FatMouse‘ Trade

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

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

按照性价比最高的排序，然后求和就行，如果m小于F[i],那么只取这个的a%,而a%=m/F[i].

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
struct point
{
	double x,y;
}a[2000];
bool cmp(point xx,point yy)
{
	if((xx.x/xx.y)>(yy.x/yy.y))
		return true;
	return false;
}
int main()
{
	int n;
	double m;
	while(~scanf("%lf%d",&m,&n))
	{
		memset(a,0,sizeof(a));
		if(m==-1&&n==-1)
			break;
		for(int i=0;i<n;i++)
		{
			scanf("%lf%lf",&a[i].x,&a[i].y);
		}
		sort(a,a+n,cmp);
		double sum=0;
		for(int i=0;i<n;i++)
		{
			if(a[i].y<=m)
			{
				m-=a[i].y;
				sum+=a[i].x;
			}
			else if(m>0)
			{
				sum+=a[i].x*(m/a[i].y);
				m-=a[i].x*(m/a[i].y);
			}
		}
		printf("%.3lf\n",sum);
	}
	return 0;
}