FatMouse （贪心算法）

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

#include 
#include 
#include 
using namespace std;

int M,N; //M-catfood N-room
struct node{
	int j;  //javabean
	int f;  //cat food
	double rate;  //两者比例 
}a[1002];

double cmp(node a,node b){
	return a.rate>b.rate;
}

int main(){
	int i,j;
	double sum;
	
	while(cin>>M>>N){
		sum=0;
		if(M==-1 && N==-1)
			break;
		for(i=0;i>a[i].j>>a[i].f;
			a[i].rate=a[i].j*1.0/a[i].f*1.0;
		}
		sort(a,a+N,cmp);
		for(i=0;i=a[i].f){
				M-=a[i].f;
				sum+=a[i].j;
			}
			else if(M0){
				sum+=M*a[i].rate;  //所需javabean/M=a[i].j/a[i].f 
				break;
			}
		}
		cout<