HDU-1009-FatMouse' Trade（贪心）

最新推荐文章于 2024-07-19 11:00:00 发布

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在一场涉及多种资源交换的游戏中，FatMouse试图通过最优策略，利用有限的猫粮换取尽可能多的JavaBeans。本篇详细解析了如何通过算法确定最大收益，包括输入输出样例与实现代码。

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

Sample Output

13.333
31.500

思路：按照性价比排好序之后，然后看是否能得到即可

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>

using namespace std;

struct node
{
	int a,b;
	double val;
}p[10005];
bool cmp(node x,node y)
{
	return x.val>y.val;
}
int main()
{
	int n,m;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		if(n==-1&&m==-1)
		{
			break;
		}
		for(int t=0;t<m;t++)
		{
			scanf("%d%d",&p[t].a,&p[t].b);
			p[t].val=p[t].a*1.0/(p[t].b*1.0);
		}
		sort(p,p+m,cmp);
		double s1=0;
		for(int t=0;t<m;t++)
		{
			if(n>=p[t].b)
			{
			n-=p[t].b;
			s1+=p[t].a;
		   }
		   else
		   {
		   	s1+=(p[t].a*1.0/p[t].b)*n;
		   	break;
		   }
		}
		printf("%.3f\n",s1);
	}
	
	return 0;
}