HDOJ 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 67166    Accepted Submission(s): 22860

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

 简单贪心

#include<stdio.h>
#include<algorithm>
using namespace std;
struct ROOM
{
	double j;
	double f;
	double p;
}rom[10100];
double cmp(ROOM a,ROOM b)
{
	return a.p>b.p;
}
int main()
{
	double m,num;
	int n;
	while(scanf("%lf%d",&m,&n)&&m>=0&&n>=0)
	{
		for(int i=0;i<n;i++)
		{
		scanf("%lf%lf",&rom[i].j,&rom[i].f);
		rom[i].p=rom[i].j/rom[i].f;
		}
		sort(rom,rom+n,cmp);
		num=0;
		for(int i=0;i<n;i++)
		{
			if(rom[i].f<=m)
			{
			num=num+rom[i].j;
			m=m-rom[i].f;
			}
			else
			{
				num=num+m*rom[i].p;
				//printf("%.3lf\n",num);
				break;
			}
		}
			printf("%.3lf\n",num);
	}
	return 0;
}