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：

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：

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

题目的意思是：有m元钱，有n种物品，每种物品有J磅，总价值f元，可以使用0到f的任意价格购买相应磅的物品。要求输出用m元钱最多能买到多少磅的物品。

输入第一行是两个数字m和n，分别代表有m元和n种物品，接下来n行分被代表每种物品的重量和价值。输入m和n为-1时程序结束。

输出为买到物品的重量。

思路：这个题就是典型的贪心算法，现在再看这个算法，其实的确十分厉害。后悔当初没有好好学习。

思路就是每次都购买剩下物品种性价比最高的物品，直到物品被买完或者钱用光了。

代码如下：

#include <iostream>
#include<algorithm>
#include<string.h>
#include<stdio.h>

using namespace std;

struct goods
{
    double j; // 重量
    double f; // 价值
    double s; // 性价比
    bool operator < (const goods &A) const
    {
        return s > A.s;
    }
}buf[1000];

int main()
{
    double m;
    int n;
    int i;
    while(scanf("%lf%d", &m, &n)!=EOF)
    {
        if(m==-1 && n==-1 ) break;
        for(i=0; i<n; i++)
        {
            scanf("%lf %lf", &buf[i].j, &buf[i].f);
            buf[i].s = buf[i].j / buf[i].f;
        }
        sort(buf, buf+n);
        int idx=0; //当前货物下标
        double ans = 0; //总重量
        while(m>0 && idx<n)  //循环条件是既有物品剩余还有钱剩余
        {
            if(m>buf[idx].f)
            {
                ans += buf[idx].j;
                m -= buf[idx].f;
            }
            else
            {
                ans += buf[idx].j * m / buf[idx].f;  //要理解这一步的含义
                m=0;
            }
            idx ++;
        }
        printf("%.3lf\n", ans);
    }
    return 0;
}