问题 E: FatMouse's Trade

题目描述

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.

 

输入

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.

 

输出

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.

 

样例输入

4 2
4 7
1 3
5 5
4 8
3 8
1 2
2 5
2 4
-1 -1

样例输出

2.286
2.500

实现代码：

#include<cstdio>
#include<algorithm>
using namespace std;
struct trade{
	double food;
	double bean;
	double percent;//豆子与猫食的比值 
}room[1010];
bool cmp(trade a,trade b){
	return a.percent>b.percent;
}

int main()
{
  int M,N;
  while(scanf("%d%d",&M,&N),M!=-1&&N!=-1)
  {
  	double max=0;
  	for(int i=0;i<N;i++)
  	{
  		scanf("%lf%lf",&room[i].bean,&room[i].food);
  		room[i].percent=room[i].bean/room[i].food;
	  }
	  sort(room,room+N,cmp);//依据比值排序 
	  for(int i=0;i<N;i++)
	  {
	    if(M>room[i].food) 
	    {
		  M-=room[i].food;
		  max+=room[i].bean;
		} 
		else//猫食小于指定房间的猫食量时，依据比值所得豆子 
		{
			max+=room[i].percent*M;
			break;
		}
	  }
	  printf("%.3lf\n",max);
  }	
  return 0;
}

 

结果如下：

注意：

此题是典型的贪心问题，应当以JavaBean/catfood的比值进行排序后进而算的最大的JavaBean的所得量。