Hangzhou Electric 1009-fatmouse ' trade (greedy)

Fatmouse ' tradeTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 54937 Accepted Submission (s): 18422

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

This is a simple greedy, first calculated that the transaction is the most worthwhile, the first to do that, until their own capital of 0

AC Code:

#include <cstdio>
#include <algorithm>
using namespace std;
struct node 
{
	int v,w;
	Double q;
} S[10000];
BOOL CMP (node X,node y)
{
	return x.q>y.q;
}
int main ()
{
	int i,n,m;
	Double a[10000];
	while (scanf ("%d%d", &m,&n), m!=-1| | N!=-1)
	{
		for (i=0;i<n;i++)
		{
			scanf ("%d%d", &S[I].V,&S[I].W);
			S[I].Q=S[I].V*1.0/S[I].W;
		}
		Sort (s,s+n,cmp);
		Double sum=0;
		for (i=0;i<n;i++)
		{
			if (s[i].w<=m)
			{
				sum+=s[i].v;
				M-=S[I].W;
			}
			else
			{
				sum+=m*s[i].q;
				break;
			}
		}
		printf ("%.3lf\n", sum);
	}
	return 0;