HDU-FatMouse' Trade （贪心）

心已赠人 2022-05-26 04:22 187阅读 0赞

## FatMouse' Trade ##

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### 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;
    }