hdu 1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 31290    Accepted Submission(s): 10104

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

题意：有 m 的猫粮，和n中交换  f[i] 能换 j[i]

求出能交换得的最大值，

按他们交换比值进行排序

然后计算出最值

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std;

#define N 1005

typedef struct numb
{
    int j,f;
    double p;
    numb()
    {
        j = 0;
        f = 0;
        p = 0.0;
    }
}numb;

numb a[N];

bool fun(numb x,numb y)
{
    if(x.p - y.p > 0.000001)
      return true;
    else return false;
}
int main()
{
    int m,n;
    while(scanf("%d%d",&m,&n) && n != -1 && m != -1)
    {
       int i;
       for(i = 0; i < n; i++)
         {
             scanf("%d%d",&a[i].j,&a[i].f);
             a[i].p = 1.0*a[i].j/a[i].f;
         }
        sort(a,a+n,fun);
        /*for(i = 0; i < n; i++)
           printf("%lf  ",a[i].p);*/

        double sum = 0.0;
        int j;
        int k = 0;
        int flag = 0;
        for(i = 0; i < n; i++)
        {
            if(m >= a[i].f)
               {
                   sum += a[i].j;
                    m -= a[i].f;
                }
            else
             {
                 sum += a[i].p*m;
                 break;
             }
        }
        printf("%0.3lf\n",sum);
    }
    return 0;
}