HDU 5190-Go to movies

最新推荐文章于 2019-08-24 12:41:41 发布

原创最新推荐文章于 2019-08-24 12:41:41 发布 · 405 阅读

0 ·

CC 4.0 BY-SA版权

ACM 专栏收录该内容

45 篇文章

订阅专栏

本博客介绍了一个利用贪心算法解决学生团体购票最小成本的问题。通过输入学生人数和影院提供的团购方案，程序帮助选择最经济的观影方案。案例分析了如何计算最少花费，适合初学者理解贪心算法的应用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Go to movies

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 259 Accepted Submission(s): 152

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.

mindicates 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

biyuan 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

3 2
2 2
3 5

Sample Output

4

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

小结：

贪心的水题....没什么需要注意的

以下是AC代码：

#include<stdio.h>
#include<string.h>
int main()
{
    int n,m,a,b,min;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        min=99999999;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&a,&b);
            int temp=n%a?(n/a+1):(n/a);
            min=temp*b<min?temp*b:min;
        }
        printf("%d\n",min);
    }
    return 0;
}