FatMouse' Trade

「爱情、让人受尽委屈。」 2021-06-24 13:58 189阅读 0赞

题目连接：[http://acm.hust.edu.cn/vjudge/contest/view.action?cid=40692\#problem/A][http_acm.hust.edu.cn_vjudge_contest_view.action_cid_40692_problem_A]

描述

FatMouse prepared M pounds of cat food, ready to trade with the cats guarding the warehouse containing his favorite food, JavaBean.    
The warehouse has N rooms. The i-th room contains J\[i\] pounds of JavaBeans and requires F\[i\] pounds of cat food. FatMouse does not have to trade for all the JavaBeans in the room, instead, he may get J\[i\]\* a% pounds of JavaBeans if he pays F\[i\]\* a% pounds of cat food. Here a is a real number. Now he is assigning this homework to you: tell him the maximum amount of JavaBeans he can obtain.

输入

The input consists of multiple test cases. Each test case begins with a line containing two non-negative integers M and N. Then N lines follow, each contains two non-negative integers J\[i\] and F\[i\] respectively. The last test case is followed by two -1's. All integers are not greater than 1000.

输出

For each test case, print in a single line a real number accurate up to 3 decimal places, which is the maximum amount of JavaBeans that FatMouse can obtain.

样例输入 样例输入

5 3
    7 2
    4 3
    5 2
    20 3
    25 18
    24 15
    15 10
    -1 -1

样例输出

13.333
    31.500

贪心思路：先交换性价比大的，在交换性价比

#include<stdio.h>
    #include<algorithm>
    using namespace std;
    struct mouse
    {
        double j,f;
        double cent;
    }a[1100];
    bool cmp(mouse x,mouse y)
    {
        return (x.cent>y.cent);
    }
    
    int main()
    {
        int n,m,i;
        double sum;
        while(scanf("%d %d",&m,&n)!=EOF)
        {
            sum=0;
            if(m==-1 && n==-1)
            break;
    
            else
            {
                for(i=0;i<n;i++)
                {
                    scanf("%lf %lf",&a[i].j,&a[i].f);
                    a[i].cent=a[i].j/a[i].f;
                }
                sort(a,a+n,cmp);
                for(i=0;i<n;i++)
                {
                    if(a[i].f<=m)
                    {
                        sum=sum+a[i].j;
                        m=m-a[i].f;
                    }
                    else
                    {
                        sum=sum+a[i].cent*m;
                        break;
                    }
                }
    
            }
            printf("%.3f\n",sum);
    
        }
        return 0;
    }

[http_acm.hust.edu.cn_vjudge_contest_view.action_cid_40692_problem_A]: http://acm.hust.edu.cn/vjudge/contest/view.action?cid=40692#problem/A