HDOJ 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 66493    Accepted Submission(s): 22603

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

 

题目一开始没看懂= =……还以为是1个单位换1个单位（迷之英语水平）。结果是1套换1套……

比方说第一组是7 5，那么用5个可以换前面的7个。只要这点看懂了就好办了。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct change
{
	double j,f,p;
}a[1111];
bool cmp(change a,change b)
{
	return a.p>b.p;
}
int main()
{
	int m,n,i;
	while(scanf("%d%d",&m,&n))
	{
		if(m==n&&m==-1)
			return 0;
		double sum=0;
		for(i=0;i<n;i++)
		{
			scanf("%lf%lf",&a[i].j,&a[i].f);
			if(a[i].f!=0)
				a[i].p=a[i].j/a[i].f;
			else
			{
				sum+=a[i].j;
				a[i].p=0;
			}
		}
		sort(a,a+n,cmp);
		for(i=0;i<n;i++)
		{
			if(m==0)
				break;
			if(a[i].p==0)
				continue;
			if(a[i].f>m)
			{
				sum+=m*a[i].p;
				m=0;
			}
			else
			{
				sum+=a[i].j;
				m-=a[i].f;
			}
		}
		printf("%.3lf\n",sum);
	}
	return 0;
}