hdu 5190 Go to movies

最新推荐文章于 2017-09-11 21:15:25 发布

黎辰

最新推荐文章于 2017-09-11 21:15:25 发布

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文章标签： C++

本文链接：https://blog.youkuaiyun.com/qq_24451605/article/details/44539365

Go to movies

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 277 Accepted Submission(s): 163

Problem Description

Winter holiday is coming!As the monitor, LeLe plans to go to the movies.
Because the winter holiday tickets are pretty expensive, LeLe decideds to try group-buying.

Input

There are multiple test cases, about

20 cases. The first line of input contains two integers

n,m(1≤n,m≤100) .

n indicates the number of the students.

m indicates how many cinemas have offered group-buying.

For the

m lines,each line contains two integers

ai,bi(1≤ai,bi≤100) , indicating the choices of the group buying cinemas offered which means you can use

bi yuan to buy

ai tickets in this cinema.

Output

For each case, please help LeLe **choose a cinema** which costs the least money. Output the total money LeLe should pay.

Sample Input

Sample Output

  
   4


   
    
     Hint
    LeLe can buy four tickets with four yuan in cinema 1.

Source

BestCoder Round #34

题目大意：好多个电影院，每个电影院有一个团购方案，问最小花费
题目分析：模拟下就行

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#define MAX 107

using namespace std;

int n,m;
int p[MAX],s[MAX];

int main ( )
{
    while ( ~scanf ( "%d%d" , &n , &m ) )
    {
        for ( int i = 1 ; i <= m ; i++ )
            scanf ( "%d%d" , &s[i] , &p[i] );
        int ans = -1;
        for ( int i = 1 ; i <= m ; i++ )
        {
            int temp = (n%s[i]?n/s[i]+1:n/s[i])*p[i];
            if ( ans == -1 ) ans = temp;
            else ans = min ( ans , temp );
        }
        printf ( "%d\n" , ans );
    }
}