1009 FatMouse' Trade

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FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 47416 Accepted Submission(s): 15987

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

Source

ZJCPC2004

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思路：首先求出J\F的值，根据JF大小排序，有技巧J，F，JF，是一体的，所以采用冒泡，要换三个都换。然后在JF从大到小往下取。

详解代码：

#include<iostream>
#include<cstdio>
using namespace std;
int J[1005],F[1005];
double JF[1005];
int main() 
{
    int i,j;
    int m,n;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        if(m==-1&&n==-1)exit(0);
        else
      {
        for(i=1;i<=n;i++)
        {
            scanf("%d",&J[i]);
            scanf("%d",&F[i]);
            JF[i]=J[i]*0.1/F[i]*10;
        }
        for(i=1;i<=n;i++)
         for(j=1;j<=n-i;j++)
           if(JF[j+1]>JF[j])
            {
            
                JF[0]=JF[j+1];
                J[0]=J[j+1];
                F[0]=F[j+1];
                
                JF[j+1]=JF[j];
                J[j+1]=J[j];
                F[j+1]=F[j];
                
                JF[j]=JF[0];
                J[j]=J[0];
                F[j]=F[0];
            }
        
            double sum=0;
            int coun=m;
            int m1;
        
            
            for(i=1;i<=n;i++)
            {
                m=m-F[i];
                
                if(m>=0)
                {
                    sum+=J[i];
                
                }
                else
                {
                m1=m+F[i]; 
                sum=sum+JF[i]*m1;
               
                break;
                }
            }
            printf("%0.3lf\n",sum);
        }
    }
    return 0;
    
}