FatMouse' Trade

Description

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.

Input

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.

Output

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.

Sample Input

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

Sample Output

13.333
31.500




题目意思：大老鼠一共有m磅的猫粮要和n个房间里面的猫做交易来换取他喜欢吃的食物，他不需要和所有的房间都进行交易，他支出F[i]，会得到J[i],
问他能得到的最大食物数。
解题思路：这算是我接触到的第一道贪心策略的题目，我用结构体来记录每个房间需要交易的猫粮和能换到的食物，以及兑换比率，贪心策略就是每次选择兑换比率大的来兑换
就能够得到最多兑换食物量。



上代码：

#include<stdio.h>
#include<algorithm>
using namespace std;
struct food
{
    int j;
    int f;
    double q;
};
int compare(food a,food b)
{
    if(a.q>b.q)
        return 1;
    else
        return 0;
}
int main()
{
    int n,m,i,k,l;
    double s;
    struct food a[1000],t;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        if(m==-1||n==-1)
            break;
        for(i=0; i<n; i++)
        {
            scanf("%d%d",&a[i].j,&a[i].f);
            a[i].q=a[i].j*1.0/a[i].f;
        }
        sort(a,a+n,compare);
        s=0.0;
        for(i=0; i<n; i++)
        {
            if(m>=a[i].f)
            {
                s=s+a[i].j;
                m=m-a[i].f;
            }
            else if(m<a[i].f&&m>=0)
            {
                s=s+(m*a[i].q);
                break;
            }
        }
        printf("%.3lf\n",s);
    }
    return 0;
}