HDU-FatMouse' Trade （贪心）

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#HDU #FatMouse' Trade （贪心）

本文介绍了一个关于FatMouse如何使用猫粮交换JavaBeans的问题，并提供了一段C++代码实现。通过输入不同房间中JavaBeans和所需猫粮的比例，计算出FatMouse能够获得的最大JavaBeans数量。

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FatMouse' Trade

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

#include<bits/stdc++.h>
using namespace std;
struct sss
{
    int j,f;
    double fre;
}q[100000];
int cmp(sss x,sss y)
{
    return x.fre>y.fre;
}
int main()
{
    double sum;
    int n,m;
    while(~scanf("%d%d",&m,&n))
    {
        if(m==-1&&n==-1)
            break;
        for(int i=0;i<n;i++)
        {
            scanf("%d %d",&q[i].j,&q[i].f);
            q[i].fre=q[i].j*1.0/q[i].f;
        }
        sort(q,q+n,cmp);
        sum=0;
        for(int i=0;i<n;i++)
        {
            m=m-q[i].f;
            if(m>0)
            {
                sum+=q[i].j;
            }
            if(m<=0)
            {
                sum+=(m+q[i].f)*q[i].j*1.0/q[i].f;
                break;
            }
        }
        printf("%.3lf\n",sum);
    }
    return 0;
}