题目描述
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.
输入描述:
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.

输出描述:
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.

示例1
输入
5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1
输出
13.333
31.500

解题思路：典型的贪心算法，我之前反复抄了很多遍还是不会做，这次竟然一次性AC.

//贪心 
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std; 
struct E{
    int J;
    int F;
    double value; //性价比 
    bool operator < (const E &b)const{
        return value > b.value;
    }
}E[1000];

int main(){
    int m,n;
    while(scanf("%d%d",&m,&n) != EOF){
        if(m == -1 && n == -1) break;
        for(int i = 0;i < n;i++){
            scanf("%d%d",&E[i].J,&E[i].F);
            E[i].value = E[i].J*1.0/E[i].F;
        }
        sort(E,E+n);
        double sum=0;
        for(int i = 0;i < n;i++){
                if(m > E[i].F){
                    sum += E[i].J;
                    m -= E[i].F;
                }else{
                    sum += E[i].value * m;
                    break; 
                }
        }
        printf("%.3f\n",sum);
    }
    return 0;
}