HDU 1009 FatMouse' Trade

Problem 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大意：老鼠准备了M磅猫食，准备拿这些猫食跟猫交换自己喜欢的食物。有N个房间，每个房间里面都有食物。你可以得到J[i]但你需要付出F[i]的猫食。要你计算你有M磅猫食可以获得最多食物的重量。思路：贪心算法，求最优解。将J[i]/F[i]的值从大到小排列，总是先取最大的，就能保证能够得出的最大值。代码：Language:C++
#include<cstdio>
#include<algorithm>
using namespace std;
struct node
{
	double get,pay,rate;
};
bool cmp(node a,node b)
{
    return b.rate<a.rate;
}
int main()
{
	node data[1001];
	int line,i;
	double ans,food;
	while(scanf("%lf%d",&food,&line))
	{
		ans=0;
		if(food==-1&&line==-1)break;
		for(i=0;i<line;i++)
		{
			scanf("%lf%lf",&data[i].get,&data[i].pay);
			data[i].rate=data[i].get/data[i].pay;
		}
		sort(data,data+line,cmp);
		for(i=0;i<line;i++)
		{
			if(food>=data[i].pay)
			{
				ans+=data[i].get;
				food-=data[i].pay;
			}
			else
			{
				ans+=(double)food*data[i].rate;
				break;
			}
		}
		printf("%.3lf\n",ans);
	}
	return 0;
}