jobdu 21 FatMouse

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

#include <stdio.h>
#include <algorithm>
using namespace std;
struct goods
{
	int j;
	int f;
	double s;
	bool operator < (const goods &A) const
	{
		return s>A.s;
	}
}buf[1000];
int main()
{
	int m,n;
	while(scanf("%d%d",&m,&n)!=EOF)
	{
		if(m==-1&&n==-1)
			break;
		else
		{
			for(int i=0;i<n;i++)
			{
				scanf("%d%d",&buf[i].j,&buf[i].f);
				buf[i].s=buf[i].j*1.0/buf[i].f;
			}
			sort(buf,buf+n);
			double ans=0;
			int index=0;
			while(m>0 && index<n)
			{
				if(m>buf[index].f)
				{
					m-=buf[index].f;
					ans+=buf[index].j;
					index++;
				}
				else
				{
					ans+=1.0*buf[index].j/buf[index].f*m;
					m=0;
				}
			}
			printf("%.3lf\n",ans);
		}
	}
	return 0;
}