HDU：1009 FatMouse' Trade

FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 68716    Accepted Submission(s): 23432

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

题目代码：
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>

using namespace std;

struct infor
{
    double value;
    double cost;
    double aver;
}a[1004];
bool cmp(infor p1,infor p2)
{
    return p1.aver < p2.aver;
}
int main()
{
    int M,N;//M磅猫粮,N个房间
    double _max;
    while(~scanf("%d%d",&M,&N))
    {
        if(M < 0 && N < 0)
            break;
        for(int i = 0; i < N; i++)
        {
            scanf("%lf%lf",&a[i].value,&a[i].cost);
            a[i].aver = a[i].cost/a[i].value;
        }
        sort(a,a+N,cmp);//按单位价格进行排序
        _max = 0;
        for(int i = 0; i < N; i++)
        {
                if(M >= a[i].cost)
                {
                    _max += a[i].value;
                    M = M - a[i].cost;
                }
                else
                {
                    _max += (M*a[i].value)/a[i].cost;
                    break;
                }
        }
        printf("%.3lf\n",_max);
    }
    return 0;
}