hdu1009

 

                                                      FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 4555 Accepted Submission(s): 1354

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

 

刚开始用了一个比较普通的算法：1循环输入数据，2根据“好坏”排序数据，3得到最优解；验证数据正确，可能是这个算法太繁琐导致超时，遗憾不能ac

 

#include<stdio.h>
int main()
{
    int m,n,i,s;
    double j[1005],f[1005],t[1005],tmpt,tmpj,tmpf,cha,max=0;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        if(m<=0&&n<=0)
            break;
        max=0;
        for(i=0;i<n;i++){
            scanf("%lf%lf",&j[i],&f[i]);
            t[i]=j[i]/f[i];
        }
        for(i=0;i<n-1;i++)
            for(s=1;s<n;s++)
                if(t[i]<t[s])
                    {
                        tmpt=t[i];  tmpj=j[i];    tmpf=f[i];
                        t[i]=t[s];    j[i]=j[s];    f[i]=f[s];
                        t[s]=tmpt;    j[s]=tmpj;    f[s]=tmpf;
                    }
        
        i=0;
        while(m)
        {
            if(m>=f[i]){
            max +=j[i];
            m -=f[i];
            i++;
            }
            else{
            cha=m/f[i];
            max +=j[i]*cha;
            m=0;
                i++;
            }
            if(i>=n)
                break;
        }
        printf("%.3f\n",max);
        i=0;max=0;
        
    }
    return 0;
}