Greedy Mouse

Greedy Mouse

时间限制：1000 ms | 内存限制：65535 KB

难度：3

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

贪心题目，但是WA了好几次，后来看别人的代码，发现别人把剩余的最后的那部分m也放在了for循环里，后来改了改AC了

#include<iostream>
#include<iomanip>
#include<stdio.h>
#include<ios>
#include<algorithm>
using namespace std;
struct mouse{
	double x,y,z;
}a[1000];
bool cmp(mouse m,mouse n){
	return m.z>n.z;
}
int main()
{
	int   n;
	double m;
	while(cin>>m>>n){
		if(m==-1&&n==-1)
		  return 0;
        for(int i =0;i!=n;i ++){
        	cin>>a[i].x>>a[i].y;
        	a[i].z=a[i].x/a[i].y;
        }
       sort(a,a+n,cmp);
       double sum=0;
       int j=0;
       for(int j=0;j!=n;j++){
       	   if(m>a[j].y){
       	      sum+=a[j].x ;
       	      m-=a[j].y ;
       	  }else{
       	  	  sum+=(m*a[j].z );
       	  	  break;
       	  }
       	   
       }
       cout<<setiosflags(ios::fixed)<<setprecision(3)<<sum<<endl;
	}
	return 0;
}