HDU1009 贪心思想

FatMouse' Trade

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  HDU 1009

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 
        
        
        
        

 

这个题的大意是：它有M猫食，N个房间，每个房间有f[i]猫食，j[i]量的 javabean，按一定的比例拿猫食来保护javabean。问一共有M猫食最多能保护多少javabean。

简单的贪心，只需按每个房间的javabean和猫食的比例从高到底排序，然后贪心即可。

代码中我处理了输入时f[i]等于0的情况，但交上去WA，我删去之后就AC了，看来没必要处理啊，，

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;

struct food
{
	int f,j;
	double ave;
}s[1010];

int cmp(const food &a,const food &b)
{
	return a.ave>b.ave;
}

int main()
{
	int i,k,n,m;
	double ans;
	while(scanf("%d%d",&m,&n))
	{
		if(m==-1 && n==-1)
			break;
		for(i=0;i<n;i++)
		{
			scanf("%d%d",&s[i].j,&s[i].f);
			s[i].ave=(double)s[i].j/s[i].f;
		}
		sort(s,s+n,cmp);
		ans=0;
		for(i=0;i<n;i++)
		{
			if(s[i].f<=m)
			{
				ans+=(double)s[i].j;
				m-=s[i].f;
			}
			else
			{
				ans+=s[i].ave*m;
				break;
			}
		}
		printf("%.3f\n",ans);
	}
	return 0;
}