HDU—1009—FatMouse' Trade

FatMouse' Trade

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  HDU 1009

Description

康康准备了 M 斤的食物, 准备跟舍长交换哲学之宝 ♂

舍长有 N 个房间. 第 i 个房间有 J[i] 的 ♂ 需要 F[i] 斤的食物. 康康可以不换完整个房间的♂ ,

他可以用 F[i]* a% 的食物 换 J[i]* a% 的 ♂ 

现在给你一个实数 M 问你康康最多能获得多少的 ♂

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 thatFatMouse 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 

#include<cstdio>
#include<algorithm>
using namespace std; 
struct node
{
    double key;
    double weight;
    double ave;
}peo[10001];
bool cmp(node x,node y)
{
    return x.ave>y.ave;
}
int main()
{
    int n;
    double m;
    while( scanf("%lf %d",&m,&n) )
    {    
        if( (m==-1) && (n==-1) )    break;
        for(int i=0;i<n;i++)
        {
            scanf("%lf %lf",&peo[i].key,&peo[i].weight);
            peo[i].ave=(double)peo[i].key/peo[i].weight;  //必须要加强转 
        }
        sort(peo,peo+n,cmp);
        double cnt=0;
        for(int i=0;i<n;i++)
        {
            if(m>=peo[i].weight)
            {
                cnt+=peo[i].key;
                m-=peo[i].weight;
            }
            else
            {
                cnt+=peo[i].ave*m;
                break;
            }
        }
        printf("%.3lf\n",cnt);
    }
    return 0;
}