Hangzhou Electric 1009 Fatmouse ' trade (greedy)

Fatmouse ' trade Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 44460 Accepted Submission (s): 14869


Problem Description Fatmouse prepared M pounds of cat food, ready-to-trade with the cats guarding the warehouse containing He favorite food, JavaBean.
The warehouse has N rooms. The i-th contains j[i] pounds of JavaBeans and requires f[i] pounds of cat food. Fatmouse does not has the to trade for all the JavaBeans in the the the same, instead, he may get j[i]* a% of pounds JavaBeans if he Pays f[i]* a% pounds of cat food. Here A is a real number. Now he's 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 is begins with a line containing the non-negative integers M and N. Then N lines follow, each contains, non-negative integers j[i] and f[i] respectively. The last test case was followed by Two-1 ' s. All integers is 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 Amou NT 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

/*
did not read test instructions
time:2014-10-19 */
#include <cstdio>
#include <cstring>
# Include<algorithm>
using namespace std;
#define MAX 1010
#define EPS 1e-7
struct warehouse{
	double f,j;
} H[max];
BOOL CMP (Warehouse A,warehouse b) {
	return a.f/a.j>=b.f/b.j;
}
int main () {
	int N;
	Double M;
	while (scanf ("%lf%d", &m,&n)!=eof) {
		if (n==-1&&m==-1) break;
		for (int i=0;i<n;i++) {
			scanf ("%lf%lf", &h[i]. F,&h[i]. J);
		}
		Sort (h,h+n,cmp);
		Double ans=0;
		for (int i=0;i<n;i++) {
			if (m>=h[i]. J) {
				m-=h[i]. J;
				Ans+=h[i]. F;
			} else{
				ans+= (M/h[i]. J) *h[i]. F);
				break;
			}
		}
		printf ("%.3lf\n", ans);
	}
return 0;