Hdoj 1009.FatMouse' Trade 题解

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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思路

题目的大意是有\(J[i]\)个豆子，要得到的话交\(F[i]\)个猫粮，给你m份猫粮，问你如何最大化受益

显然是根据\(J[i]/F[i]\)的高低排序，也就是我们通俗意义上的性价比

代码

#include<bits/stdc++.h>
using namespace std;
struct node
{
	double j;	//豆子
	double f;	//猫粮
	double single; 		//豆子/猫粮
} a[1010];
int main()
{
	int m,n;
	while(cin>>m>>n)
	{
		if(m==-1 && n==-1) break;
		for(int i=1;i<=n;i++)
		{
			cin >> a[i].j >> a[i].f;
			a[i].single = a[i].j/a[i].f;
		}	//读入
		sort(a+1,a+n+1,[=](node x,node y) -> bool {return x.single > y.single;});//lamba表达式作为cmp函数

		double ans = 0.0;
		for(int i=1;i<=n;i++)
		{
			if(m>a[i].f)
			{
				ans += a[i].j;
				m -= a[i].f;
			}else
			{
				ans += m/a[i].f*a[i].j;
				break;	//不够只能按百分比买
			}
		}
		printf("%.3lf\n",ans);
	}
	return 0;
}