HDU 1009 FatMouse' Trade

最新推荐文章于 2021-05-09 21:32:11 发布

LZK1997

最新推荐文章于 2021-05-09 21:32:11 发布

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解决FatMouse如何通过最优交易策略，利用有限的猫粮换取仓库中最大数量的JavaBeans。采用贪心算法，按单位猫粮可换得JavaBeans的比例排序进行交易。

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FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 67262 Accepted Submission(s): 22899

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

Sample Output

13.333
31.500

Author

CHEN, Yue

典型的贪心算法

用sort排序一下就好了；

#include <stdio.h>
#include <iostream>
#include <algorithm>
using namespace std;

struct asd{
	int j;
	int f;
	double ave;
}mou[3005];
bool cmp(asd a,asd b)
{
	return a.ave>b.ave;
}
int main()
{
	int m,n;
	
	while(~scanf("%d%d",&m,&n)&&(m!=-1||n!=-1))
	{
		int i;
		for(i=0;i<n;i++)
		{
			scanf("%d%d",&mou[i].j,&mou[i].f);
			mou[i].ave=(1.0*mou[i].j)/(1.0*mou[i].f);
		}
		sort(mou,mou+n,cmp);
		
		double sum=0;
		for(i=0;i<n;i++)
		{

            if(m>mou[i].f)
			{
				sum+=mou[i].j;
				m=m-mou[i].f;
			}

            else
			{
				sum+=mou[i].ave*m;
				m=0;

                break;

            }
		}
		printf("%.3lf\n",sum);
	}
	
	return 0;
}