HDU_1009——老鼠的交易,性价比排序,最大化收益

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

/*
算性价比j[i]/f[i],然后按性价比对每个房间的数据重新排序
再累加最大化兑换比例的老鼠粮 
*/
#include <cstdio>
#include <cstdlib>
const int MAX = 1005;
/*
int compare(const void *a,const void *b)
{
    //return *(double*)a - *(double*)b; //小——大 
    return *(double*)b - *(double*)a;    //大——小 
}
*/
int main()
{
    int m,n,j[MAX]={0},f[MAX]={0};
    while(~scanf("%d%d",&m,&n))
        {
            if(m==-1 && n==-1)
                break;
            double j_f[MAX]={0},temp=0;
            for(int i=1;i<=n;i++)
                {
                    scanf("%d%d",&j[i],&f[i]);
                    j_f[i]=(double)j[i]/f[i];
                }
            //快速排序：数组首地址，元素个数，一个元素大小，指向比较函数的指针 
            //qsort(j_f+1,n,sizeof(double),compare);
            //好吧，写到一半发现排序不能这样用- -排序函数还是留着吧- -
            for(int i=1;i<=n-1;i++)
                {
                    for(int k=i+1;k<=n;k++)
                        {
                            if(j_f[i]<j_f[k])
                                {
                                    temp=j_f[k];
                                    j_f[k]=j_f[i];
                                    j_f[i]=temp;
                                    
                                    temp=j[k];
                                    j[k]=j[i];
                                    j[i]=(int)temp;
                                    
                                    temp=f[k];
                                    f[k]=f[i];
                                    f[i]=(int)temp;
                                }
                        }
                }
            double ans=0;
            for(int i=0;i<=n;i++)
                {
                    if(m>=f[i])
                        {
                            ans+=j[i];
                            m-=f[i];
                        }
                    else
                        {
                            ans+=m*j_f[i];
                            break;
                        }
                }
            printf("%.3lf\n",ans);
        }    
    return 0;
}