay(i)
if j < 0: continue
ans = min(ans, j + k[i-p])
print(ans)
``` | output | 1 | 17,637 | 10 | 35,275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro, a boy who hates any inefficiencies, pays coins so that the number of coins to be returned as change is minimized in order to do smoothly when he buys something.
One day, however, he doubt if this way is really efficient. When he pays more number of coins, a clerk consumes longer time to find the total value. Maybe he should pay with least possible number of coins.
Thinking for a while, he has decided to take the middle course. So he tries to minimize total number of paid coins and returned coins as change.
Now he is going to buy a product of P yen having several coins. Since he is not good at calculation, please write a program that computes the minimal number of coins.
You may assume following things:
* There are 6 kinds of coins, 1 yen, 5 yen, 10 yen, 50 yen, 100 yen and 500 yen.
* The total value of coins he has is at least P yen.
* A clerk will return the change with least number of coins.
Constraints
* Judge data contains at most 100 data sets.
* 0 ≤ Ni ≤ 1000
Input
Input file contains several data sets. One data set has following format:
P N1 N5 N10 N50 N100 N500
Ni is an integer and is the number of coins of i yen that he have.
The end of input is denoted by a case where P = 0. You should output nothing for this data set.
Output
Output total number of coins that are paid and are returned.
Example
Input
123 3 0 2 0 1 1
999 9 9 9 9 9 9
0 0 0 0 0 0 0
Output
6
3
Submitted Solution:
```
price = [1, 5, 10, 50, 100, 500]
while True:
values = list(map(int, input().split()))
p, n = values[0], values[1:]
ans = 1e100
if p == 0:
break
p_sum = sum(map(lambda s: s[0] * s[1], zip(price, n)))
for change in range(p_sum - p + 1):
total = p + change
pay = [0] * 6
#for i in reversed(range(6)):
# if total >= price[i]:
# pay[i] = min(n[i], int(total / price[i]))
# total -= pay[i] * price[i]
#if total > 0:
# break
coins = sum(pay)
_change = change
for i in [5, 4, 3, 2, 1, 0]:
if _change >= price[i]:
coins += int(_change / price[i])
_change %= price[i]
ans = min(ans, coins)
print(ans)
``` | instruction | 0 | 17,638 | 10 | 35,276 |
No | output | 1 | 17,638 | 10 | 35,277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro, a boy who hates any inefficiencies, pays coins so that the number of coins to be returned as change is minimized in order to do smoothly when he buys something.
One day, however, he doubt if this way is really efficient. When he pays more number of coins, a clerk consumes longer time to find the total value. Maybe he should pay with least possible number of coins.
Thinking for a while, he has decided to take the middle course. So he tries to minimize total number of paid coins and returned coins as change.
Now he is going to buy a product of P yen having several coins. Since he is not good at calculation, please write a program that computes the minimal number of coins.
You may assume following things:
* There are 6 kinds of coins, 1 yen, 5 yen, 10 yen, 50 yen, 100 yen and 500 yen.
* The total value of coins he has is at least P yen.
* A clerk will return the change with least number of coins.
Constraints
* Judge data contains at most 100 data sets.
* 0 ≤ Ni ≤ 1000
Input
Input file contains several data sets. One data set has following format:
P N1 N5 N10 N50 N100 N500
Ni is an integer and is the number of coins of i yen that he have.
The end of input is denoted by a case where P = 0. You should output nothing for this data set.
Output
Output total number of coins that are paid and are returned.
Example
Input
123 3 0 2 0 1 1
999 9 9 9 9 9 9
0 0 0 0 0 0 0
Output
6
3
Submitted Solution:
```
price = [1, 5, 10, 50, 100, 500]
while True:
values = list(map(int, input().split()))
p, n = values[0], values[1:]
ans = 1e100
if p == 0:
break
p_sum = sum(map(lambda s: s[0] * s[1], zip(price, n)))
for change in range(p_sum - p + 1):
coins = 0
_change = change
for i in reversed(range(6)):
if _change >= price[i]:
coins += int(_change / price[i])
_change %= price[i]
total = p + change
pay = [0] * 6
for i in reversed(range(6)):
if total >= price[i]:
pay[i] = min(n[i], int(total / price[i]))
total -= pay[i] * price[i]
if total > 0:
break
coins += sum(pay)
ans = min(ans, coins)
print(ans)
``` | instruction | 0 | 17,639 | 10 | 35,278 |
No | output | 1 | 17,639 | 10 | 35,279 |
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