HDU 1009 FatMouse' Trade (贪心）

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 66476    Accepted Submission(s): 22593

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
  

 

Author

CHEN, Yue

 

Source

ZJCPC2004

 

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 题意：

康康准备了 M 斤的食物, 准备跟舍长交换哲学之宝 

舍长有 N 个房间. 第 i 个房间有 J[i] 的 需要 F[i] 斤的食物. 康康可以不换完整个房间的 ,

他可以用 F[i]* a% 的食物 换 J[i]* a% 的 

现在给你一个实数 M 问你康康最多能获得多少的 

#include<string.h>
#include<stdio.h>
#include<algorithm>
using namespace std;
struct node
{
	double a,b,c;
}t[11000];
bool cmp(node x,node y)
{
	if(x.c==y.c)
	return x.a>y.a;
	return x.c>y.c;
}
int main()
{
	int n,m,i,j,k;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		if(n==-1&&m==-1)
		break;
		for(i=0;i<m;i++)
		{
			scanf("%lf%lf",&t[i].a,&t[i].b);
			t[i].c=t[i].a/t[i].b;
		}
		sort(t,t+m,cmp);
		double sum=0.0;
		for(i=0;i<m;i++)
		{
			if(n>=t[i].b)
			{
				sum+=t[i].a;
				n=n-t[i].b;
			}
			else
			{
				sum+=n*t[i].c;
				n=0;
			}
		}
		printf("%.3lf\n",sum);
	}
	return 0;
}