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.
'Jeopardy!' is an intellectual game where players answer questions and earn points. Company Q conducts a simplified 'Jeopardy!' tournament among the best IT companies. By a lucky coincidence, the old rivals made it to the finals: company R1 and company R2.
The finals will have n questions, m of them are auction questions and n - m of them are regular questions. Each question has a price. The price of the i-th question is ai points. During the game the players chose the questions. At that, if the question is an auction, then the player who chose it can change the price if the number of his current points is strictly larger than the price of the question. The new price of the question cannot be less than the original price and cannot be greater than the current number of points of the player who chose the question. The correct answer brings the player the points equal to the price of the question. The wrong answer to the question reduces the number of the player's points by the value of the question price.
The game will go as follows. First, the R2 company selects a question, then the questions are chosen by the one who answered the previous question correctly. If no one answered the question, then the person who chose last chooses again.
All R2 employees support their team. They want to calculate what maximum possible number of points the R2 team can get if luck is on their side during the whole game (they will always be the first to correctly answer questions). Perhaps you are not going to be surprised, but this problem was again entrusted for you to solve.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100; m ≤ min(n, 30)) — the total number of questions and the number of auction questions, correspondingly. The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 107) — the prices of the questions. The third line contains m distinct integers bi (1 ≤ bi ≤ n) — the numbers of auction questions. Assume that the questions are numbered from 1 to n.
Output
In the single line, print the answer to the problem — the maximum points the R2 company can get if it plays optimally well. It is guaranteed that the answer fits into the integer 64-bit signed type.
Examples
Input
4 1
1 3 7 5
3
Output
18
Input
3 2
10 3 8
2 3
Output
40
Input
2 2
100 200
1 2
Output
400 | instruction | 0 | 68,483 | 10 | 136,966 |
Tags: greedy, math
Correct Solution:
```
n,m = map(int,input().split())
mat = list(map(int,input().split()))
auction = list(map(int,input().split()))
aucpoint = [mat[i-1] for i in auction]
aucpoint.sort(reverse = True)
for i in auction:
mat[i-1] = 0
s = sum(mat)
for point in aucpoint:
if s > point:
s *= 2
else: s += point
print(s)
``` | output | 1 | 68,483 | 10 | 136,967 |
Provide tags and a correct Python 3 solution for this coding contest problem.
'Jeopardy!' is an intellectual game where players answer questions and earn points. Company Q conducts a simplified 'Jeopardy!' tournament among the best IT companies. By a lucky coincidence, the old rivals made it to the finals: company R1 and company R2.
The finals will have n questions, m of them are auction questions and n - m of them are regular questions. Each question has a price. The price of the i-th question is ai points. During the game the players chose the questions. At that, if the question is an auction, then the player who chose it can change the price if the number of his current points is strictly larger than the price of the question. The new price of the question cannot be less than the original price and cannot be greater than the current number of points of the player who chose the question. The correct answer brings the player the points equal to the price of the question. The wrong answer to the question reduces the number of the player's points by the value of the question price.
The game will go as follows. First, the R2 company selects a question, then the questions are chosen by the one who answered the previous question correctly. If no one answered the question, then the person who chose last chooses again.
All R2 employees support their team. They want to calculate what maximum possible number of points the R2 team can get if luck is on their side during the whole game (they will always be the first to correctly answer questions). Perhaps you are not going to be surprised, but this problem was again entrusted for you to solve.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100; m ≤ min(n, 30)) — the total number of questions and the number of auction questions, correspondingly. The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 107) — the prices of the questions. The third line contains m distinct integers bi (1 ≤ bi ≤ n) — the numbers of auction questions. Assume that the questions are numbered from 1 to n.
Output
In the single line, print the answer to the problem — the maximum points the R2 company can get if it plays optimally well. It is guaranteed that the answer fits into the integer 64-bit signed type.
Examples
Input
4 1
1 3 7 5
3
Output
18
Input
3 2
10 3 8
2 3
Output
40
Input
2 2
100 200
1 2
Output
400 | instruction | 0 | 68,484 | 10 | 136,968 |
Tags: greedy, math
Correct Solution:
```
# your code goes herefrom collections import defaultdict
import bisect
from itertools import accumulate, count
import os
import sys
import math
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
def input(): return sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
n,m=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
r2=0
a=[0]+a
auction=[]
for i in range(0,len(b)):
auction.append(a[b[i]])
a[b[i]]=-1
for i in range(1,len(a)):
if a[i]!=-1:
r2+=a[i]
r1=r2
auction.sort()
for i in range(0,len(auction)):
if auction[i]<r2:
r2=r2*2
else:
r2+=auction[i]
for i in range(len(auction)-1,-1,-1):
if auction[i]<r1:
r1+=r1
else:
r1+=auction[i]
print(max(r2,r1))
``` | output | 1 | 68,484 | 10 | 136,969 |
Provide tags and a correct Python 3 solution for this coding contest problem.
'Jeopardy!' is an intellectual game where players answer questions and earn points. Company Q conducts a simplified 'Jeopardy!' tournament among the best IT companies. By a lucky coincidence, the old rivals made it to the finals: company R1 and company R2.
The finals will have n questions, m of them are auction questions and n - m of them are regular questions. Each question has a price. The price of the i-th question is ai points. During the game the players chose the questions. At that, if the question is an auction, then the player who chose it can change the price if the number of his current points is strictly larger than the price of the question. The new price of the question cannot be less than the original price and cannot be greater than the current number of points of the player who chose the question. The correct answer brings the player the points equal to the price of the question. The wrong answer to the question reduces the number of the player's points by the value of the question price.
The game will go as follows. First, the R2 company selects a question, then the questions are chosen by the one who answered the previous question correctly. If no one answered the question, then the person who chose last chooses again.
All R2 employees support their team. They want to calculate what maximum possible number of points the R2 team can get if luck is on their side during the whole game (they will always be the first to correctly answer questions). Perhaps you are not going to be surprised, but this problem was again entrusted for you to solve.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100; m ≤ min(n, 30)) — the total number of questions and the number of auction questions, correspondingly. The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 107) — the prices of the questions. The third line contains m distinct integers bi (1 ≤ bi ≤ n) — the numbers of auction questions. Assume that the questions are numbered from 1 to n.
Output
In the single line, print the answer to the problem — the maximum points the R2 company can get if it plays optimally well. It is guaranteed that the answer fits into the integer 64-bit signed type.
Examples
Input
4 1
1 3 7 5
3
Output
18
Input
3 2
10 3 8
2 3
Output
40
Input
2 2
100 200
1 2
Output
400 | instruction | 0 | 68,485 | 10 | 136,970 |
Tags: greedy, math
Correct Solution:
```
n, m = list(map(int, input().split()))
lst = list(map(int, input().split()))
au = list(map(int, input().split()))
res = sum(lst)
ls = []
for i in range(m):
res -= lst[au[i] - 1]
ls.append(lst[au[i] - 1])
ls.sort(reverse = True)
for i in range(len(ls)):
res += max(ls[i], res)
print(res)
``` | output | 1 | 68,485 | 10 | 136,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Appleman has n cards. Each card has an uppercase letter written on it. Toastman must choose k cards from Appleman's cards. Then Appleman should give Toastman some coins depending on the chosen cards. Formally, for each Toastman's card i you should calculate how much Toastman's cards have the letter equal to letter on ith, then sum up all these quantities, such a number of coins Appleman should give to Toastman.
Given the description of Appleman's cards. What is the maximum number of coins Toastman can get?
Input
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 105). The next line contains n uppercase letters without spaces — the i-th letter describes the i-th card of the Appleman.
Output
Print a single integer – the answer to the problem.
Examples
Input
15 10
DZFDFZDFDDDDDDF
Output
82
Input
6 4
YJSNPI
Output
4
Note
In the first test example Toastman can choose nine cards with letter D and one additional card with any letter. For each card with D he will get 9 coins and for the additional card he will get 1 coin.
Submitted Solution:
```
def spliter():
d = input()
a = d.split()
r = []
for i in a:
k = int(i)
r.append(k)
return r
n = spliter()
k=input()
lst=[]
counter=[]
count=1
maxcoin=0
noremain=n[1]
count1=0
for i in range(n[0]):
lst.append(k[i])
lst.sort()
for i in range(1,n[0],1):
if lst[i]==lst[i-1]:
count+=1
count1+=1
else:
counter.append(count)
count=1
count1+=1
if i==n[0]-1:
counter.append(count)
counter.sort(reverse=True)
if count1==0:
counter.append(1)
for i in counter:
if i>noremain:
maxcoin+=noremain*noremain
break
else:
maxcoin+=i*i
noremain-=i
print(maxcoin)
``` | instruction | 0 | 68,504 | 10 | 137,008 |
Yes | output | 1 | 68,504 | 10 | 137,009 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mikhail the Freelancer dreams of two things: to become a cool programmer and to buy a flat in Moscow. To become a cool programmer, he needs at least p experience points, and a desired flat in Moscow costs q dollars. Mikhail is determined to follow his dreams and registered at a freelance site.
He has suggestions to work on n distinct projects. Mikhail has already evaluated that the participation in the i-th project will increase his experience by ai per day and bring bi dollars per day. As freelance work implies flexible working hours, Mikhail is free to stop working on one project at any time and start working on another project. Doing so, he receives the respective share of experience and money. Mikhail is only trying to become a cool programmer, so he is able to work only on one project at any moment of time.
Find the real value, equal to the minimum number of days Mikhail needs to make his dream come true.
For example, suppose Mikhail is suggested to work on three projects and a1 = 6, b1 = 2, a2 = 1, b2 = 3, a3 = 2, b3 = 6. Also, p = 20 and q = 20. In order to achieve his aims Mikhail has to work for 2.5 days on both first and third projects. Indeed, a1·2.5 + a2·0 + a3·2.5 = 6·2.5 + 1·0 + 2·2.5 = 20 and b1·2.5 + b2·0 + b3·2.5 = 2·2.5 + 3·0 + 6·2.5 = 20.
Input
The first line of the input contains three integers n, p and q (1 ≤ n ≤ 100 000, 1 ≤ p, q ≤ 1 000 000) — the number of projects and the required number of experience and money.
Each of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 1 000 000) — the daily increase in experience and daily income for working on the i-th project.
Output
Print a real value — the minimum number of days Mikhail needs to get the required amount of experience and money. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
3 20 20
6 2
1 3
2 6
Output
5.000000000000000
Input
4 1 1
2 3
3 2
2 3
3 2
Output
0.400000000000000
Note
First sample corresponds to the example in the problem statement. | instruction | 0 | 68,570 | 10 | 137,140 |
Tags: geometry
Correct Solution:
```
from fractions import Fraction
def higher(x1, y1, x2, y2):
if x1 == 0:
if x2 == 0:
return
def min_days(p, q, pr):
ma = max(a for a, b in pr)
mb = max(b for a, b in pr)
pr.sort(key=lambda t: (t[0], -t[1]))
ch = [(0, mb)]
for a, b in pr:
if a == ch[-1][0]:
continue
while (
len(ch) >= 2
and ((b-ch[-2][1])*(ch[-1][0]-ch[-2][0])
>= (a-ch[-2][0])*(ch[-1][1]-ch[-2][1]))
):
ch.pop()
ch.append((a, b))
ch.append((ma, 0))
a1, b1 = None, None
for a2, b2 in ch:
if a1 is not None:
d = (a2-a1)*q + (b1-b2)*p
s = Fraction(b1*p-a1*q, d)
if 0 <= s <= 1:
return Fraction(d, a2*b1 - a1*b2)
a1, b1 = a2, b2
if __name__ == '__main__':
n, p, q = map(int, input().split())
pr = []
for _ in range(n):
a, b = map(int, input().split())
pr.append((a, b))
print("{:.7f}".format(float(min_days(p, q, pr))))
``` | output | 1 | 68,570 | 10 | 137,141 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mikhail the Freelancer dreams of two things: to become a cool programmer and to buy a flat in Moscow. To become a cool programmer, he needs at least p experience points, and a desired flat in Moscow costs q dollars. Mikhail is determined to follow his dreams and registered at a freelance site.
He has suggestions to work on n distinct projects. Mikhail has already evaluated that the participation in the i-th project will increase his experience by ai per day and bring bi dollars per day. As freelance work implies flexible working hours, Mikhail is free to stop working on one project at any time and start working on another project. Doing so, he receives the respective share of experience and money. Mikhail is only trying to become a cool programmer, so he is able to work only on one project at any moment of time.
Find the real value, equal to the minimum number of days Mikhail needs to make his dream come true.
For example, suppose Mikhail is suggested to work on three projects and a1 = 6, b1 = 2, a2 = 1, b2 = 3, a3 = 2, b3 = 6. Also, p = 20 and q = 20. In order to achieve his aims Mikhail has to work for 2.5 days on both first and third projects. Indeed, a1·2.5 + a2·0 + a3·2.5 = 6·2.5 + 1·0 + 2·2.5 = 20 and b1·2.5 + b2·0 + b3·2.5 = 2·2.5 + 3·0 + 6·2.5 = 20.
Input
The first line of the input contains three integers n, p and q (1 ≤ n ≤ 100 000, 1 ≤ p, q ≤ 1 000 000) — the number of projects and the required number of experience and money.
Each of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 1 000 000) — the daily increase in experience and daily income for working on the i-th project.
Output
Print a real value — the minimum number of days Mikhail needs to get the required amount of experience and money. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
3 20 20
6 2
1 3
2 6
Output
5.000000000000000
Input
4 1 1
2 3
3 2
2 3
3 2
Output
0.400000000000000
Note
First sample corresponds to the example in the problem statement.
Submitted Solution:
```
from fractions import Fraction
def higher(x1, y1, x2, y2):
if x1 == 0:
if x2 == 0:
return
def min_days(p, q, pr):
ma = max(a for a, b in pr)
mb = max(b for a, b in pr)
pr.sort(key=lambda t: (t[0], -t[1]))
ch = [(0, mb)]
for a, b in pr:
if a == ch[-1][0]:
continue
while (
len(ch) >= 2
and ((b-ch[-2][1])*(ch[-1][1]-ch[-2][1])
>= (a-ch[-2][0])*(ch[-1][0]-ch[-2][0]))
):
ch.pop()
ch.append((a, b))
ch.append((ma, 0))
a1, b1 = None, None
for a2, b2 in ch:
if a1 is not None:
d = (a2-a1)*q + (b1-b2)*p
s = Fraction(b1*p-a1*q, d)
if 0 <= s <= 1:
return Fraction(d, a2*b1 - a1*b2)
a1, b1 = a2, b2
if __name__ == '__main__':
n, p, q = map(int, input().split())
pr = []
for _ in range(n):
a, b = map(int, input().split())
pr.append((a, b))
print("{:.7f}".format(float(min_days(p, q, pr))))
``` | instruction | 0 | 68,571 | 10 | 137,142 |
No | output | 1 | 68,571 | 10 | 137,143 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mikhail the Freelancer dreams of two things: to become a cool programmer and to buy a flat in Moscow. To become a cool programmer, he needs at least p experience points, and a desired flat in Moscow costs q dollars. Mikhail is determined to follow his dreams and registered at a freelance site.
He has suggestions to work on n distinct projects. Mikhail has already evaluated that the participation in the i-th project will increase his experience by ai per day and bring bi dollars per day. As freelance work implies flexible working hours, Mikhail is free to stop working on one project at any time and start working on another project. Doing so, he receives the respective share of experience and money. Mikhail is only trying to become a cool programmer, so he is able to work only on one project at any moment of time.
Find the real value, equal to the minimum number of days Mikhail needs to make his dream come true.
For example, suppose Mikhail is suggested to work on three projects and a1 = 6, b1 = 2, a2 = 1, b2 = 3, a3 = 2, b3 = 6. Also, p = 20 and q = 20. In order to achieve his aims Mikhail has to work for 2.5 days on both first and third projects. Indeed, a1·2.5 + a2·0 + a3·2.5 = 6·2.5 + 1·0 + 2·2.5 = 20 and b1·2.5 + b2·0 + b3·2.5 = 2·2.5 + 3·0 + 6·2.5 = 20.
Input
The first line of the input contains three integers n, p and q (1 ≤ n ≤ 100 000, 1 ≤ p, q ≤ 1 000 000) — the number of projects and the required number of experience and money.
Each of the next n lines contains two integers ai and bi (1 ≤ ai, bi ≤ 1 000 000) — the daily increase in experience and daily income for working on the i-th project.
Output
Print a real value — the minimum number of days Mikhail needs to get the required amount of experience and money. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.
Examples
Input
3 20 20
6 2
1 3
2 6
Output
5.000000000000000
Input
4 1 1
2 3
3 2
2 3
3 2
Output
0.400000000000000
Note
First sample corresponds to the example in the problem statement.
Submitted Solution:
```
def higher(x1, y1, x2, y2):
if x1 == 0:
if x2 == 0:
return
def min_days(p, q, pr):
ma = max(a for a, b in pr)
mb = max(b for a, b in pr)
pr.sort(key=lambda t: (t[0], -t[1]))
ch = [(0, mb)]
for a, b in pr:
if a == ch[-1][0]:
continue
while (
len(ch) >= 2
and ((b-ch[-2][1])*(ch[-1][1]-ch[-2][1])
>= (a-ch[-2][0])*(ch[-1][0]-ch[-2][0]))
):
ch.pop()
ch.append((a, b))
ch.append((ma, 0))
a1, b1 = None, None
for a2, b2 in ch:
if a1 is not None:
d = (a2-a1)*q + (b1-b2)*p
if d != 0:
s = (b1*p-a1*q)/float(d)
if 0 <= s <= 1:
return float(d)/(a2*b1 - a1*b2)
a1, b1 = a2, b2
if __name__ == '__main__':
n, p, q = map(int, input().split())
pr = []
for _ in range(n):
a, b = map(int, input().split())
pr.append((a, b))
print(min_days(p, q, pr))
``` | instruction | 0 | 68,572 | 10 | 137,144 |
No | output | 1 | 68,572 | 10 | 137,145 |
Provide a correct Python 3 solution for this coding contest problem.
As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloomed with an advertisement for Seven-Eleven. Since this sale was called "Customer Thanksgiving Day", the highlight is that the cheapest vegetables in the bag are free. When I read the advertisement, it looks like the following sale.
* Up to m vegetables can be packed in one bag.
* For bags packed with m vegetables, the cheapest vegetables are free.
* Bags with less than m vegetables are not eligible for the discount.
The three went shopping at Seven Mart immediately.
When the three people who met outside the store after shopping were satisfied with the fact that they were able to buy a lot at a low price, they apparently found that they were all buying the same vegetables. One person said, "It's really cheap, isn't it? This is XXX yen!", And the other was surprised, "What? I was ** yen higher than that! Why?", And the rest One of them saw the receipt and noticed that he bought the cheapest one.
Now, how can you pack the bag at the lowest purchase price? Enter the number of vegetables to purchase, the number of vegetables in the bag, and the price of each vegetable, and create a program that outputs the minimum purchase price.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
p1 p2 ... pn
The first line gives the number of vegetables to buy n (1 ≤ n ≤ 1000) and the number of vegetables to put in the bag m (1 ≤ m ≤ 1000). The second line gives the price pi (10 ≤ pi ≤ 10000) for each vegetable.
The number of datasets does not exceed 100.
Output
Prints the minimum purchase price on one line for each input dataset.
Example
Input
4 2
50 40 100 80
7 3
400 300 100 700 200 600 500
0 0
Output
150
2100 | instruction | 0 | 69,708 | 10 | 139,416 |
"Correct Solution:
```
for i in range(1, 101):
count = 0
temp = 0
SetCount = 0
try:
str1 = input()
list1 = str1.split(" ")
str2 = input()
list2 = str2.split(" ")
list2.sort(key=int, reverse=True)
for m in range(0, int(list1[0])):
if (m+1) % int(list1[1]) == 0:
SetCount = SetCount + 1
#elif (m+1) % int(list1[1]) != 0 | SetCount == int(list1[0])//int(list1[1]):
# count = count + int(list2[m])
else:
count = count + int(list2[m])
print(count)
except:
break
``` | output | 1 | 69,708 | 10 | 139,417 |
Provide a correct Python 3 solution for this coding contest problem.
As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloomed with an advertisement for Seven-Eleven. Since this sale was called "Customer Thanksgiving Day", the highlight is that the cheapest vegetables in the bag are free. When I read the advertisement, it looks like the following sale.
* Up to m vegetables can be packed in one bag.
* For bags packed with m vegetables, the cheapest vegetables are free.
* Bags with less than m vegetables are not eligible for the discount.
The three went shopping at Seven Mart immediately.
When the three people who met outside the store after shopping were satisfied with the fact that they were able to buy a lot at a low price, they apparently found that they were all buying the same vegetables. One person said, "It's really cheap, isn't it? This is XXX yen!", And the other was surprised, "What? I was ** yen higher than that! Why?", And the rest One of them saw the receipt and noticed that he bought the cheapest one.
Now, how can you pack the bag at the lowest purchase price? Enter the number of vegetables to purchase, the number of vegetables in the bag, and the price of each vegetable, and create a program that outputs the minimum purchase price.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
p1 p2 ... pn
The first line gives the number of vegetables to buy n (1 ≤ n ≤ 1000) and the number of vegetables to put in the bag m (1 ≤ m ≤ 1000). The second line gives the price pi (10 ≤ pi ≤ 10000) for each vegetable.
The number of datasets does not exceed 100.
Output
Prints the minimum purchase price on one line for each input dataset.
Example
Input
4 2
50 40 100 80
7 3
400 300 100 700 200 600 500
0 0
Output
150
2100 | instruction | 0 | 69,709 | 10 | 139,418 |
"Correct Solution:
```
# Aizu Problem 0227: Thanksgiving
import sys, math, os, bisect
# read input:
PYDEV = os.environ.get('PYDEV')
if PYDEV=="True":
sys.stdin = open("sample-input.txt", "rt")
while True:
n, m = [int(_) for _ in input().split()]
if n == m == 0:
break
prices = sorted([int(_) for _ in input().split()], reverse=True)
for i in range(m - 1, n, m):
prices[i] = 0
print(sum(prices))
``` | output | 1 | 69,709 | 10 | 139,419 |
Provide a correct Python 3 solution for this coding contest problem.
As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloomed with an advertisement for Seven-Eleven. Since this sale was called "Customer Thanksgiving Day", the highlight is that the cheapest vegetables in the bag are free. When I read the advertisement, it looks like the following sale.
* Up to m vegetables can be packed in one bag.
* For bags packed with m vegetables, the cheapest vegetables are free.
* Bags with less than m vegetables are not eligible for the discount.
The three went shopping at Seven Mart immediately.
When the three people who met outside the store after shopping were satisfied with the fact that they were able to buy a lot at a low price, they apparently found that they were all buying the same vegetables. One person said, "It's really cheap, isn't it? This is XXX yen!", And the other was surprised, "What? I was ** yen higher than that! Why?", And the rest One of them saw the receipt and noticed that he bought the cheapest one.
Now, how can you pack the bag at the lowest purchase price? Enter the number of vegetables to purchase, the number of vegetables in the bag, and the price of each vegetable, and create a program that outputs the minimum purchase price.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
p1 p2 ... pn
The first line gives the number of vegetables to buy n (1 ≤ n ≤ 1000) and the number of vegetables to put in the bag m (1 ≤ m ≤ 1000). The second line gives the price pi (10 ≤ pi ≤ 10000) for each vegetable.
The number of datasets does not exceed 100.
Output
Prints the minimum purchase price on one line for each input dataset.
Example
Input
4 2
50 40 100 80
7 3
400 300 100 700 200 600 500
0 0
Output
150
2100 | instruction | 0 | 69,710 | 10 | 139,420 |
"Correct Solution:
```
while 1:
_,m=map(int, input().split())
if m==0:break
p=sorted(map(int,input().split()),reverse=1)
print(sum(p)-sum(p[m-1::m]))
``` | output | 1 | 69,710 | 10 | 139,421 |
Provide a correct Python 3 solution for this coding contest problem.
As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloomed with an advertisement for Seven-Eleven. Since this sale was called "Customer Thanksgiving Day", the highlight is that the cheapest vegetables in the bag are free. When I read the advertisement, it looks like the following sale.
* Up to m vegetables can be packed in one bag.
* For bags packed with m vegetables, the cheapest vegetables are free.
* Bags with less than m vegetables are not eligible for the discount.
The three went shopping at Seven Mart immediately.
When the three people who met outside the store after shopping were satisfied with the fact that they were able to buy a lot at a low price, they apparently found that they were all buying the same vegetables. One person said, "It's really cheap, isn't it? This is XXX yen!", And the other was surprised, "What? I was ** yen higher than that! Why?", And the rest One of them saw the receipt and noticed that he bought the cheapest one.
Now, how can you pack the bag at the lowest purchase price? Enter the number of vegetables to purchase, the number of vegetables in the bag, and the price of each vegetable, and create a program that outputs the minimum purchase price.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
p1 p2 ... pn
The first line gives the number of vegetables to buy n (1 ≤ n ≤ 1000) and the number of vegetables to put in the bag m (1 ≤ m ≤ 1000). The second line gives the price pi (10 ≤ pi ≤ 10000) for each vegetable.
The number of datasets does not exceed 100.
Output
Prints the minimum purchase price on one line for each input dataset.
Example
Input
4 2
50 40 100 80
7 3
400 300 100 700 200 600 500
0 0
Output
150
2100 | instruction | 0 | 69,711 | 10 | 139,422 |
"Correct Solution:
```
while True:
n, m = map(int, input().split())
if n == 0:
break
price = sorted(map(int, input().split()), reverse=True)
print(sum(price) - sum(price[m - 1::m]))
``` | output | 1 | 69,711 | 10 | 139,423 |
Provide a correct Python 3 solution for this coding contest problem.
As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloomed with an advertisement for Seven-Eleven. Since this sale was called "Customer Thanksgiving Day", the highlight is that the cheapest vegetables in the bag are free. When I read the advertisement, it looks like the following sale.
* Up to m vegetables can be packed in one bag.
* For bags packed with m vegetables, the cheapest vegetables are free.
* Bags with less than m vegetables are not eligible for the discount.
The three went shopping at Seven Mart immediately.
When the three people who met outside the store after shopping were satisfied with the fact that they were able to buy a lot at a low price, they apparently found that they were all buying the same vegetables. One person said, "It's really cheap, isn't it? This is XXX yen!", And the other was surprised, "What? I was ** yen higher than that! Why?", And the rest One of them saw the receipt and noticed that he bought the cheapest one.
Now, how can you pack the bag at the lowest purchase price? Enter the number of vegetables to purchase, the number of vegetables in the bag, and the price of each vegetable, and create a program that outputs the minimum purchase price.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
p1 p2 ... pn
The first line gives the number of vegetables to buy n (1 ≤ n ≤ 1000) and the number of vegetables to put in the bag m (1 ≤ m ≤ 1000). The second line gives the price pi (10 ≤ pi ≤ 10000) for each vegetable.
The number of datasets does not exceed 100.
Output
Prints the minimum purchase price on one line for each input dataset.
Example
Input
4 2
50 40 100 80
7 3
400 300 100 700 200 600 500
0 0
Output
150
2100 | instruction | 0 | 69,712 | 10 | 139,424 |
"Correct Solution:
```
while True:
try:
n,m = list(map(int,input().split()))
if n == 0 and m == 0:
break
amari = n % m
yasai = list(map(int,input().split()))
yasai.sort()
s = 0
if amari > 0:
s = sum(yasai[0:amari])
i = 0
for y in yasai[amari:]:
if i % m > 0:
s += y
i+=1
print(s)
except:
pass
``` | output | 1 | 69,712 | 10 | 139,425 |
Provide a correct Python 3 solution for this coding contest problem.
As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloomed with an advertisement for Seven-Eleven. Since this sale was called "Customer Thanksgiving Day", the highlight is that the cheapest vegetables in the bag are free. When I read the advertisement, it looks like the following sale.
* Up to m vegetables can be packed in one bag.
* For bags packed with m vegetables, the cheapest vegetables are free.
* Bags with less than m vegetables are not eligible for the discount.
The three went shopping at Seven Mart immediately.
When the three people who met outside the store after shopping were satisfied with the fact that they were able to buy a lot at a low price, they apparently found that they were all buying the same vegetables. One person said, "It's really cheap, isn't it? This is XXX yen!", And the other was surprised, "What? I was ** yen higher than that! Why?", And the rest One of them saw the receipt and noticed that he bought the cheapest one.
Now, how can you pack the bag at the lowest purchase price? Enter the number of vegetables to purchase, the number of vegetables in the bag, and the price of each vegetable, and create a program that outputs the minimum purchase price.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
p1 p2 ... pn
The first line gives the number of vegetables to buy n (1 ≤ n ≤ 1000) and the number of vegetables to put in the bag m (1 ≤ m ≤ 1000). The second line gives the price pi (10 ≤ pi ≤ 10000) for each vegetable.
The number of datasets does not exceed 100.
Output
Prints the minimum purchase price on one line for each input dataset.
Example
Input
4 2
50 40 100 80
7 3
400 300 100 700 200 600 500
0 0
Output
150
2100 | instruction | 0 | 69,713 | 10 | 139,426 |
"Correct Solution:
```
while True :
n, m = map(int, input().split())
if n == 0 and m == 0 :
break
p = list(map(int, input().split()))
p.sort(reverse=True)
cst = 0
for i in range(n) :
if (i+1) % m != 0 :
cst += p[i]
print(cst)
``` | output | 1 | 69,713 | 10 | 139,427 |
Provide a correct Python 3 solution for this coding contest problem.
As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloomed with an advertisement for Seven-Eleven. Since this sale was called "Customer Thanksgiving Day", the highlight is that the cheapest vegetables in the bag are free. When I read the advertisement, it looks like the following sale.
* Up to m vegetables can be packed in one bag.
* For bags packed with m vegetables, the cheapest vegetables are free.
* Bags with less than m vegetables are not eligible for the discount.
The three went shopping at Seven Mart immediately.
When the three people who met outside the store after shopping were satisfied with the fact that they were able to buy a lot at a low price, they apparently found that they were all buying the same vegetables. One person said, "It's really cheap, isn't it? This is XXX yen!", And the other was surprised, "What? I was ** yen higher than that! Why?", And the rest One of them saw the receipt and noticed that he bought the cheapest one.
Now, how can you pack the bag at the lowest purchase price? Enter the number of vegetables to purchase, the number of vegetables in the bag, and the price of each vegetable, and create a program that outputs the minimum purchase price.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
p1 p2 ... pn
The first line gives the number of vegetables to buy n (1 ≤ n ≤ 1000) and the number of vegetables to put in the bag m (1 ≤ m ≤ 1000). The second line gives the price pi (10 ≤ pi ≤ 10000) for each vegetable.
The number of datasets does not exceed 100.
Output
Prints the minimum purchase price on one line for each input dataset.
Example
Input
4 2
50 40 100 80
7 3
400 300 100 700 200 600 500
0 0
Output
150
2100 | instruction | 0 | 69,714 | 10 | 139,428 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Thanksgiving
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0227
"""
import sys
def solve(n, m, prices):
ans = 0
for i in range(0, n, m):
t = prices[i:i+m]
if len(t) == m:
t[-1] = 0
ans += sum(t)
return ans
def main(args):
while True:
n, m = map(int, input().split())
if n == 0 and m == 0:
break
prices = sorted([int(p) for p in input().split()], reverse=True)
ans = solve(n, m, prices)
print(ans)
if __name__ == '__main__':
main(sys.argv[1:])
``` | output | 1 | 69,714 | 10 | 139,429 |
Provide a correct Python 3 solution for this coding contest problem.
As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloomed with an advertisement for Seven-Eleven. Since this sale was called "Customer Thanksgiving Day", the highlight is that the cheapest vegetables in the bag are free. When I read the advertisement, it looks like the following sale.
* Up to m vegetables can be packed in one bag.
* For bags packed with m vegetables, the cheapest vegetables are free.
* Bags with less than m vegetables are not eligible for the discount.
The three went shopping at Seven Mart immediately.
When the three people who met outside the store after shopping were satisfied with the fact that they were able to buy a lot at a low price, they apparently found that they were all buying the same vegetables. One person said, "It's really cheap, isn't it? This is XXX yen!", And the other was surprised, "What? I was ** yen higher than that! Why?", And the rest One of them saw the receipt and noticed that he bought the cheapest one.
Now, how can you pack the bag at the lowest purchase price? Enter the number of vegetables to purchase, the number of vegetables in the bag, and the price of each vegetable, and create a program that outputs the minimum purchase price.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
p1 p2 ... pn
The first line gives the number of vegetables to buy n (1 ≤ n ≤ 1000) and the number of vegetables to put in the bag m (1 ≤ m ≤ 1000). The second line gives the price pi (10 ≤ pi ≤ 10000) for each vegetable.
The number of datasets does not exceed 100.
Output
Prints the minimum purchase price on one line for each input dataset.
Example
Input
4 2
50 40 100 80
7 3
400 300 100 700 200 600 500
0 0
Output
150
2100 | instruction | 0 | 69,715 | 10 | 139,430 |
"Correct Solution:
```
while True:
n, m = map(int, input().split())
if n == 0:
break
plst = sorted(list(map(int, input().split())), reverse=True)
s = sum(plst)
for i in range(m - 1, n, m):
s -= plst[i]
print(s)
``` | output | 1 | 69,715 | 10 | 139,431 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloomed with an advertisement for Seven-Eleven. Since this sale was called "Customer Thanksgiving Day", the highlight is that the cheapest vegetables in the bag are free. When I read the advertisement, it looks like the following sale.
* Up to m vegetables can be packed in one bag.
* For bags packed with m vegetables, the cheapest vegetables are free.
* Bags with less than m vegetables are not eligible for the discount.
The three went shopping at Seven Mart immediately.
When the three people who met outside the store after shopping were satisfied with the fact that they were able to buy a lot at a low price, they apparently found that they were all buying the same vegetables. One person said, "It's really cheap, isn't it? This is XXX yen!", And the other was surprised, "What? I was ** yen higher than that! Why?", And the rest One of them saw the receipt and noticed that he bought the cheapest one.
Now, how can you pack the bag at the lowest purchase price? Enter the number of vegetables to purchase, the number of vegetables in the bag, and the price of each vegetable, and create a program that outputs the minimum purchase price.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
p1 p2 ... pn
The first line gives the number of vegetables to buy n (1 ≤ n ≤ 1000) and the number of vegetables to put in the bag m (1 ≤ m ≤ 1000). The second line gives the price pi (10 ≤ pi ≤ 10000) for each vegetable.
The number of datasets does not exceed 100.
Output
Prints the minimum purchase price on one line for each input dataset.
Example
Input
4 2
50 40 100 80
7 3
400 300 100 700 200 600 500
0 0
Output
150
2100
Submitted Solution:
```
while 1:
qua,bag = input().split()
qua = int(qua)
bag = int(bag)
if qua == 0:
break
price = list(map(int,input().split()))
price.sort(reverse=True)
for i in range(qua):
if (i+1)%bag == 0:
price[i]=-1
for i in range(qua//bag):
price.remove(-1)
pay = sum(price)
print(pay)
``` | instruction | 0 | 69,716 | 10 | 139,432 |
Yes | output | 1 | 69,716 | 10 | 139,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloomed with an advertisement for Seven-Eleven. Since this sale was called "Customer Thanksgiving Day", the highlight is that the cheapest vegetables in the bag are free. When I read the advertisement, it looks like the following sale.
* Up to m vegetables can be packed in one bag.
* For bags packed with m vegetables, the cheapest vegetables are free.
* Bags with less than m vegetables are not eligible for the discount.
The three went shopping at Seven Mart immediately.
When the three people who met outside the store after shopping were satisfied with the fact that they were able to buy a lot at a low price, they apparently found that they were all buying the same vegetables. One person said, "It's really cheap, isn't it? This is XXX yen!", And the other was surprised, "What? I was ** yen higher than that! Why?", And the rest One of them saw the receipt and noticed that he bought the cheapest one.
Now, how can you pack the bag at the lowest purchase price? Enter the number of vegetables to purchase, the number of vegetables in the bag, and the price of each vegetable, and create a program that outputs the minimum purchase price.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
p1 p2 ... pn
The first line gives the number of vegetables to buy n (1 ≤ n ≤ 1000) and the number of vegetables to put in the bag m (1 ≤ m ≤ 1000). The second line gives the price pi (10 ≤ pi ≤ 10000) for each vegetable.
The number of datasets does not exceed 100.
Output
Prints the minimum purchase price on one line for each input dataset.
Example
Input
4 2
50 40 100 80
7 3
400 300 100 700 200 600 500
0 0
Output
150
2100
Submitted Solution:
```
# -*- coding: utf-8 -*-
while 1:
try:
n, m = map(int, input().split())
if n == 0:
break
p = sorted(list(map(int, input().split())))
p.reverse()
for i in range(m - 1, n, m):
p[i] = 0
print(sum(p))
except:
break
``` | instruction | 0 | 69,717 | 10 | 139,434 |
Yes | output | 1 | 69,717 | 10 | 139,435 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloomed with an advertisement for Seven-Eleven. Since this sale was called "Customer Thanksgiving Day", the highlight is that the cheapest vegetables in the bag are free. When I read the advertisement, it looks like the following sale.
* Up to m vegetables can be packed in one bag.
* For bags packed with m vegetables, the cheapest vegetables are free.
* Bags with less than m vegetables are not eligible for the discount.
The three went shopping at Seven Mart immediately.
When the three people who met outside the store after shopping were satisfied with the fact that they were able to buy a lot at a low price, they apparently found that they were all buying the same vegetables. One person said, "It's really cheap, isn't it? This is XXX yen!", And the other was surprised, "What? I was ** yen higher than that! Why?", And the rest One of them saw the receipt and noticed that he bought the cheapest one.
Now, how can you pack the bag at the lowest purchase price? Enter the number of vegetables to purchase, the number of vegetables in the bag, and the price of each vegetable, and create a program that outputs the minimum purchase price.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
p1 p2 ... pn
The first line gives the number of vegetables to buy n (1 ≤ n ≤ 1000) and the number of vegetables to put in the bag m (1 ≤ m ≤ 1000). The second line gives the price pi (10 ≤ pi ≤ 10000) for each vegetable.
The number of datasets does not exceed 100.
Output
Prints the minimum purchase price on one line for each input dataset.
Example
Input
4 2
50 40 100 80
7 3
400 300 100 700 200 600 500
0 0
Output
150
2100
Submitted Solution:
```
while 1:
n, m = map(int, input().split())
if n == 0:
break
veg = list(map(int, input().split()))
veg.sort(reverse=True)
cnt = 0
ans = 0
while veg != []:
cnt += 1
tmp = veg.pop(0)
if cnt % m == 0:
pass
else:
ans += tmp
print(ans)
``` | instruction | 0 | 69,718 | 10 | 139,436 |
Yes | output | 1 | 69,718 | 10 | 139,437 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloomed with an advertisement for Seven-Eleven. Since this sale was called "Customer Thanksgiving Day", the highlight is that the cheapest vegetables in the bag are free. When I read the advertisement, it looks like the following sale.
* Up to m vegetables can be packed in one bag.
* For bags packed with m vegetables, the cheapest vegetables are free.
* Bags with less than m vegetables are not eligible for the discount.
The three went shopping at Seven Mart immediately.
When the three people who met outside the store after shopping were satisfied with the fact that they were able to buy a lot at a low price, they apparently found that they were all buying the same vegetables. One person said, "It's really cheap, isn't it? This is XXX yen!", And the other was surprised, "What? I was ** yen higher than that! Why?", And the rest One of them saw the receipt and noticed that he bought the cheapest one.
Now, how can you pack the bag at the lowest purchase price? Enter the number of vegetables to purchase, the number of vegetables in the bag, and the price of each vegetable, and create a program that outputs the minimum purchase price.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
p1 p2 ... pn
The first line gives the number of vegetables to buy n (1 ≤ n ≤ 1000) and the number of vegetables to put in the bag m (1 ≤ m ≤ 1000). The second line gives the price pi (10 ≤ pi ≤ 10000) for each vegetable.
The number of datasets does not exceed 100.
Output
Prints the minimum purchase price on one line for each input dataset.
Example
Input
4 2
50 40 100 80
7 3
400 300 100 700 200 600 500
0 0
Output
150
2100
Submitted Solution:
```
while True:
try:
n,m = list(map(int,input().split()))
if n == 0 and m == 0:
break
amari = n % m
yasai = list(map(int,input().split()))
yasai.sort()
s = 0
for i in range(n):
if (i+1) % m > 0:
s += yasai[-i-1]
print(s)
except:
pass
``` | instruction | 0 | 69,719 | 10 | 139,438 |
Yes | output | 1 | 69,719 | 10 | 139,439 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloomed with an advertisement for Seven-Eleven. Since this sale was called "Customer Thanksgiving Day", the highlight is that the cheapest vegetables in the bag are free. When I read the advertisement, it looks like the following sale.
* Up to m vegetables can be packed in one bag.
* For bags packed with m vegetables, the cheapest vegetables are free.
* Bags with less than m vegetables are not eligible for the discount.
The three went shopping at Seven Mart immediately.
When the three people who met outside the store after shopping were satisfied with the fact that they were able to buy a lot at a low price, they apparently found that they were all buying the same vegetables. One person said, "It's really cheap, isn't it? This is XXX yen!", And the other was surprised, "What? I was ** yen higher than that! Why?", And the rest One of them saw the receipt and noticed that he bought the cheapest one.
Now, how can you pack the bag at the lowest purchase price? Enter the number of vegetables to purchase, the number of vegetables in the bag, and the price of each vegetable, and create a program that outputs the minimum purchase price.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
p1 p2 ... pn
The first line gives the number of vegetables to buy n (1 ≤ n ≤ 1000) and the number of vegetables to put in the bag m (1 ≤ m ≤ 1000). The second line gives the price pi (10 ≤ pi ≤ 10000) for each vegetable.
The number of datasets does not exceed 100.
Output
Prints the minimum purchase price on one line for each input dataset.
Example
Input
4 2
50 40 100 80
7 3
400 300 100 700 200 600 500
0 0
Output
150
2100
Submitted Solution:
```
for i in range(1, 101):
count = 0
SetCount = 0
try:
str1 = input()
list1 = str1.split(" ")
str2 = input()
list2 = str2.split(" ")
list2.sort(reverse=True)
for m in range(0, int(list1[0])):
if (m+1) % int(list1[1]) == 0:
count = count + int(list2[m])
SetCount = SetCount + 1
elif (m+1) % int(list1[1]) != 0 | SetCount == int(list1[0])//int(list1[1]):
count = count + int(list2[m])
print(count)
except:
break
``` | instruction | 0 | 69,720 | 10 | 139,440 |
No | output | 1 | 69,720 | 10 | 139,441 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloomed with an advertisement for Seven-Eleven. Since this sale was called "Customer Thanksgiving Day", the highlight is that the cheapest vegetables in the bag are free. When I read the advertisement, it looks like the following sale.
* Up to m vegetables can be packed in one bag.
* For bags packed with m vegetables, the cheapest vegetables are free.
* Bags with less than m vegetables are not eligible for the discount.
The three went shopping at Seven Mart immediately.
When the three people who met outside the store after shopping were satisfied with the fact that they were able to buy a lot at a low price, they apparently found that they were all buying the same vegetables. One person said, "It's really cheap, isn't it? This is XXX yen!", And the other was surprised, "What? I was ** yen higher than that! Why?", And the rest One of them saw the receipt and noticed that he bought the cheapest one.
Now, how can you pack the bag at the lowest purchase price? Enter the number of vegetables to purchase, the number of vegetables in the bag, and the price of each vegetable, and create a program that outputs the minimum purchase price.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
p1 p2 ... pn
The first line gives the number of vegetables to buy n (1 ≤ n ≤ 1000) and the number of vegetables to put in the bag m (1 ≤ m ≤ 1000). The second line gives the price pi (10 ≤ pi ≤ 10000) for each vegetable.
The number of datasets does not exceed 100.
Output
Prints the minimum purchase price on one line for each input dataset.
Example
Input
4 2
50 40 100 80
7 3
400 300 100 700 200 600 500
0 0
Output
150
2100
Submitted Solution:
```
while True:
n,m = list(map(int,input().split()))
if n == 0 and m == 0:
break
amari = n % m
yasai = list(map(int,input().split()))
yasai.sort()
s = 0
if amari > 0:
s = sum(yasai[0:amari])
i = 0
for y in yasai[amari:]:
if i % m > 0:
s += y
i+=1
print(s)
``` | instruction | 0 | 69,721 | 10 | 139,442 |
No | output | 1 | 69,721 | 10 | 139,443 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloomed with an advertisement for Seven-Eleven. Since this sale was called "Customer Thanksgiving Day", the highlight is that the cheapest vegetables in the bag are free. When I read the advertisement, it looks like the following sale.
* Up to m vegetables can be packed in one bag.
* For bags packed with m vegetables, the cheapest vegetables are free.
* Bags with less than m vegetables are not eligible for the discount.
The three went shopping at Seven Mart immediately.
When the three people who met outside the store after shopping were satisfied with the fact that they were able to buy a lot at a low price, they apparently found that they were all buying the same vegetables. One person said, "It's really cheap, isn't it? This is XXX yen!", And the other was surprised, "What? I was ** yen higher than that! Why?", And the rest One of them saw the receipt and noticed that he bought the cheapest one.
Now, how can you pack the bag at the lowest purchase price? Enter the number of vegetables to purchase, the number of vegetables in the bag, and the price of each vegetable, and create a program that outputs the minimum purchase price.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
p1 p2 ... pn
The first line gives the number of vegetables to buy n (1 ≤ n ≤ 1000) and the number of vegetables to put in the bag m (1 ≤ m ≤ 1000). The second line gives the price pi (10 ≤ pi ≤ 10000) for each vegetable.
The number of datasets does not exceed 100.
Output
Prints the minimum purchase price on one line for each input dataset.
Example
Input
4 2
50 40 100 80
7 3
400 300 100 700 200 600 500
0 0
Output
150
2100
Submitted Solution:
```
for i in range(1, 101):
count = 0
temp = 0
SetCount = 0
try:
str1 = input()
list1 = str1.split(" ")
str2 = input()
list2 = str2.split(" ")
list2.sort(reverse=True)
for m in range(0, int(list1[0])):
if (m+1) % int(list1[1]) == 0:
SetCount = SetCount + 1
#elif (m+1) % int(list1[1]) != 0 | SetCount == int(list1[0])//int(list1[1]):
# count = count + int(list2[m])
else:
count = count + int(list2[m])
print(count)
except:
break
``` | instruction | 0 | 69,722 | 10 | 139,444 |
No | output | 1 | 69,722 | 10 | 139,445 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloomed with an advertisement for Seven-Eleven. Since this sale was called "Customer Thanksgiving Day", the highlight is that the cheapest vegetables in the bag are free. When I read the advertisement, it looks like the following sale.
* Up to m vegetables can be packed in one bag.
* For bags packed with m vegetables, the cheapest vegetables are free.
* Bags with less than m vegetables are not eligible for the discount.
The three went shopping at Seven Mart immediately.
When the three people who met outside the store after shopping were satisfied with the fact that they were able to buy a lot at a low price, they apparently found that they were all buying the same vegetables. One person said, "It's really cheap, isn't it? This is XXX yen!", And the other was surprised, "What? I was ** yen higher than that! Why?", And the rest One of them saw the receipt and noticed that he bought the cheapest one.
Now, how can you pack the bag at the lowest purchase price? Enter the number of vegetables to purchase, the number of vegetables in the bag, and the price of each vegetable, and create a program that outputs the minimum purchase price.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n m
p1 p2 ... pn
The first line gives the number of vegetables to buy n (1 ≤ n ≤ 1000) and the number of vegetables to put in the bag m (1 ≤ m ≤ 1000). The second line gives the price pi (10 ≤ pi ≤ 10000) for each vegetable.
The number of datasets does not exceed 100.
Output
Prints the minimum purchase price on one line for each input dataset.
Example
Input
4 2
50 40 100 80
7 3
400 300 100 700 200 600 500
0 0
Output
150
2100
Submitted Solution:
```
while True:
a = input()
print('a')
``` | instruction | 0 | 69,723 | 10 | 139,446 |
No | output | 1 | 69,723 | 10 | 139,447 |
Provide a correct Python 3 solution for this coding contest problem.
Princess'Marriage
Marriage of a princess
English text is not available in this practice contest.
A brave princess in a poor country, knowing that gambling payouts are determined by the parimutuel method, felt more familiar with gambling and was convinced of her victory in gambling. As a result, he spent more money than ever before and lost enough to lose all the taxes paid by the people. The King, who took this situation seriously, decided to marry the princess to the neighboring country. By doing this, I thought that I would like the princess to reflect on her daily activities and at the same time deepen her friendship with neighboring countries and receive financial assistance.
The princess and the prince of the neighboring country liked each other, and the kings of both countries agreed on a political marriage. The princess triumphantly went to the neighboring country with a little money in her hand. On the other hand, the motive for the princess to marry is to pursue the unilateral interests of the king, and the aide of the prince of the neighboring country who thinks that it is not pleasant shoots countless thugs along the way to make the princess dead. It was.
The path the princess will take has already been decided. There are a total of L post stations on the path of the princess. For convenience, the departure and arrival points are also set as post stations, and each post station is called S1, S2, ... SL. The princess shall be in S1 first, visit the post station in ascending order (S2, S3 ... in that order), and finally go to SL. At the post station, you can pay money to hire an escort, and as long as you have the money, you can contract for as long as you like to protect the princess. The cost of hiring an escort is 1 gold per distance. Note that the princess can also partially protect the section through which she passes. The distance between Si and Si + 1 is given by Di, and the expected value of the number of times a thug is attacked per distance between Si and Si + 1 is given by Pi.
Find the expected number of thugs to reach your destination when the princess has a budget of M and hires an escort to minimize the expected number of thugs.
Input
The input consists of multiple datasets. Each dataset has the following format.
> N M
> D1 P1
> D2 P2
> ...
> DN PN
Two integers are given in the first row of each dataset, representing the number of intervals N (1 ≤ N ≤ 10,000) and the budget M (0 ≤ M ≤ 1,000,000,000) of the princess difference, respectively. The next N lines show information about the path the princess takes. Each line contains two integers, and the i-th line is the expected value of the interval Di (1 ≤ Di ≤ 10,000) and the number of attacks when moving one unit distance between them Pi (0 ≤ Pi <= 10) ). The end of the input is represented by a data set with N = 0 and M = 0. Do not output the calculation result for this data set.
Output
For each dataset, output the expected number of times the princess will be attacked by thugs to your destination.
Sample Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output for the Sample Input
Five
140
Example
Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output
5
140 | instruction | 0 | 69,760 | 10 | 139,520 |
"Correct Solution:
```
while True:
N,M = map(int,input().split())
if N == 0: break
src = []
for i in range(N):
d,p = map(int,input().split())
src.append((p,d))
money = M
ans = 0
for p,d in sorted(src,reverse=True):
guard = min(d,money)
money -= guard
ans += (d - guard) * p
print(ans)
``` | output | 1 | 69,760 | 10 | 139,521 |
Provide a correct Python 3 solution for this coding contest problem.
Princess'Marriage
Marriage of a princess
English text is not available in this practice contest.
A brave princess in a poor country, knowing that gambling payouts are determined by the parimutuel method, felt more familiar with gambling and was convinced of her victory in gambling. As a result, he spent more money than ever before and lost enough to lose all the taxes paid by the people. The King, who took this situation seriously, decided to marry the princess to the neighboring country. By doing this, I thought that I would like the princess to reflect on her daily activities and at the same time deepen her friendship with neighboring countries and receive financial assistance.
The princess and the prince of the neighboring country liked each other, and the kings of both countries agreed on a political marriage. The princess triumphantly went to the neighboring country with a little money in her hand. On the other hand, the motive for the princess to marry is to pursue the unilateral interests of the king, and the aide of the prince of the neighboring country who thinks that it is not pleasant shoots countless thugs along the way to make the princess dead. It was.
The path the princess will take has already been decided. There are a total of L post stations on the path of the princess. For convenience, the departure and arrival points are also set as post stations, and each post station is called S1, S2, ... SL. The princess shall be in S1 first, visit the post station in ascending order (S2, S3 ... in that order), and finally go to SL. At the post station, you can pay money to hire an escort, and as long as you have the money, you can contract for as long as you like to protect the princess. The cost of hiring an escort is 1 gold per distance. Note that the princess can also partially protect the section through which she passes. The distance between Si and Si + 1 is given by Di, and the expected value of the number of times a thug is attacked per distance between Si and Si + 1 is given by Pi.
Find the expected number of thugs to reach your destination when the princess has a budget of M and hires an escort to minimize the expected number of thugs.
Input
The input consists of multiple datasets. Each dataset has the following format.
> N M
> D1 P1
> D2 P2
> ...
> DN PN
Two integers are given in the first row of each dataset, representing the number of intervals N (1 ≤ N ≤ 10,000) and the budget M (0 ≤ M ≤ 1,000,000,000) of the princess difference, respectively. The next N lines show information about the path the princess takes. Each line contains two integers, and the i-th line is the expected value of the interval Di (1 ≤ Di ≤ 10,000) and the number of attacks when moving one unit distance between them Pi (0 ≤ Pi <= 10) ). The end of the input is represented by a data set with N = 0 and M = 0. Do not output the calculation result for this data set.
Output
For each dataset, output the expected number of times the princess will be attacked by thugs to your destination.
Sample Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output for the Sample Input
Five
140
Example
Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output
5
140 | instruction | 0 | 69,761 | 10 | 139,522 |
"Correct Solution:
```
da = list(map(int,input().split()))
k = [[0 for i in range(2)]for j in range(10001)]
i = 0
su = 0
j = 0
while da[0] + da[1] != 0:
i = 0
su = 0
j = 0
for i in range(da[0]):
k[i][0],k[i][1] = map(int,input().split())
k = sorted(k,key = lambda x: x[1],reverse = True)
while j != i + 1:
if da[1] > k[j][0]:
da[1] -= k[j][0]
j += 1
elif da[1] <= k[j][0] and da[1] > 0:
su += (k[j][0] - da[1]) * k[j][1]
da[1] = 0
j += 1
else:
su += k[j][0] * k[j][1]
j += 1
print(su)
da = list(map(int,input().split()))
k = [[0 for i in range(2)]for j in range(10001)]
``` | output | 1 | 69,761 | 10 | 139,523 |
Provide a correct Python 3 solution for this coding contest problem.
Princess'Marriage
Marriage of a princess
English text is not available in this practice contest.
A brave princess in a poor country, knowing that gambling payouts are determined by the parimutuel method, felt more familiar with gambling and was convinced of her victory in gambling. As a result, he spent more money than ever before and lost enough to lose all the taxes paid by the people. The King, who took this situation seriously, decided to marry the princess to the neighboring country. By doing this, I thought that I would like the princess to reflect on her daily activities and at the same time deepen her friendship with neighboring countries and receive financial assistance.
The princess and the prince of the neighboring country liked each other, and the kings of both countries agreed on a political marriage. The princess triumphantly went to the neighboring country with a little money in her hand. On the other hand, the motive for the princess to marry is to pursue the unilateral interests of the king, and the aide of the prince of the neighboring country who thinks that it is not pleasant shoots countless thugs along the way to make the princess dead. It was.
The path the princess will take has already been decided. There are a total of L post stations on the path of the princess. For convenience, the departure and arrival points are also set as post stations, and each post station is called S1, S2, ... SL. The princess shall be in S1 first, visit the post station in ascending order (S2, S3 ... in that order), and finally go to SL. At the post station, you can pay money to hire an escort, and as long as you have the money, you can contract for as long as you like to protect the princess. The cost of hiring an escort is 1 gold per distance. Note that the princess can also partially protect the section through which she passes. The distance between Si and Si + 1 is given by Di, and the expected value of the number of times a thug is attacked per distance between Si and Si + 1 is given by Pi.
Find the expected number of thugs to reach your destination when the princess has a budget of M and hires an escort to minimize the expected number of thugs.
Input
The input consists of multiple datasets. Each dataset has the following format.
> N M
> D1 P1
> D2 P2
> ...
> DN PN
Two integers are given in the first row of each dataset, representing the number of intervals N (1 ≤ N ≤ 10,000) and the budget M (0 ≤ M ≤ 1,000,000,000) of the princess difference, respectively. The next N lines show information about the path the princess takes. Each line contains two integers, and the i-th line is the expected value of the interval Di (1 ≤ Di ≤ 10,000) and the number of attacks when moving one unit distance between them Pi (0 ≤ Pi <= 10) ). The end of the input is represented by a data set with N = 0 and M = 0. Do not output the calculation result for this data set.
Output
For each dataset, output the expected number of times the princess will be attacked by thugs to your destination.
Sample Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output for the Sample Input
Five
140
Example
Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output
5
140 | instruction | 0 | 69,762 | 10 | 139,524 |
"Correct Solution:
```
#月曜五時間目 実践的プログラミング
#ソートして襲われてはいけない時には、できるだけ護衛をつけるようにすることが最適であるので
#ソートした後に左から貪欲に護衛をつけていって、お金がなくなったらそのまま何もしないと言う戦略で正解した。
#時間:15分
#Princess' Marriage
#学んだこと：ソートして貪欲に考える
#URL: https://onlinejudge.u-aizu.ac.jp/challenges/search/titles/2019
def solve():
N,M=map(int,input().split())
if N==0 and M==0:
#終了条件
exit()
else:
d=[tuple(map(int,input().split())) for _ in range(N)]
d.sort(key=lambda x:-x[1])
#襲われたやすいところから優先的に保護するためにソート
cnt=0
money=M
#money=所持金、cnt=襲われる期待値の累計和
for que in d:
distance,value=que
if money>=distance:
money-=distance
elif money<distance:
cnt+=value*(distance-money)
money=0
print(cnt)
return solve()
if __name__=="__main__":
solve()
``` | output | 1 | 69,762 | 10 | 139,525 |
Provide a correct Python 3 solution for this coding contest problem.
Princess'Marriage
Marriage of a princess
English text is not available in this practice contest.
A brave princess in a poor country, knowing that gambling payouts are determined by the parimutuel method, felt more familiar with gambling and was convinced of her victory in gambling. As a result, he spent more money than ever before and lost enough to lose all the taxes paid by the people. The King, who took this situation seriously, decided to marry the princess to the neighboring country. By doing this, I thought that I would like the princess to reflect on her daily activities and at the same time deepen her friendship with neighboring countries and receive financial assistance.
The princess and the prince of the neighboring country liked each other, and the kings of both countries agreed on a political marriage. The princess triumphantly went to the neighboring country with a little money in her hand. On the other hand, the motive for the princess to marry is to pursue the unilateral interests of the king, and the aide of the prince of the neighboring country who thinks that it is not pleasant shoots countless thugs along the way to make the princess dead. It was.
The path the princess will take has already been decided. There are a total of L post stations on the path of the princess. For convenience, the departure and arrival points are also set as post stations, and each post station is called S1, S2, ... SL. The princess shall be in S1 first, visit the post station in ascending order (S2, S3 ... in that order), and finally go to SL. At the post station, you can pay money to hire an escort, and as long as you have the money, you can contract for as long as you like to protect the princess. The cost of hiring an escort is 1 gold per distance. Note that the princess can also partially protect the section through which she passes. The distance between Si and Si + 1 is given by Di, and the expected value of the number of times a thug is attacked per distance between Si and Si + 1 is given by Pi.
Find the expected number of thugs to reach your destination when the princess has a budget of M and hires an escort to minimize the expected number of thugs.
Input
The input consists of multiple datasets. Each dataset has the following format.
> N M
> D1 P1
> D2 P2
> ...
> DN PN
Two integers are given in the first row of each dataset, representing the number of intervals N (1 ≤ N ≤ 10,000) and the budget M (0 ≤ M ≤ 1,000,000,000) of the princess difference, respectively. The next N lines show information about the path the princess takes. Each line contains two integers, and the i-th line is the expected value of the interval Di (1 ≤ Di ≤ 10,000) and the number of attacks when moving one unit distance between them Pi (0 ≤ Pi <= 10) ). The end of the input is represented by a data set with N = 0 and M = 0. Do not output the calculation result for this data set.
Output
For each dataset, output the expected number of times the princess will be attacked by thugs to your destination.
Sample Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output for the Sample Input
Five
140
Example
Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output
5
140 | instruction | 0 | 69,763 | 10 | 139,526 |
"Correct Solution:
```
while True:
n,m=map(int,input().split())
matrix=[]
if n==m==0:
break
else:
for i in range(n):
D,P=map(int,input().split())
matrix.append([D,P])
matrix.sort(key = lambda x:x[1],reverse=True)
i=0
while m>0 and i<=n-1:
pp=matrix[i][1]
dd=matrix[i][0]
if dd > m:
matrix[i][0]=dd-m
m=0
else:
matrix[i][1]=0
m=m-dd
i+=1
sum_matrix=[mm[0]*mm[1] for mm in matrix]
print(sum(sum_matrix))
``` | output | 1 | 69,763 | 10 | 139,527 |
Provide a correct Python 3 solution for this coding contest problem.
Princess'Marriage
Marriage of a princess
English text is not available in this practice contest.
A brave princess in a poor country, knowing that gambling payouts are determined by the parimutuel method, felt more familiar with gambling and was convinced of her victory in gambling. As a result, he spent more money than ever before and lost enough to lose all the taxes paid by the people. The King, who took this situation seriously, decided to marry the princess to the neighboring country. By doing this, I thought that I would like the princess to reflect on her daily activities and at the same time deepen her friendship with neighboring countries and receive financial assistance.
The princess and the prince of the neighboring country liked each other, and the kings of both countries agreed on a political marriage. The princess triumphantly went to the neighboring country with a little money in her hand. On the other hand, the motive for the princess to marry is to pursue the unilateral interests of the king, and the aide of the prince of the neighboring country who thinks that it is not pleasant shoots countless thugs along the way to make the princess dead. It was.
The path the princess will take has already been decided. There are a total of L post stations on the path of the princess. For convenience, the departure and arrival points are also set as post stations, and each post station is called S1, S2, ... SL. The princess shall be in S1 first, visit the post station in ascending order (S2, S3 ... in that order), and finally go to SL. At the post station, you can pay money to hire an escort, and as long as you have the money, you can contract for as long as you like to protect the princess. The cost of hiring an escort is 1 gold per distance. Note that the princess can also partially protect the section through which she passes. The distance between Si and Si + 1 is given by Di, and the expected value of the number of times a thug is attacked per distance between Si and Si + 1 is given by Pi.
Find the expected number of thugs to reach your destination when the princess has a budget of M and hires an escort to minimize the expected number of thugs.
Input
The input consists of multiple datasets. Each dataset has the following format.
> N M
> D1 P1
> D2 P2
> ...
> DN PN
Two integers are given in the first row of each dataset, representing the number of intervals N (1 ≤ N ≤ 10,000) and the budget M (0 ≤ M ≤ 1,000,000,000) of the princess difference, respectively. The next N lines show information about the path the princess takes. Each line contains two integers, and the i-th line is the expected value of the interval Di (1 ≤ Di ≤ 10,000) and the number of attacks when moving one unit distance between them Pi (0 ≤ Pi <= 10) ). The end of the input is represented by a data set with N = 0 and M = 0. Do not output the calculation result for this data set.
Output
For each dataset, output the expected number of times the princess will be attacked by thugs to your destination.
Sample Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output for the Sample Input
Five
140
Example
Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output
5
140 | instruction | 0 | 69,764 | 10 | 139,528 |
"Correct Solution:
```
while True:
n,m=map(int,input().split())
if n==0 and m==0:
break
pd=[list(map(int,input().split()))[::-1] for i in range(n)]
pd.sort()
pd.reverse()
ans=0
for i in pd:
if m>i[1]:
m-=i[1]
elif m>0:
ans+=i[0]*(i[1]-m)
m=0
else:
ans+=i[0]*i[1]
print(ans)
``` | output | 1 | 69,764 | 10 | 139,529 |
Provide a correct Python 3 solution for this coding contest problem.
Princess'Marriage
Marriage of a princess
English text is not available in this practice contest.
A brave princess in a poor country, knowing that gambling payouts are determined by the parimutuel method, felt more familiar with gambling and was convinced of her victory in gambling. As a result, he spent more money than ever before and lost enough to lose all the taxes paid by the people. The King, who took this situation seriously, decided to marry the princess to the neighboring country. By doing this, I thought that I would like the princess to reflect on her daily activities and at the same time deepen her friendship with neighboring countries and receive financial assistance.
The princess and the prince of the neighboring country liked each other, and the kings of both countries agreed on a political marriage. The princess triumphantly went to the neighboring country with a little money in her hand. On the other hand, the motive for the princess to marry is to pursue the unilateral interests of the king, and the aide of the prince of the neighboring country who thinks that it is not pleasant shoots countless thugs along the way to make the princess dead. It was.
The path the princess will take has already been decided. There are a total of L post stations on the path of the princess. For convenience, the departure and arrival points are also set as post stations, and each post station is called S1, S2, ... SL. The princess shall be in S1 first, visit the post station in ascending order (S2, S3 ... in that order), and finally go to SL. At the post station, you can pay money to hire an escort, and as long as you have the money, you can contract for as long as you like to protect the princess. The cost of hiring an escort is 1 gold per distance. Note that the princess can also partially protect the section through which she passes. The distance between Si and Si + 1 is given by Di, and the expected value of the number of times a thug is attacked per distance between Si and Si + 1 is given by Pi.
Find the expected number of thugs to reach your destination when the princess has a budget of M and hires an escort to minimize the expected number of thugs.
Input
The input consists of multiple datasets. Each dataset has the following format.
> N M
> D1 P1
> D2 P2
> ...
> DN PN
Two integers are given in the first row of each dataset, representing the number of intervals N (1 ≤ N ≤ 10,000) and the budget M (0 ≤ M ≤ 1,000,000,000) of the princess difference, respectively. The next N lines show information about the path the princess takes. Each line contains two integers, and the i-th line is the expected value of the interval Di (1 ≤ Di ≤ 10,000) and the number of attacks when moving one unit distance between them Pi (0 ≤ Pi <= 10) ). The end of the input is represented by a data set with N = 0 and M = 0. Do not output the calculation result for this data set.
Output
For each dataset, output the expected number of times the princess will be attacked by thugs to your destination.
Sample Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output for the Sample Input
Five
140
Example
Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output
5
140 | instruction | 0 | 69,765 | 10 | 139,530 |
"Correct Solution:
```
while 1:
N,M=map(int,input().split())
if not N and not M:break
s=0
t=[]
for _ in range(N):
a,b=map(int,input().split())
s+=a*b
t.extend([(b,a)])
for i in sorted(t)[::-1]:
if i[1]<=M:
s-=i[0]*i[1]
M-=i[1]
elif 0<M<i[1]:
s-=M*i[0]
M=0
break
print(s)
``` | output | 1 | 69,765 | 10 | 139,531 |
Provide a correct Python 3 solution for this coding contest problem.
Princess'Marriage
Marriage of a princess
English text is not available in this practice contest.
A brave princess in a poor country, knowing that gambling payouts are determined by the parimutuel method, felt more familiar with gambling and was convinced of her victory in gambling. As a result, he spent more money than ever before and lost enough to lose all the taxes paid by the people. The King, who took this situation seriously, decided to marry the princess to the neighboring country. By doing this, I thought that I would like the princess to reflect on her daily activities and at the same time deepen her friendship with neighboring countries and receive financial assistance.
The princess and the prince of the neighboring country liked each other, and the kings of both countries agreed on a political marriage. The princess triumphantly went to the neighboring country with a little money in her hand. On the other hand, the motive for the princess to marry is to pursue the unilateral interests of the king, and the aide of the prince of the neighboring country who thinks that it is not pleasant shoots countless thugs along the way to make the princess dead. It was.
The path the princess will take has already been decided. There are a total of L post stations on the path of the princess. For convenience, the departure and arrival points are also set as post stations, and each post station is called S1, S2, ... SL. The princess shall be in S1 first, visit the post station in ascending order (S2, S3 ... in that order), and finally go to SL. At the post station, you can pay money to hire an escort, and as long as you have the money, you can contract for as long as you like to protect the princess. The cost of hiring an escort is 1 gold per distance. Note that the princess can also partially protect the section through which she passes. The distance between Si and Si + 1 is given by Di, and the expected value of the number of times a thug is attacked per distance between Si and Si + 1 is given by Pi.
Find the expected number of thugs to reach your destination when the princess has a budget of M and hires an escort to minimize the expected number of thugs.
Input
The input consists of multiple datasets. Each dataset has the following format.
> N M
> D1 P1
> D2 P2
> ...
> DN PN
Two integers are given in the first row of each dataset, representing the number of intervals N (1 ≤ N ≤ 10,000) and the budget M (0 ≤ M ≤ 1,000,000,000) of the princess difference, respectively. The next N lines show information about the path the princess takes. Each line contains two integers, and the i-th line is the expected value of the interval Di (1 ≤ Di ≤ 10,000) and the number of attacks when moving one unit distance between them Pi (0 ≤ Pi <= 10) ). The end of the input is represented by a data set with N = 0 and M = 0. Do not output the calculation result for this data set.
Output
For each dataset, output the expected number of times the princess will be attacked by thugs to your destination.
Sample Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output for the Sample Input
Five
140
Example
Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output
5
140 | instruction | 0 | 69,766 | 10 | 139,532 |
"Correct Solution:
```
while True:
n, m = (int(s) for s in input().split())
if not n:
break
sections = sorted(([int(s) for s in input().split()] for i in range(n)),
key=lambda x: x[1], reverse=True)
for i, (d, p) in enumerate(sections):
m -= d
if m <= 0:
print(sum(s[0] * s[1] for s in sections[i+1:]) - p * m)
break
else:
print(0)
``` | output | 1 | 69,766 | 10 | 139,533 |
Provide a correct Python 3 solution for this coding contest problem.
Princess'Marriage
Marriage of a princess
English text is not available in this practice contest.
A brave princess in a poor country, knowing that gambling payouts are determined by the parimutuel method, felt more familiar with gambling and was convinced of her victory in gambling. As a result, he spent more money than ever before and lost enough to lose all the taxes paid by the people. The King, who took this situation seriously, decided to marry the princess to the neighboring country. By doing this, I thought that I would like the princess to reflect on her daily activities and at the same time deepen her friendship with neighboring countries and receive financial assistance.
The princess and the prince of the neighboring country liked each other, and the kings of both countries agreed on a political marriage. The princess triumphantly went to the neighboring country with a little money in her hand. On the other hand, the motive for the princess to marry is to pursue the unilateral interests of the king, and the aide of the prince of the neighboring country who thinks that it is not pleasant shoots countless thugs along the way to make the princess dead. It was.
The path the princess will take has already been decided. There are a total of L post stations on the path of the princess. For convenience, the departure and arrival points are also set as post stations, and each post station is called S1, S2, ... SL. The princess shall be in S1 first, visit the post station in ascending order (S2, S3 ... in that order), and finally go to SL. At the post station, you can pay money to hire an escort, and as long as you have the money, you can contract for as long as you like to protect the princess. The cost of hiring an escort is 1 gold per distance. Note that the princess can also partially protect the section through which she passes. The distance between Si and Si + 1 is given by Di, and the expected value of the number of times a thug is attacked per distance between Si and Si + 1 is given by Pi.
Find the expected number of thugs to reach your destination when the princess has a budget of M and hires an escort to minimize the expected number of thugs.
Input
The input consists of multiple datasets. Each dataset has the following format.
> N M
> D1 P1
> D2 P2
> ...
> DN PN
Two integers are given in the first row of each dataset, representing the number of intervals N (1 ≤ N ≤ 10,000) and the budget M (0 ≤ M ≤ 1,000,000,000) of the princess difference, respectively. The next N lines show information about the path the princess takes. Each line contains two integers, and the i-th line is the expected value of the interval Di (1 ≤ Di ≤ 10,000) and the number of attacks when moving one unit distance between them Pi (0 ≤ Pi <= 10) ). The end of the input is represented by a data set with N = 0 and M = 0. Do not output the calculation result for this data set.
Output
For each dataset, output the expected number of times the princess will be attacked by thugs to your destination.
Sample Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output for the Sample Input
Five
140
Example
Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output
5
140 | instruction | 0 | 69,767 | 10 | 139,534 |
"Correct Solution:
```
b=True
while b:
n,m=map(int,input().split())
if n==0 and m==0:
b=False
exit
else:
l=[]
for i in range(n):
l1=list(map(int,input().split()))
l1=l1[::-1]
l.append(l1)
l=sorted(l)
l=l[::-1]
s=0
for i in range(n):
s+=l[i][0]*l[i][1]
for i in range(n):
if m-l[i][1]>=0:
s-=l[i][0]*l[i][1]
m-=l[i][1]
else:
s-=l[i][0]*m
break
print(s)
``` | output | 1 | 69,767 | 10 | 139,535 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Princess'Marriage
Marriage of a princess
English text is not available in this practice contest.
A brave princess in a poor country, knowing that gambling payouts are determined by the parimutuel method, felt more familiar with gambling and was convinced of her victory in gambling. As a result, he spent more money than ever before and lost enough to lose all the taxes paid by the people. The King, who took this situation seriously, decided to marry the princess to the neighboring country. By doing this, I thought that I would like the princess to reflect on her daily activities and at the same time deepen her friendship with neighboring countries and receive financial assistance.
The princess and the prince of the neighboring country liked each other, and the kings of both countries agreed on a political marriage. The princess triumphantly went to the neighboring country with a little money in her hand. On the other hand, the motive for the princess to marry is to pursue the unilateral interests of the king, and the aide of the prince of the neighboring country who thinks that it is not pleasant shoots countless thugs along the way to make the princess dead. It was.
The path the princess will take has already been decided. There are a total of L post stations on the path of the princess. For convenience, the departure and arrival points are also set as post stations, and each post station is called S1, S2, ... SL. The princess shall be in S1 first, visit the post station in ascending order (S2, S3 ... in that order), and finally go to SL. At the post station, you can pay money to hire an escort, and as long as you have the money, you can contract for as long as you like to protect the princess. The cost of hiring an escort is 1 gold per distance. Note that the princess can also partially protect the section through which she passes. The distance between Si and Si + 1 is given by Di, and the expected value of the number of times a thug is attacked per distance between Si and Si + 1 is given by Pi.
Find the expected number of thugs to reach your destination when the princess has a budget of M and hires an escort to minimize the expected number of thugs.
Input
The input consists of multiple datasets. Each dataset has the following format.
> N M
> D1 P1
> D2 P2
> ...
> DN PN
Two integers are given in the first row of each dataset, representing the number of intervals N (1 ≤ N ≤ 10,000) and the budget M (0 ≤ M ≤ 1,000,000,000) of the princess difference, respectively. The next N lines show information about the path the princess takes. Each line contains two integers, and the i-th line is the expected value of the interval Di (1 ≤ Di ≤ 10,000) and the number of attacks when moving one unit distance between them Pi (0 ≤ Pi <= 10) ). The end of the input is represented by a data set with N = 0 and M = 0. Do not output the calculation result for this data set.
Output
For each dataset, output the expected number of times the princess will be attacked by thugs to your destination.
Sample Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output for the Sample Input
Five
140
Example
Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output
5
140
Submitted Solution:
```
while True:
n, m = map(int, input().split())
ex = [0] * 11
if n == 0: break
for i in range(n):
d, p = map(int, input().split())
ex[p] += d
ex.reverse()
for i in range(len(ex)):
m -= ex[i]
if m <= 0:
ex[i] = abs(m)
break
else:
ex[i] = 0
ex.reverse()
ans = 0
for i in range(len(ex)):
ans += ex[i] * i
print(ans)
``` | instruction | 0 | 69,768 | 10 | 139,536 |
Yes | output | 1 | 69,768 | 10 | 139,537 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Princess'Marriage
Marriage of a princess
English text is not available in this practice contest.
A brave princess in a poor country, knowing that gambling payouts are determined by the parimutuel method, felt more familiar with gambling and was convinced of her victory in gambling. As a result, he spent more money than ever before and lost enough to lose all the taxes paid by the people. The King, who took this situation seriously, decided to marry the princess to the neighboring country. By doing this, I thought that I would like the princess to reflect on her daily activities and at the same time deepen her friendship with neighboring countries and receive financial assistance.
The princess and the prince of the neighboring country liked each other, and the kings of both countries agreed on a political marriage. The princess triumphantly went to the neighboring country with a little money in her hand. On the other hand, the motive for the princess to marry is to pursue the unilateral interests of the king, and the aide of the prince of the neighboring country who thinks that it is not pleasant shoots countless thugs along the way to make the princess dead. It was.
The path the princess will take has already been decided. There are a total of L post stations on the path of the princess. For convenience, the departure and arrival points are also set as post stations, and each post station is called S1, S2, ... SL. The princess shall be in S1 first, visit the post station in ascending order (S2, S3 ... in that order), and finally go to SL. At the post station, you can pay money to hire an escort, and as long as you have the money, you can contract for as long as you like to protect the princess. The cost of hiring an escort is 1 gold per distance. Note that the princess can also partially protect the section through which she passes. The distance between Si and Si + 1 is given by Di, and the expected value of the number of times a thug is attacked per distance between Si and Si + 1 is given by Pi.
Find the expected number of thugs to reach your destination when the princess has a budget of M and hires an escort to minimize the expected number of thugs.
Input
The input consists of multiple datasets. Each dataset has the following format.
> N M
> D1 P1
> D2 P2
> ...
> DN PN
Two integers are given in the first row of each dataset, representing the number of intervals N (1 ≤ N ≤ 10,000) and the budget M (0 ≤ M ≤ 1,000,000,000) of the princess difference, respectively. The next N lines show information about the path the princess takes. Each line contains two integers, and the i-th line is the expected value of the interval Di (1 ≤ Di ≤ 10,000) and the number of attacks when moving one unit distance between them Pi (0 ≤ Pi <= 10) ). The end of the input is represented by a data set with N = 0 and M = 0. Do not output the calculation result for this data set.
Output
For each dataset, output the expected number of times the princess will be attacked by thugs to your destination.
Sample Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output for the Sample Input
Five
140
Example
Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output
5
140
Submitted Solution:
```
while True:
N, M = map(int, input().split())
if N == 0 and M == 0: break
travel = [list(map(int, input().split())) for _ in range(N)]
travel.sort(reverse=True, key=lambda x:x[1])
for i, (d, _) in enumerate(travel):
if M == 0: break
travel[i][0] = max(0, d-M)
M = max(0, M-d)
print(sum(d*p for d, p in travel))
``` | instruction | 0 | 69,769 | 10 | 139,538 |
Yes | output | 1 | 69,769 | 10 | 139,539 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Princess'Marriage
Marriage of a princess
English text is not available in this practice contest.
A brave princess in a poor country, knowing that gambling payouts are determined by the parimutuel method, felt more familiar with gambling and was convinced of her victory in gambling. As a result, he spent more money than ever before and lost enough to lose all the taxes paid by the people. The King, who took this situation seriously, decided to marry the princess to the neighboring country. By doing this, I thought that I would like the princess to reflect on her daily activities and at the same time deepen her friendship with neighboring countries and receive financial assistance.
The princess and the prince of the neighboring country liked each other, and the kings of both countries agreed on a political marriage. The princess triumphantly went to the neighboring country with a little money in her hand. On the other hand, the motive for the princess to marry is to pursue the unilateral interests of the king, and the aide of the prince of the neighboring country who thinks that it is not pleasant shoots countless thugs along the way to make the princess dead. It was.
The path the princess will take has already been decided. There are a total of L post stations on the path of the princess. For convenience, the departure and arrival points are also set as post stations, and each post station is called S1, S2, ... SL. The princess shall be in S1 first, visit the post station in ascending order (S2, S3 ... in that order), and finally go to SL. At the post station, you can pay money to hire an escort, and as long as you have the money, you can contract for as long as you like to protect the princess. The cost of hiring an escort is 1 gold per distance. Note that the princess can also partially protect the section through which she passes. The distance between Si and Si + 1 is given by Di, and the expected value of the number of times a thug is attacked per distance between Si and Si + 1 is given by Pi.
Find the expected number of thugs to reach your destination when the princess has a budget of M and hires an escort to minimize the expected number of thugs.
Input
The input consists of multiple datasets. Each dataset has the following format.
> N M
> D1 P1
> D2 P2
> ...
> DN PN
Two integers are given in the first row of each dataset, representing the number of intervals N (1 ≤ N ≤ 10,000) and the budget M (0 ≤ M ≤ 1,000,000,000) of the princess difference, respectively. The next N lines show information about the path the princess takes. Each line contains two integers, and the i-th line is the expected value of the interval Di (1 ≤ Di ≤ 10,000) and the number of attacks when moving one unit distance between them Pi (0 ≤ Pi <= 10) ). The end of the input is represented by a data set with N = 0 and M = 0. Do not output the calculation result for this data set.
Output
For each dataset, output the expected number of times the princess will be attacked by thugs to your destination.
Sample Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output for the Sample Input
Five
140
Example
Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output
5
140
Submitted Solution:
```
while 1:
N, M = map(int, input().split())
if N == M == 0:
break
S = []
for i in range(N):
S.append(list(map(int, input().split())))
S.sort(key=lambda x:x[1], reverse=True)
ans = 0
for d in S:
if M >= d[0]:
M -= d[0]
else:
d[0] -= M
M = 0
ans += d[0] * d[1]
print(ans)
``` | instruction | 0 | 69,770 | 10 | 139,540 |
Yes | output | 1 | 69,770 | 10 | 139,541 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Princess'Marriage
Marriage of a princess
English text is not available in this practice contest.
A brave princess in a poor country, knowing that gambling payouts are determined by the parimutuel method, felt more familiar with gambling and was convinced of her victory in gambling. As a result, he spent more money than ever before and lost enough to lose all the taxes paid by the people. The King, who took this situation seriously, decided to marry the princess to the neighboring country. By doing this, I thought that I would like the princess to reflect on her daily activities and at the same time deepen her friendship with neighboring countries and receive financial assistance.
The princess and the prince of the neighboring country liked each other, and the kings of both countries agreed on a political marriage. The princess triumphantly went to the neighboring country with a little money in her hand. On the other hand, the motive for the princess to marry is to pursue the unilateral interests of the king, and the aide of the prince of the neighboring country who thinks that it is not pleasant shoots countless thugs along the way to make the princess dead. It was.
The path the princess will take has already been decided. There are a total of L post stations on the path of the princess. For convenience, the departure and arrival points are also set as post stations, and each post station is called S1, S2, ... SL. The princess shall be in S1 first, visit the post station in ascending order (S2, S3 ... in that order), and finally go to SL. At the post station, you can pay money to hire an escort, and as long as you have the money, you can contract for as long as you like to protect the princess. The cost of hiring an escort is 1 gold per distance. Note that the princess can also partially protect the section through which she passes. The distance between Si and Si + 1 is given by Di, and the expected value of the number of times a thug is attacked per distance between Si and Si + 1 is given by Pi.
Find the expected number of thugs to reach your destination when the princess has a budget of M and hires an escort to minimize the expected number of thugs.
Input
The input consists of multiple datasets. Each dataset has the following format.
> N M
> D1 P1
> D2 P2
> ...
> DN PN
Two integers are given in the first row of each dataset, representing the number of intervals N (1 ≤ N ≤ 10,000) and the budget M (0 ≤ M ≤ 1,000,000,000) of the princess difference, respectively. The next N lines show information about the path the princess takes. Each line contains two integers, and the i-th line is the expected value of the interval Di (1 ≤ Di ≤ 10,000) and the number of attacks when moving one unit distance between them Pi (0 ≤ Pi <= 10) ). The end of the input is represented by a data set with N = 0 and M = 0. Do not output the calculation result for this data set.
Output
For each dataset, output the expected number of times the princess will be attacked by thugs to your destination.
Sample Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output for the Sample Input
Five
140
Example
Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output
5
140
Submitted Solution:
```
readline = open(0).readline
while 1:
N, M = map(int, readline().split())
if N == M == 0:
break
Q = []
for i in range(N):
d, p = map(int, readline().split())
Q.append((p, d))
Q.sort(reverse=1)
ans = 0
for p, d in Q:
if M >= 0:
x = min(M, d)
M -= x; d -= x
ans += d * p
print(ans)
``` | instruction | 0 | 69,771 | 10 | 139,542 |
Yes | output | 1 | 69,771 | 10 | 139,543 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Princess'Marriage
Marriage of a princess
English text is not available in this practice contest.
A brave princess in a poor country, knowing that gambling payouts are determined by the parimutuel method, felt more familiar with gambling and was convinced of her victory in gambling. As a result, he spent more money than ever before and lost enough to lose all the taxes paid by the people. The King, who took this situation seriously, decided to marry the princess to the neighboring country. By doing this, I thought that I would like the princess to reflect on her daily activities and at the same time deepen her friendship with neighboring countries and receive financial assistance.
The princess and the prince of the neighboring country liked each other, and the kings of both countries agreed on a political marriage. The princess triumphantly went to the neighboring country with a little money in her hand. On the other hand, the motive for the princess to marry is to pursue the unilateral interests of the king, and the aide of the prince of the neighboring country who thinks that it is not pleasant shoots countless thugs along the way to make the princess dead. It was.
The path the princess will take has already been decided. There are a total of L post stations on the path of the princess. For convenience, the departure and arrival points are also set as post stations, and each post station is called S1, S2, ... SL. The princess shall be in S1 first, visit the post station in ascending order (S2, S3 ... in that order), and finally go to SL. At the post station, you can pay money to hire an escort, and as long as you have the money, you can contract for as long as you like to protect the princess. The cost of hiring an escort is 1 gold per distance. Note that the princess can also partially protect the section through which she passes. The distance between Si and Si + 1 is given by Di, and the expected value of the number of times a thug is attacked per distance between Si and Si + 1 is given by Pi.
Find the expected number of thugs to reach your destination when the princess has a budget of M and hires an escort to minimize the expected number of thugs.
Input
The input consists of multiple datasets. Each dataset has the following format.
> N M
> D1 P1
> D2 P2
> ...
> DN PN
Two integers are given in the first row of each dataset, representing the number of intervals N (1 ≤ N ≤ 10,000) and the budget M (0 ≤ M ≤ 1,000,000,000) of the princess difference, respectively. The next N lines show information about the path the princess takes. Each line contains two integers, and the i-th line is the expected value of the interval Di (1 ≤ Di ≤ 10,000) and the number of attacks when moving one unit distance between them Pi (0 ≤ Pi <= 10) ). The end of the input is represented by a data set with N = 0 and M = 0. Do not output the calculation result for this data set.
Output
For each dataset, output the expected number of times the princess will be attacked by thugs to your destination.
Sample Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output for the Sample Input
Five
140
Example
Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output
5
140
Submitted Solution:
```
while True:
n, m = map(int, input().split())
if (n, m) == (0, 0):
break
dp = [list(map(int, input().split())) for i in range(n)]
dp = sorted(dp, key=lambda kv:kv[1], reverse=True)
r = 0
for d,p in dp:
if m >= d:
m -= d
else:
r += p*(d-m)
m = 0
print(r
``` | instruction | 0 | 69,772 | 10 | 139,544 |
No | output | 1 | 69,772 | 10 | 139,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Princess'Marriage
Marriage of a princess
English text is not available in this practice contest.
A brave princess in a poor country, knowing that gambling payouts are determined by the parimutuel method, felt more familiar with gambling and was convinced of her victory in gambling. As a result, he spent more money than ever before and lost enough to lose all the taxes paid by the people. The King, who took this situation seriously, decided to marry the princess to the neighboring country. By doing this, I thought that I would like the princess to reflect on her daily activities and at the same time deepen her friendship with neighboring countries and receive financial assistance.
The princess and the prince of the neighboring country liked each other, and the kings of both countries agreed on a political marriage. The princess triumphantly went to the neighboring country with a little money in her hand. On the other hand, the motive for the princess to marry is to pursue the unilateral interests of the king, and the aide of the prince of the neighboring country who thinks that it is not pleasant shoots countless thugs along the way to make the princess dead. It was.
The path the princess will take has already been decided. There are a total of L post stations on the path of the princess. For convenience, the departure and arrival points are also set as post stations, and each post station is called S1, S2, ... SL. The princess shall be in S1 first, visit the post station in ascending order (S2, S3 ... in that order), and finally go to SL. At the post station, you can pay money to hire an escort, and as long as you have the money, you can contract for as long as you like to protect the princess. The cost of hiring an escort is 1 gold per distance. Note that the princess can also partially protect the section through which she passes. The distance between Si and Si + 1 is given by Di, and the expected value of the number of times a thug is attacked per distance between Si and Si + 1 is given by Pi.
Find the expected number of thugs to reach your destination when the princess has a budget of M and hires an escort to minimize the expected number of thugs.
Input
The input consists of multiple datasets. Each dataset has the following format.
> N M
> D1 P1
> D2 P2
> ...
> DN PN
Two integers are given in the first row of each dataset, representing the number of intervals N (1 ≤ N ≤ 10,000) and the budget M (0 ≤ M ≤ 1,000,000,000) of the princess difference, respectively. The next N lines show information about the path the princess takes. Each line contains two integers, and the i-th line is the expected value of the interval Di (1 ≤ Di ≤ 10,000) and the number of attacks when moving one unit distance between them Pi (0 ≤ Pi <= 10) ). The end of the input is represented by a data set with N = 0 and M = 0. Do not output the calculation result for this data set.
Output
For each dataset, output the expected number of times the princess will be attacked by thugs to your destination.
Sample Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output for the Sample Input
Five
140
Example
Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output
5
140
Submitted Solution:
```
while True:
N,M = map(int,input().strip().split(" "))
if [N,M] == [0,0]:
break
L = []
S = 0 #Sが襲われる回数の期待値
for i in range(N):
l = list(map(int,input().strip().split(" ")))
l.reverse() #ソートするとき具合がわるいので[P,D]の順に入れ替え
L.append(l)
S = S + l[0]*l[1]
L.sort()
j = N
while M > 0:
if M >= L[j-1][1]:
S = S - L[j][0]*L[j][1]
M = M - L[j][1]
j = j - 1
else:
S = S - L[j][0]*M
M = M - L[j][1]
print(S)
``` | instruction | 0 | 69,773 | 10 | 139,546 |
No | output | 1 | 69,773 | 10 | 139,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Princess'Marriage
Marriage of a princess
English text is not available in this practice contest.
A brave princess in a poor country, knowing that gambling payouts are determined by the parimutuel method, felt more familiar with gambling and was convinced of her victory in gambling. As a result, he spent more money than ever before and lost enough to lose all the taxes paid by the people. The King, who took this situation seriously, decided to marry the princess to the neighboring country. By doing this, I thought that I would like the princess to reflect on her daily activities and at the same time deepen her friendship with neighboring countries and receive financial assistance.
The princess and the prince of the neighboring country liked each other, and the kings of both countries agreed on a political marriage. The princess triumphantly went to the neighboring country with a little money in her hand. On the other hand, the motive for the princess to marry is to pursue the unilateral interests of the king, and the aide of the prince of the neighboring country who thinks that it is not pleasant shoots countless thugs along the way to make the princess dead. It was.
The path the princess will take has already been decided. There are a total of L post stations on the path of the princess. For convenience, the departure and arrival points are also set as post stations, and each post station is called S1, S2, ... SL. The princess shall be in S1 first, visit the post station in ascending order (S2, S3 ... in that order), and finally go to SL. At the post station, you can pay money to hire an escort, and as long as you have the money, you can contract for as long as you like to protect the princess. The cost of hiring an escort is 1 gold per distance. Note that the princess can also partially protect the section through which she passes. The distance between Si and Si + 1 is given by Di, and the expected value of the number of times a thug is attacked per distance between Si and Si + 1 is given by Pi.
Find the expected number of thugs to reach your destination when the princess has a budget of M and hires an escort to minimize the expected number of thugs.
Input
The input consists of multiple datasets. Each dataset has the following format.
> N M
> D1 P1
> D2 P2
> ...
> DN PN
Two integers are given in the first row of each dataset, representing the number of intervals N (1 ≤ N ≤ 10,000) and the budget M (0 ≤ M ≤ 1,000,000,000) of the princess difference, respectively. The next N lines show information about the path the princess takes. Each line contains two integers, and the i-th line is the expected value of the interval Di (1 ≤ Di ≤ 10,000) and the number of attacks when moving one unit distance between them Pi (0 ≤ Pi <= 10) ). The end of the input is represented by a data set with N = 0 and M = 0. Do not output the calculation result for this data set.
Output
For each dataset, output the expected number of times the princess will be attacked by thugs to your destination.
Sample Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output for the Sample Input
Five
140
Example
Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output
5
140
Submitted Solution:
```
da = list(map(int,input().split()))
k = [[0 for i in range(2)]for j in range(10001)]
i = 0
su = 0
j = 0
while da[0] != 0 and da[1] != 0:
i = 0
su = 0
j = 0
for i in range(da[0]):
k[i][0],k[i][1] = list(map(int,input().split()))
su += k[i][0] * k[i][1]
sorted(k,key = lambda x: x[1],reverse = True)
while da[1] != 0:
if da[1] >= k[j][0]:
su -= k[j][0] * k[j][1]
da[1] -= k[j][0]
j += 1
else:
su -= da[1] * k[j][1]
da[1] = 0
print(su)
da = list(map(int,input().split()))
if da[0] == 0 and da[1] == 0:
break
k = [[0 for i in range(2)]for j in range(10001)]
``` | instruction | 0 | 69,774 | 10 | 139,548 |
No | output | 1 | 69,774 | 10 | 139,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Princess'Marriage
Marriage of a princess
English text is not available in this practice contest.
A brave princess in a poor country, knowing that gambling payouts are determined by the parimutuel method, felt more familiar with gambling and was convinced of her victory in gambling. As a result, he spent more money than ever before and lost enough to lose all the taxes paid by the people. The King, who took this situation seriously, decided to marry the princess to the neighboring country. By doing this, I thought that I would like the princess to reflect on her daily activities and at the same time deepen her friendship with neighboring countries and receive financial assistance.
The princess and the prince of the neighboring country liked each other, and the kings of both countries agreed on a political marriage. The princess triumphantly went to the neighboring country with a little money in her hand. On the other hand, the motive for the princess to marry is to pursue the unilateral interests of the king, and the aide of the prince of the neighboring country who thinks that it is not pleasant shoots countless thugs along the way to make the princess dead. It was.
The path the princess will take has already been decided. There are a total of L post stations on the path of the princess. For convenience, the departure and arrival points are also set as post stations, and each post station is called S1, S2, ... SL. The princess shall be in S1 first, visit the post station in ascending order (S2, S3 ... in that order), and finally go to SL. At the post station, you can pay money to hire an escort, and as long as you have the money, you can contract for as long as you like to protect the princess. The cost of hiring an escort is 1 gold per distance. Note that the princess can also partially protect the section through which she passes. The distance between Si and Si + 1 is given by Di, and the expected value of the number of times a thug is attacked per distance between Si and Si + 1 is given by Pi.
Find the expected number of thugs to reach your destination when the princess has a budget of M and hires an escort to minimize the expected number of thugs.
Input
The input consists of multiple datasets. Each dataset has the following format.
> N M
> D1 P1
> D2 P2
> ...
> DN PN
Two integers are given in the first row of each dataset, representing the number of intervals N (1 ≤ N ≤ 10,000) and the budget M (0 ≤ M ≤ 1,000,000,000) of the princess difference, respectively. The next N lines show information about the path the princess takes. Each line contains two integers, and the i-th line is the expected value of the interval Di (1 ≤ Di ≤ 10,000) and the number of attacks when moving one unit distance between them Pi (0 ≤ Pi <= 10) ). The end of the input is represented by a data set with N = 0 and M = 0. Do not output the calculation result for this data set.
Output
For each dataset, output the expected number of times the princess will be attacked by thugs to your destination.
Sample Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output for the Sample Input
Five
140
Example
Input
2 8
5 6
4 5
3 1
5 10
5 10
5 10
0 0
Output
5
140
Submitted Solution:
```
da = list(map(int,input().split()))
k = [[0 for i in range(2)]for j in range(10001)]
i = 0
su = 0
j = 0
while da[0] != 0 and da[1] != 0:
i = 0
su = 0
j = 0
for i in range(da[0]):
k[i][0],k[i][1] = list(map(int,input().split()))
su += k[i][0] * k[i][1]
sorted(k,key = lambda x: x[1],reverse = True)
while da[1] > 0 and j != i - 1:
if da[1] >= k[j][0] and j != i:
su -= k[j][0] * k[j][1]
da[1] -= k[j][0]
j += 1
else:
su -= da[1] * k[j][1]
da[1] = 0
print(su)
da = list(map(int,input().split()))
if da[0] == 0 and da[1] == 0:
break
k = [[0 for i in range(2)]for j in range(10001)]
``` | instruction | 0 | 69,775 | 10 | 139,550 |
No | output | 1 | 69,775 | 10 | 139,551 |
Provide a correct Python 3 solution for this coding contest problem.
You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsack.
* You can select at most $m_i$ items for $i$-th item.
Find the maximum total value of items in the knapsack.
Constraints
* $1 \le N \le 50$
* $1 \le v_i \le 50$
* $1 \le w_i \le 10^9$
* $1 \le m_i \le 10^9$
* $1 \le W \le 10^9$
Input
$N$ $W$
$v_1$ $w_1$ $m_1$
$v_2$ $w_2$ $m_2$
:
$v_N$ $w_N$ $m_N$
The first line consists of the integers $N$ and $W$. In the following $N$ lines, the value, weight and limitation of the $i$-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 8
4 3 2
2 1 1
1 2 4
3 2 2
Output
12
Input
2 100
1 1 100
2 1 50
Output
150
Input
5 1000000000
3 5 1000000000
7 6 1000000000
4 4 1000000000
6 8 1000000000
2 5 1000000000
Output
1166666666 | instruction | 0 | 69,787 | 10 | 139,574 |
"Correct Solution:
```
import sys
from collections import deque
readline = sys.stdin.readline
write = sys.stdout.write
def solve():
N, W = map(int, readline().split())
vs = [0]*N; ws = [0]*N; ms = [0]*N
for i in range(N):
vs[i], ws[i], ms[i] = map(int, readline().split())
V0 = max(vs)
V = sum(v * min(V0, m) for v, m in zip(vs, ms))
dp = [W+1]*(V + 1)
dp[0] = 0
for i in range(N):
v = vs[i]; w = ws[i]; m = ms[i]
c = min(V0, m)
ms[i] -= c
for k in range(v):
que = deque()
push = que.append
popf = que.popleft; popb = que.pop
for j in range((V-k)//v+1):
a = dp[k + j*v] - j * w
while que and a <= que[-1][1]:
popb()
push((j, a))
p, b = que[0]
dp[k + j*v] = b + j * w
if que and p <= j-c:
popf()
*I, = range(N)
I.sort(key=lambda x: ws[x]/vs[x])
*S, = [(vs[i], ws[i], ms[i]) for i in I]
ans = 0
def greedy():
yield 0
for i in range(V + 1):
if dp[i] > W:
continue
rest = W - dp[i]
r = i
for v, w, m in S:
m = min(m, rest // w)
r += m * v
rest -= m * w
yield r
write("%d\n" % max(greedy()))
solve()
``` | output | 1 | 69,787 | 10 | 139,575 |
Provide a correct Python 3 solution for this coding contest problem.
You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsack.
* You can select at most $m_i$ items for $i$-th item.
Find the maximum total value of items in the knapsack.
Constraints
* $1 \le N \le 50$
* $1 \le v_i \le 50$
* $1 \le w_i \le 10^9$
* $1 \le m_i \le 10^9$
* $1 \le W \le 10^9$
Input
$N$ $W$
$v_1$ $w_1$ $m_1$
$v_2$ $w_2$ $m_2$
:
$v_N$ $w_N$ $m_N$
The first line consists of the integers $N$ and $W$. In the following $N$ lines, the value, weight and limitation of the $i$-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 8
4 3 2
2 1 1
1 2 4
3 2 2
Output
12
Input
2 100
1 1 100
2 1 50
Output
150
Input
5 1000000000
3 5 1000000000
7 6 1000000000
4 4 1000000000
6 8 1000000000
2 5 1000000000
Output
1166666666 | instruction | 0 | 69,788 | 10 | 139,576 |
"Correct Solution:
```
#!python3
iim = lambda: map(int, input().rstrip().split())
from heapq import heappush, heappop
def resolve():
N, W = iim()
S = [list(iim()) for i in range(N)]
def f1(v, w, m):
mm = []
i = 1
while i <= m:
yield i
m -= i
i <<= 1
if m:
yield m
SS = list(sorted(((v*i, w*i) for v, w, m in S for i in f1(v, w, m)), key=lambda x: (x[0]/x[1], x[1]), reverse=True))
NN = len(SS)
def ubound(v, w, i):
for j in range(i, NN):
vj, wj = SS[j]
if w + wj > W:
return (-v, -v - (W - w) * vj / wj)
w += wj
v += vj
return (-v, -v)
u0, u1 = ubound(0, 0, 0)
um = u0
ans = 0
q = []
heappush(q, (u1, u0, 0, 0, 0))
#print(S, NN)
#print(W, SS)
while q:
#print(q)
uq0, uq1, vq, wq, i = heappop(q)
if uq0 > um:
break
if i >= NN: continue
vi, wi = SS[i]
if wq + wi < W:
heappush(q, (uq0, uq1, vq+vi, wq+wi, i + 1))
u0, u1 = ubound(vq, wq, i + 1)
if u1 <= um:
if u0 < um:
um = u0
heappush(q, (u1, u0, vq, wq, i + 1))
print(-um)
if __name__ == "__main__":
resolve()
``` | output | 1 | 69,788 | 10 | 139,577 |
Provide a correct Python 3 solution for this coding contest problem.
You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsack.
* You can select at most $m_i$ items for $i$-th item.
Find the maximum total value of items in the knapsack.
Constraints
* $1 \le N \le 50$
* $1 \le v_i \le 50$
* $1 \le w_i \le 10^9$
* $1 \le m_i \le 10^9$
* $1 \le W \le 10^9$
Input
$N$ $W$
$v_1$ $w_1$ $m_1$
$v_2$ $w_2$ $m_2$
:
$v_N$ $w_N$ $m_N$
The first line consists of the integers $N$ and $W$. In the following $N$ lines, the value, weight and limitation of the $i$-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 8
4 3 2
2 1 1
1 2 4
3 2 2
Output
12
Input
2 100
1 1 100
2 1 50
Output
150
Input
5 1000000000
3 5 1000000000
7 6 1000000000
4 4 1000000000
6 8 1000000000
2 5 1000000000
Output
1166666666 | instruction | 0 | 69,789 | 10 | 139,578 |
"Correct Solution:
```
from typing import Tuple, List
from heapq import heappush, heappop
def max_value(max_weight: int, items: List[Tuple[int, int]]) -> float:
def _cost(weight: int, value: int, i: int) -> Tuple[int, float]:
# Use value as negative cost
for j in range(i, n):
v, w = items[j]
if (weight + w > max_weight):
return (-value, -value - (v / w) * (max_weight - weight))
weight += w
value += v
return (-value, -value * 1.0)
n = len(items)
items.sort(key=lambda x: x[1], reverse=True)
items.sort(key=lambda x: x[0] / x[1], reverse=True)
u, c = _cost(0, 0, 0)
heap = [(c, u, 0, 0, 0)]
maxcost = u
while heap:
cost, upper, weight, value, i = heappop(heap)
if (cost > maxcost):
break
if (i < n):
v, w = items[i]
if (weight + w <= max_weight):
heappush(heap, (cost, upper, weight + w, value + v, i + 1))
u, c = _cost(weight, value, i + 1)
if (c < maxcost):
if (u < maxcost):
maxcost = u
heappush(heap, (c, u, weight, value, i + 1))
return -maxcost
if __name__ == '__main__':
N, W = map(lambda x: int(x), input().split())
items: List[Tuple[int, int]] = []
for _ in range(N):
v, w, m = map(lambda x: int(x), input().split())
for i in range(m.bit_length() - 1):
h = 2 ** i
m -= h
items.append((v * h, w * h))
if m > 0:
items.append((v * m, w * m))
print(max_value(W, items))
``` | output | 1 | 69,789 | 10 | 139,579 |
Provide a correct Python 3 solution for this coding contest problem.
You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsack.
* You can select at most $m_i$ items for $i$-th item.
Find the maximum total value of items in the knapsack.
Constraints
* $1 \le N \le 50$
* $1 \le v_i \le 50$
* $1 \le w_i \le 10^9$
* $1 \le m_i \le 10^9$
* $1 \le W \le 10^9$
Input
$N$ $W$
$v_1$ $w_1$ $m_1$
$v_2$ $w_2$ $m_2$
:
$v_N$ $w_N$ $m_N$
The first line consists of the integers $N$ and $W$. In the following $N$ lines, the value, weight and limitation of the $i$-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 8
4 3 2
2 1 1
1 2 4
3 2 2
Output
12
Input
2 100
1 1 100
2 1 50
Output
150
Input
5 1000000000
3 5 1000000000
7 6 1000000000
4 4 1000000000
6 8 1000000000
2 5 1000000000
Output
1166666666 | instruction | 0 | 69,790 | 10 | 139,580 |
"Correct Solution:
```
def main():
from collections import deque
from functools import cmp_to_key
N, W, *A = map(int, open(0).read().split())
A = [list(a) for a in zip(*[iter(A)]*3)]
Vmax = max(v for v, w, m in A)
Vsum = 0
B = []
for i, (v, w, m) in enumerate(A):
m_upper = min(m, Vmax-1)
m_rest = m - m_upper
A[i][2] = m_upper
if m_rest:
B.append((v, w, m_rest))
Vsum += v*m_upper
INF = 1<<60
dp = [INF]*(Vsum+1)
dp[0] = 0
for v, w, m in A:
for ofs in range(v):
q = deque()
for j in range(Vsum+1):
vj = ofs + j*v
if vj > Vsum:
break
src = dp[vj] - j*w
while q and q[-1][1] >= src:
q.pop()
q.append((j, src))
if q:
ni, nw = q[0]
dp[vj] = nw + j*w
if ni == j - m:
q.popleft()
B.sort(key=cmp_to_key(lambda a, b: -1 if a[0]*b[1] > a[1]*b[0] else 1))
ans = 0
for V, d in enumerate(dp):
if d > W:
continue
W_rest = W - d
for v, w, m in B:
n_item = min(m, W_rest//w)
if n_item:
V += n_item * v
W_rest -= n_item * w
if ans < V:
ans = V
print(ans)
if __name__ == '__main__':
main()
``` | output | 1 | 69,790 | 10 | 139,581 |
Provide a correct Python 3 solution for this coding contest problem.
You have $N$ items that you want to put them into a knapsack. Item $i$ has value $v_i$, weight $w_i$ and limitation $m_i$.
You want to find a subset of items to put such that:
* The total value of the items is as large as possible.
* The items have combined weight at most $W$, that is capacity of the knapsack.
* You can select at most $m_i$ items for $i$-th item.
Find the maximum total value of items in the knapsack.
Constraints
* $1 \le N \le 50$
* $1 \le v_i \le 50$
* $1 \le w_i \le 10^9$
* $1 \le m_i \le 10^9$
* $1 \le W \le 10^9$
Input
$N$ $W$
$v_1$ $w_1$ $m_1$
$v_2$ $w_2$ $m_2$
:
$v_N$ $w_N$ $m_N$
The first line consists of the integers $N$ and $W$. In the following $N$ lines, the value, weight and limitation of the $i$-th item are given.
Output
Print the maximum total values of the items in a line.
Examples
Input
4 8
4 3 2
2 1 1
1 2 4
3 2 2
Output
12
Input
2 100
1 1 100
2 1 50
Output
150
Input
5 1000000000
3 5 1000000000
7 6 1000000000
4 4 1000000000
6 8 1000000000
2 5 1000000000
Output
1166666666 | instruction | 0 | 69,791 | 10 | 139,582 |
"Correct Solution:
```
#!/usr/bin/env python3
# DPL_1_I: Combinatorial - Knapsack Problem with Limitations II
from heapq import heappush, heappop
def max_value(t, xs):
def _cost(weight, value, i):
for j in range(i, n):
v, w = xs[j]
if weight + w > t:
return (-value, -value - (v/w)*(t-weight))
weight += w
value += v
return (-value, -value*1.0)
n = len(xs)
xs.sort(key=lambda x: x[1], reverse=True)
xs.sort(key=lambda x: x[0]/x[1], reverse=True)
u, c = _cost(0, 0, 0)
heap = [(c, u, 0, 0, 0)]
maxcost = u
while heap:
cost, upper, weight, value, i = heappop(heap)
if cost > maxcost:
break
if i < n:
v, w = xs[i]
if weight+w <= t:
heappush(heap, (cost, upper, weight+w, value+v, i+1))
u, c = _cost(weight, value, i+1)
if c < maxcost:
if u < maxcost:
maxcost = u
heappush(heap, (c, u, weight, value, i+1))
return -maxcost
def run():
n, t = [int(i) for i in input().split()]
xs = []
for _ in range(n):
v, w, m = [int(i) for i in input().split()]
for i in range(m.bit_length()-1):
h = 2**i
m -= h
xs.append((v*h, w*h))
if m > 0:
xs.append((v*m, w*m))
print(max_value(t, xs))
if __name__ == '__main__':
run()
``` | output | 1 | 69,791 | 10 | 139,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties — n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study — you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 ≤ n, m ≤ 3000) — the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 ≤ p_i ≤ m, 1 ≤ c_i ≤ 10^9) — the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number — the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party. | instruction | 0 | 70,730 | 10 | 141,460 |
Tags: brute force, greedy
Correct Solution:
```
from sys import stdin
from collections import deque
n,m = map(int, stdin.readline().split())
pc=dict()
ph=dict()
costs=[]
class Cost:
def __init__(self, id, cost, party) -> None:
self.id=id
self.cost=cost
self.party=party
self.removed=False
for i in range(n):
p, c = map(int,stdin.readline().split())
if p != 1:
cost = Cost(i, c, p)
costs.append(cost)
pc.setdefault(p,[]).append(cost)
ph[p] = ph.setdefault(p, 0) + 1
max_h=0
for p, pa in pc.items():
pa.sort(reverse=True, key=lambda x: x.cost)
max_h = max(max_h, len(pa))
hp = [[] for _ in range(max_h+2)]
for p, h in ph.items():
if p != 1:
hp[h].append(p)
ans=[0]*(max_h+2)
costs.sort(key=lambda x:x.cost)
dq = deque(costs)
p_set = set()
height_p1 = ph[1] if 1 in ph else 0
top_sum = 0
top_count = 0
for i in range(max_h+1, -1, -1):
if len(costs) < i:
ans[i] = float('inf')
continue
for p in hp[i]:
p_set.add(p)
for p in p_set:
if len(pc[p]) > 0:
min_cost = pc[p].pop()
min_cost.removed=True
top_sum += min_cost.cost
top_count += 1
ans[i] += top_sum
if height_p1+top_count < i:
for j in range(i-height_p1-top_count):
while dq[0].removed: dq.popleft()
ans[i] += dq[0].cost
dq.rotate(-1)
dq.rotate(i-height_p1-top_count)
print(min(ans))
``` | output | 1 | 70,730 | 10 | 141,461 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties — n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study — you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 ≤ n, m ≤ 3000) — the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 ≤ p_i ≤ m, 1 ≤ c_i ≤ 10^9) — the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number — the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party. | instruction | 0 | 70,731 | 10 | 141,462 |
Tags: brute force, greedy
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
import io
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
from collections import Counter
import threading
#sys.setrecursionlimit(300000)
#threading.stack_size(10**8)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
#mod = 9223372036854775807
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: math.gcd(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
class SegmentTree1:
def __init__(self, data, default=0, func=lambda a, b: a+b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
MOD=10**9+7
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
mod=10**9+7
omod=998244353
#-------------------------------------------------------------------------
prime = [True for i in range(11)]
prime[0]=prime[1]=False
#pp=[0]*10
def SieveOfEratosthenes(n=10):
p = 2
c=0
while (p <= n):
if (prime[p] == True):
c+=1
for i in range(p, n+1, p):
#pp[i]=1
prime[i] = False
p += 1
#-----------------------------------DSU--------------------------------------------------
class DSU:
def __init__(self, R, C):
#R * C is the source, and isn't a grid square
self.par = range(R*C + 1)
self.rnk = [0] * (R*C + 1)
self.sz = [1] * (R*C + 1)
def find(self, x):
if self.par[x] != x:
self.par[x] = self.find(self.par[x])
return self.par[x]
def union(self, x, y):
xr, yr = self.find(x), self.find(y)
if xr == yr: return
if self.rnk[xr] < self.rnk[yr]:
xr, yr = yr, xr
if self.rnk[xr] == self.rnk[yr]:
self.rnk[xr] += 1
self.par[yr] = xr
self.sz[xr] += self.sz[yr]
def size(self, x):
return self.sz[self.find(x)]
def top(self):
# Size of component at ephemeral "source" node at index R*C,
# minus 1 to not count the source itself in the size
return self.size(len(self.sz) - 1) - 1
#---------------------------------Lazy Segment Tree--------------------------------------
# https://github.com/atcoder/ac-library/blob/master/atcoder/lazysegtree.hpp
class LazySegTree:
def __init__(self, _op, _e, _mapping, _composition, _id, v):
def set(p, x):
assert 0 <= p < _n
p += _size
for i in range(_log, 0, -1):
_push(p >> i)
_d[p] = x
for i in range(1, _log + 1):
_update(p >> i)
def get(p):
assert 0 <= p < _n
p += _size
for i in range(_log, 0, -1):
_push(p >> i)
return _d[p]
def prod(l, r):
assert 0 <= l <= r <= _n
if l == r:
return _e
l += _size
r += _size
for i in range(_log, 0, -1):
if ((l >> i) << i) != l:
_push(l >> i)
if ((r >> i) << i) != r:
_push(r >> i)
sml = _e
smr = _e
while l < r:
if l & 1:
sml = _op(sml, _d[l])
l += 1
if r & 1:
r -= 1
smr = _op(_d[r], smr)
l >>= 1
r >>= 1
return _op(sml, smr)
def apply(l, r, f):
assert 0 <= l <= r <= _n
if l == r:
return
l += _size
r += _size
for i in range(_log, 0, -1):
if ((l >> i) << i) != l:
_push(l >> i)
if ((r >> i) << i) != r:
_push((r - 1) >> i)
l2 = l
r2 = r
while l < r:
if l & 1:
_all_apply(l, f)
l += 1
if r & 1:
r -= 1
_all_apply(r, f)
l >>= 1
r >>= 1
l = l2
r = r2
for i in range(1, _log + 1):
if ((l >> i) << i) != l:
_update(l >> i)
if ((r >> i) << i) != r:
_update((r - 1) >> i)
def _update(k):
_d[k] = _op(_d[2 * k], _d[2 * k + 1])
def _all_apply(k, f):
_d[k] = _mapping(f, _d[k])
if k < _size:
_lz[k] = _composition(f, _lz[k])
def _push(k):
_all_apply(2 * k, _lz[k])
_all_apply(2 * k + 1, _lz[k])
_lz[k] = _id
_n = len(v)
_log = _n.bit_length()
_size = 1 << _log
_d = [_e] * (2 * _size)
_lz = [_id] * _size
for i in range(_n):
_d[_size + i] = v[i]
for i in range(_size - 1, 0, -1):
_update(i)
self.set = set
self.get = get
self.prod = prod
self.apply = apply
MIL = 1 << 20
def makeNode(total, count):
# Pack a pair into a float
return (total * MIL) + count
def getTotal(node):
return math.floor(node / MIL)
def getCount(node):
return node - getTotal(node) * MIL
nodeIdentity = makeNode(0.0, 0.0)
def nodeOp(node1, node2):
return node1 + node2
# Equivalent to the following:
return makeNode(
getTotal(node1) + getTotal(node2), getCount(node1) + getCount(node2)
)
identityMapping = -1
def mapping(tag, node):
if tag == identityMapping:
return node
# If assigned, new total is the number assigned times count
count = getCount(node)
return makeNode(tag * count, count)
def composition(mapping1, mapping2):
# If assigned multiple times, take first non-identity assignment
return mapping1 if mapping1 != identityMapping else mapping2
#---------------------------------Pollard rho--------------------------------------------
def memodict(f):
"""memoization decorator for a function taking a single argument"""
class memodict(dict):
def __missing__(self, key):
ret = self[key] = f(key)
return ret
return memodict().__getitem__
def pollard_rho(n):
"""returns a random factor of n"""
if n & 1 == 0:
return 2
if n % 3 == 0:
return 3
s = ((n - 1) & (1 - n)).bit_length() - 1
d = n >> s
for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
p = pow(a, d, n)
if p == 1 or p == n - 1 or a % n == 0:
continue
for _ in range(s):
prev = p
p = (p * p) % n
if p == 1:
return math.gcd(prev - 1, n)
if p == n - 1:
break
else:
for i in range(2, n):
x, y = i, (i * i + 1) % n
f = math.gcd(abs(x - y), n)
while f == 1:
x, y = (x * x + 1) % n, (y * y + 1) % n
y = (y * y + 1) % n
f = math.gcd(abs(x - y), n)
if f != n:
return f
return n
@memodict
def prime_factors(n):
"""returns a Counter of the prime factorization of n"""
if n <= 1:
return Counter()
f = pollard_rho(n)
return Counter([n]) if f == n else prime_factors(f) + prime_factors(n // f)
def distinct_factors(n):
"""returns a list of all distinct factors of n"""
factors = [1]
for p, exp in prime_factors(n).items():
factors += [p**i * factor for factor in factors for i in range(1, exp + 1)]
return factors
def all_factors(n):
"""returns a sorted list of all distinct factors of n"""
small, large = [], []
for i in range(1, int(n**0.5) + 1, 2 if n & 1 else 1):
if not n % i:
small.append(i)
large.append(n // i)
if small[-1] == large[-1]:
large.pop()
large.reverse()
small.extend(large)
return small
#---------------------------------Binary Search------------------------------------------
def binarySearch(arr, n, key):
left = 0
right = n-1
mid = 0
res = n
while (left <= right):
mid = (right + left)//2
if (arr[mid] > key):
res=mid
right = mid-1
else:
left = mid + 1
return res
def binarySearch1(arr, n, key):
left = 0
right = n-1
mid = 0
res=-1
while (left <= right):
mid = (right + left)//2
if (arr[mid] > key):
right = mid-1
else:
res=mid
left = mid + 1
return res
#---------------------------------running code------------------------------------------
t=1
#t=int(input())
for _ in range (t):
#n=int(input())
n,m=map(int,input().split())
#a=list(map(int,input().split()))
#b=list(map(int,input().split()))
#s=input()
#n=len(s)
res=10**13
votes=[[] for i in range (3001)]
for i in range (n):
p,cost=map(int,input().split())
votes[p].append(cost)
for i in range (3001):
votes[i].sort()
for i in range (max(len(votes[1]),1),n+1):
present=len(votes[1])
spare=[]
ans=0
for j in range (2,3001):
ind=0
if len(votes[j])>=i:
for k in range (len(votes[j])-i+1):
ans+=votes[j][k]
present+=1
ind=len(votes[j])-i+1
for k in range (ind,len(votes[j])):
spare.append(votes[j][k])
#print(i,spare,present,ans)
if present>i:
continue
spare.sort()
for j in range (min(i-present,len(spare))):
ans+=spare[j]
present+=1
if present!=i:
continue
res=min(res,ans)
#print(i,ans)
print(res)
``` | output | 1 | 70,731 | 10 | 141,463 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties — n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study — you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 ≤ n, m ≤ 3000) — the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 ≤ p_i ≤ m, 1 ≤ c_i ≤ 10^9) — the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number — the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party. | instruction | 0 | 70,732 | 10 | 141,464 |
Tags: brute force, greedy
Correct Solution:
```
n, m = map(int, input().split())
pc = [(0, 0) for _ in range(n)]
party_votes = [0 for _ in range(m)]
for i in range(n):
p, c = map(int, input().split())
pc[i] = (p - 1, c)
party_votes[p - 1] += 1
pc.sort(key=lambda x: x[1])
min_cost = 10**20
for votes in range(n + 1):
_party_votes = party_votes[:]
dangerous = list(map(lambda party: _party_votes[party] >= votes, range(0, m)))
used = list(map(lambda i: pc[i][0] == 0, range(n)))
cur_cost = 0
for i in range(n):
if dangerous[pc[i][0]] and pc[i][0] != 0:
cur_cost += pc[i][1]
_party_votes[0] += 1
_party_votes[pc[i][0]] -= 1
dangerous[pc[i][0]] = _party_votes[pc[i][0]] >= votes
used[i] = True
for i in range(n):
if _party_votes[0] >= votes:
break
if not used[i]:
_party_votes[0] += 1
cur_cost += pc[i][1]
min_cost = min(min_cost, cur_cost)
print(min_cost)
``` | output | 1 | 70,732 | 10 | 141,465 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties — n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study — you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 ≤ n, m ≤ 3000) — the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 ≤ p_i ≤ m, 1 ≤ c_i ≤ 10^9) — the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number — the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party. | instruction | 0 | 70,733 | 10 | 141,466 |
Tags: brute force, greedy
Correct Solution:
```
from sys import stdin
n,m = map(int, stdin.readline().split())
pc=dict()
ph=dict()
prices = []
for i in range(n):
p,c=map(int,stdin.readline().split())
if p != 1:
prices.append(c)
pc.setdefault(p,[]).append(len(prices)-1)
ph[p] = ph.setdefault(p, 0) + 1
max_h=0
for p, pa in pc.items():
pa.sort(reverse=True, key=lambda x: prices[x])
max_h = max(max_h, len(pa))
hp = [[] for _ in range(max_h+2)]
for p, h in ph.items():
if p != 1:
hp[h].append(p)
ans=[0]*(max_h+2)
sprices_ind = sorted(range(len(prices)), key=lambda x: prices[x])
p_set = set()
height_p1 = ph[1] if 1 in ph else 0
top_sum = 0
top_count = 0
for i in range(max_h+1, -1, -1):
if len(prices) < i:
ans[i] = float('inf')
continue
for p in hp[i]:
p_set.add(p)
for p in p_set:
if len(pc[p]) > 0:
top_ind = pc[p].pop()
c = prices[top_ind]
prices[top_ind] = -1
top_sum += c
top_count += 1
ans[i] += top_sum
if height_p1+top_count < i:
k = 0
for j in range(i-height_p1-top_count):
while prices[sprices_ind[k]] == -1: k += 1
ans[i] += prices[sprices_ind[k]]
k += 1
print(min(ans))
``` | output | 1 | 70,733 | 10 | 141,467 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties — n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study — you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 ≤ n, m ≤ 3000) — the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 ≤ p_i ≤ m, 1 ≤ c_i ≤ 10^9) — the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number — the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party. | instruction | 0 | 70,734 | 10 | 141,468 |
Tags: brute force, greedy
Correct Solution:
```
import sys
import os
def solve(m, candidates):
n = len(candidates)
candidates.sort(key=lambda x: x[1])
party = dict()
granted = 0
for i in range(len(candidates)):
p = candidates[i][0]
c = candidates[i][1]
if p == 1:
granted += 1
continue
if p in party:
party[p].append((i, c))
else:
party[p] = [(i, c)]
result = None
for t in range(1, n // 2 + 2):
total = 0
chosen = set()
for k, v in party.items():
if len(v) >= t:
for i in range(len(v) + 1 - t):
chosen.add(v[i][0])
total += v[i][1]
for i in range(n):
if len(chosen) + granted >= t:
break
if i not in chosen and candidates[i][0] != 1:
chosen.add(i)
total += candidates[i][1]
if result is None:
result = total
else:
result = min(result, total)
return result
def main():
n, m = map(int, input().split())
candidates = []
for i in range(n):
p, c = map(int, input().split())
candidates.append((p, c))
print(solve(m, candidates))
if __name__ == '__main__':
main()
``` | output | 1 | 70,734 | 10 | 141,469 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties — n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study — you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 ≤ n, m ≤ 3000) — the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 ≤ p_i ≤ m, 1 ≤ c_i ≤ 10^9) — the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number — the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party. | instruction | 0 | 70,735 | 10 | 141,470 |
Tags: brute force, greedy
Correct Solution:
```
import sys
f=sys.stdin
out=sys.stdout
n,m=map(int,f.readline().rstrip('\r\n').split())
cos={}
cost=[]
nvot=[0 for i in range(m+1)]
party=[[] for i in range(m+1)]
for i in range(n):
p,c=map(int,f.readline().rstrip('\r\n').split())
if p!=1:
if c in cos:
cos[c]+=1
else:
cos[c]=1
cost.append(c)
party[p].append(c)
nvot[p]+=1
cost.sort()
for i in party:
i.sort()
mi=float('inf')
for x in range(1,n+1):
dcos=dict(cos)
tmp=0
vot=nvot[1]
for j in range(2,m+1):
if nvot[j]>=x:
for k in range(nvot[j]-x+1):
vot+=1
tmp+=party[j][k]
dcos[party[j][k]]-=1
j=0
while vot<x:
if(dcos[cost[j]]>0):
dcos[cost[j]]-=1
tmp+=cost[j]
vot+=1
j+=1
#print(tmp)
mi=min(mi,tmp)
out.write(str(mi)+"\n")
``` | output | 1 | 70,735 | 10 | 141,471 |
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