HDU-1009(贪心)

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 91424    Accepted Submission(s): 31633

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

• 先排序。排序依据是j/f。将其从大到小进行排序。后对数据进行处理。
• AC Code

#include <cstdio>
#include <cmath> 
#include <cstring>
#include <algorithm>
using namespace std;

struct Data
{
	double j;
	double f;
	double id;
};
bool cmp(Data x, Data y)
{
//	if(x.f < y.f) return 1;
//	else if(x.f == y.f) 
//	{
//		return x.j < y.j;
//	}
//	else return 0;
	return x.id > y.id;
}
int main()
{
	double m;
	int n;
	while(~scanf("%lf %d",&m ,&n))
	{
		if(m==-1 || n==-1) break;
		Data st[n+1];
		for(int i=1; i<=n; i++)
		{
			scanf("%lf %lf",&st[i].j,&st[i].f);
			st[i].id = st[i].j / st[i].f;
		}
		sort(st+1, st+1+n, cmp);
		double sum = 0;
		int i=0;
		while(m > 0)
		{
			i++;
			if(m > st[i].f)
			{
				m = m - st[i].f;
				sum = sum + st[i].j;
			}
			else
			{
				sum = sum + m / st[i].f * st[i].j;
				break;
			}		
		} 
		printf("%.3lf\n",sum);
		
		
		
	}	
	
	
	return 0;
}