描述

A fat mouse prepared M pounds of cat food,ready to trade with the cats guarding the warehouse containing his

favorite food:peanut. The warehouse has N rooms.The ith room containsW[i] pounds of peanut and requires 

F[i] pounds of cat food. Fatmouse does not have to trade for all the peanut in the room,instead,he may get 

 W[i]*a% pounds of peanut if he pays F[i]*a% pounds of cat food.The mouse is a stupid mouse,so can you tell 

him the maximum amount of peanut 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 W[i] and F[i] respectively. The test case is terminated by two -1. 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 penaut 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：猫粮换花生。有n个仓库，每个仓库有w的花生，需要用f的猫粮交换。可以按照比例换，问m的猫粮最多能换多少花生。

思路：用猫粮换花生。猫粮是背包容量，w是价值，f是重量，物品可以分割。那么就是尽可能放单位重量价值最大的物品。按照w/f进行排序，从单位价值最大开始装，直到背包装满或者物品装完。

#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
struct Room{
	double w,f;
}r[10000];
bool cmp(Room r1,Room r2){
	return (r1.w/r1.f)>(r2.w/r2.f);
}
int m,n;
int main(){
	while(scanf("%d%d",&m,&n)!=EOF&&(m!=-1&&n!=-1)){
		for(int i=0 ;i<n ;i++){
			scanf("%lf%lf",&r[i].w,&r[i].f);
		}
		sort(r,r+n,cmp);
		double ans = 0;
		int i=0;
		while(m&&i<n){//背包装完或者物品装完 
			if(r[i].f<=m){
				ans += r[i].w;
				m-=r[i].f;
			}
			else{
				ans += m*(r[i].w/r[i].f);
				m = 0;
			}
			i++;
			
		}

		printf("%.3lf\n",ans);
		
	}
	
	return 0;
}