HDU 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25744    Accepted Submission(s): 8172

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

  
  

   
   5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1
  
  

 

Sample Output

  
  

   
   13.333
31.500
  
  

 

Author

CHEN, Yue

 

Source

ZJCPC2004

 

Recommend

JGShining

 

分析：对J,F按J/F大小排序即可，注意数据N=0时，要输出0.000

代码：

#include<stdio.h>

typedef struct node{
    int j;
    int f;
}Node;

Node a[1010];

int main()
{
    int m,n;
    Node temp;
    while(scanf("%d%d",&m,&n))
    {
        if(m==-1&&n==-1)
            break;
        if(n==0)
        {
            printf("0.000\n");
            continue;
        }

        int i,k;
        double sum=0;
        for(i=0;i<n;i++)
            scanf("%d%d",&a[i].f,&a[i].j);
        for(i=0;i<n;i++)
        {
            double s1,s2;
            for(k=i;k<n;k++)
            {    
                s1=a[i].f*1.0/a[i].j;
                s2=a[k].f*1.0/a[k].j;
                if(s1<s2)
                {
                    temp=a[i];
                    a[i]=a[k];
                    a[k]=temp;
                }
            }
        }
        i=0;
        while(m>=a[i].j)
        {
            sum+=a[i].f;
            m-=a[i].j;
            i++;
        }
        sum+=((double)m/a[i].j*a[i].f);
        printf("%.3lf\n",sum);
    }
    return 0;
}