杭电 acm 1009 FatMouse' Trade

FatMouse' Trade

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

Total Submission(s): 48150    Accepted Submission(s): 16204

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  

//一开始看不懂题意，其实就是有n个猫住的小房子；n列数据即可理解为价位表，
//如几个猫食可以兑换几个豆子，用老鼠持有的猫食总数m从兑换率最高的房间开始依次兑换：贪心思想；
#include<iostream>
#include<algorithm>
using namespace std;

struct food
{
	int beans;
	int catfood;
	double rate;
};

bool cmp(const food a,const food b)//对结构中某一数据进行排序；
{
	return a.rate>b.rate;
}

int main()
{
	int m,n,i;
	double s;

	while(cin>>m>>n)
	{	
		if(m==-1&&n==-1)
			break;
		food *p=new food[n];
		s=0;
		for(i=0;i<n;i++)
		
		{
			cin>>p[i].beans>>p[i].catfood;
			p[i].rate=(double)p[i].beans/p[i].catfood;
		}
	
	sort(p,p+n,cmp);
	for(i=0;i<n;i++)
	{
		if(p[i].catfood<m)//所需猫食小于原持有则执行
		{
			m-=p[i].catfood;
			s+=p[i].beans;
		}
		else
		{
			s+=p[i].rate*m;
			break;//不要忘记break否则 循环继续，继续兑换豆子；
		}

	}
	printf("%.3lf\n",s);
	delete []p;
	}
	return 0;
}