A - FatMouse' Trade

题目如下：

FatMouse prepared M pounds of cat food, ready to trade with thecats guarding the warehouse containing his favorite food, JavaBean. 
Thewarehouse has N rooms. The i-th room contains Jii pounds of JavaBeans and requires Fii pounds of cat food. FatMouse does not have to trade for all theJavaBeans in the room, instead, he may get Jii*a% pounds of JavaBeans if he pays Fii 。

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输入如下：

The input consists of multiple test cases. Each test casebegins with a line containing two non-negative integers M and N. Then N linesfollow, each contains two non-negative integers Jii and Fii respectively. The last test case isfollowed by two -1's. All integers are not greater than 1000. 

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输出如下：

For each test case, print in a single linea real number accurate up to 3 decimal places, which is the maximum amount ofJavaBeans that FatMouse can obtain. 

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思路就是简单的贪心算法就可以了。

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代码如下：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>

using namespace std;

struct node
{
	int m;
	int n;
};

bool cmp(node n,node m)
{
	return n.n*1.0/n.m>m.n*1.0/m.m;
}

struct node a[1010];
int n;
int m;

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