学步园

A fat mouse prepared M pounds of cat food,ready to trade with the cats guarding the warehouse containing his

favorite food:peanut. The warehouse has N rooms.The ith room containsW[i] pounds of peanut and requires 

F[i] pounds of cat food. Fatmouse does not have to trade for all the peanut in the room,instead,he may get 

 W[i]*a% pounds of peanut if he pays F[i]*a% pounds of cat food.The mouse is a stupid mouse,so can you tell 

him the maximum amount of peanut 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 W[i] and F[i] respectively.
The test case is terminated by two -1. 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 penaut that FatMouse can obtain.

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

上传者

TC_李远航

思路：翻译：胖老鼠准备英镑的猫粮,准备与猫的贸易保护仓库包含他最喜欢的食物:花生。仓库有N个房间。第i个房间containsW[我]磅的花生,需要F[我]磅的猫粮。Fatmouse不必对所有房间里的花生贸易,相反,他可能会得到W[我]*
%磅的花生如果他支付F[我]* %磅的猫粮。鼠标是一个愚蠢的鼠标,所以你能告诉他可以获得最大的花生。

”先把性价比高的保住，嗯，就是这样的“ 耗子如是说=-=

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
#include<cmath>
using namespace std;

struct mouse
{
    int w,f;
    double rate;
}p[1005];

int cmp(mouse a,mouse b)
{
    return a.rate>=b.rate;
}

int main()
{
    int m,n;
    while(scanf("%d%d",&m,&n))
    {
        if(m==-1&&n==-1)
        break;

        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&p[i].w,&p[i].f);
            p[i].rate=p[i].w*1.0/p[i].f;
        }

        sort(p,p+n,cmp);

//        for(int i=0;i<n;i++)
//        {
//            printf("%d %d %lf\n",p[i].w,p[i].f,p[i].rate);
//        }

        double sum=0;
        for(int i=0;m>0&&i<n;i++)
        {
            if(p[i].f<=m)
            {
                sum+=p[i].w;
                m-=p[i].f;
            }
            else
            {
                sum += m*p[i].rate;
//                cout<<"**"<<p[i].rate<<endl;;
                break;
            }
        }
        printf("%.3lf\n",sum);
    }
}