FatMouse' Trade

最新推荐文章于 2021-02-23 22:53:14 发布

林_沈

最新推荐文章于 2021-02-23 22:53:14 发布

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CC 4.0 BY-SA版权

文章标签：贪心

本文链接：https://blog.youkuaiyun.com/u013246445/article/details/19347511

题目：

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

这个问题和背包问题几乎一样，只是有点细微差别。

要注意的是，有可能某个房间里有JavaBean但是需要的猫粮是零，这种情况要考虑。

代码：

#include<stdio.h>
main()
{
	int i,j;
	int a[1100],b[1100],tempx;
	double sum;
	double c[1100],temp;
	int m,n;
	while(scanf("%d%d",&m,&n)&&(m!=-1&&n!=-1))
	{
		sum=0;
		for(i=0;i<n;i++)
		{
			scanf("%d%d",&a[i],&b[i]);
		}
		for(i=0;i<n;i++)
		{
			if(b[i]==0)
			{
				sum=sum+a[i];
				for(j=i;j<n-1;j++)
				{
					a[j]=a[j+1];
					b[j]=b[j+1];

				}
				n--;
				i--;
			}
		}

		for(i=0;i<n;i++)
		{
			c[i]=1.0*a[i]/b[i];
		}
		for(i=0;i<n;i++)
		{
			for(j=0;j<n-1;j++)
			{
				if(c[j]<c[j+1])
				{
					temp=c[j];
					c[j]=c[j+1];
					c[j+1]=temp;
					tempx=a[j];
					a[j]=a[j+1];
					a[j+1]=tempx;
					tempx=b[j];
					b[j]=b[j+1];
					b[j+1]=tempx;
				}
			}
		}
		
		for(i=0;i<n;i++)
		{
			if(m>b[i])
			{
				sum=sum+a[i];
				m=m-b[i];
			}
			else
			{
				sum=sum+m*c[i];
                break;
			}

		}

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

还有同学的一个：

#include<stdio.h>
#include<algorithm>
using namespace std;
struct node
{
	double a;
	double b;
	double c;
}s[1000];
bool cmp(node a,node b)
{
	return a.c>b.c;
}
int main()
{ 
   double p,q;
   int i,m,n;
  while(scanf("%d%d",&m,&n),(m!=-1)&&(n!=-1))
{
 p=0;
 q=0;
 for(i=0;i<n;i++)
{
	scanf("%lf%lf",&s[i].a,&s[i].b);
	s[i].c=s[i].a/s[i].b;
}
sort(s,s+n,cmp);
i=0;
while(p<m&&i<n)
{
	p+=s[i].b;
	q+=s[i].a;
	i++;
}
printf("%.3f\n",q-(p-m)*(s[i-1].a/s[i-1].b));

}
}