A - FatMouse' Trade-优快云博客

探讨了FatMouse如何通过最优策略用猫粮交换仓库中的JavaBean，利用算法解决资源分配问题，实现最大收益。

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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 

#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
struct p
{
	double a;
	double b;
	double c;
};
bool compare(p A,p B)
{
	return A.c>B.c;
}
int main()
{
	int m,n,i,j;
	while(scanf("%d%d",&m,&n)!=EOF)
	{
		if(m==-1&&n==-1)
			break;
	    p t[1001];
        for(i=0;i<n;i++)
		{
			cin>>t[i].a>>t[i].b;
			t[i].c=t[i].a/t[i].b;
		}
		sort(t,t+n,compare);	
		double count=0;
		for(i=0;i<n;i++)
		{
			count+=t[i].b;
			if(count>m)
				break;
		}
		double sum=0;
		//if(i==0)
			//printf("%.3lf\n",t[0].c*m);
        if(count<m)
		{
			for(j=0;j<n;j++)
				sum+=t[j].a;
			printf("%.3lf\n",sum);
		}
		else
		{
			for(j=0;j<i;j++)
			{
				sum+=t[j].a;
				m-=t[j].b;
			}
			printf("%.3lf\n",sum+m*t[i].c);
		}
	}
	return 0;
}

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
	int m,n;
	int i,j;
	double a[1002],b[1002];
	double c[1002];
	while(scanf("%d%d",&m,&n)!=EOF)
	{
		if(m==-1&&n==-1)
			break;
		for(i=0;i<n;i++)
		{
			cin>>a[i]>>b[i];
			c[i]=a[i]/b[i];
		}
		for(i=0;i<n;i++)
		{
			for(j=0;j<n-i-1;j++)
			{
				if(c[j+1]>c[j])
				{
					swap(b[j],b[j+1]);
					swap(a[j],a[j+1]);
					swap(c[j],c[j+1]);
				}
			}
		}
		double count=0;
		for(i=0;i<n;i++)
		{
			count+=b[i];
			if(count>m)
				break;
		}
		double sum=0;
		if(count<m)
		{
			for(j=0;j<n;j++)
				sum+=a[j];
			printf("%.3lf\n",sum);
		}
		else
		{
			for(j=0;j<i;j++)
			{
				sum+=a[j];
				m-=b[j];
			}
			printf("%.3lf\n",sum+m*c[i]);
		}
	}
	return 0;
}

转载于:https://www.cnblogs.com/NYNU-ACM/p/4237306.html