HDU 1009 FatMouse' Trade(贪心)

1009 FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19485    Accepted Submission(s): 6041

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

Author

CHEN, Yue

Source

ZJCPC2004

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解题报告：先按J[i]/F[i]从大到小排序，再贪心就可以了！

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
const int N =1010;
struct node
{
    double f;
    double j;
    double r;
}a[N];
int cmp(const void *a,const void *b)
{
    return (*(struct node *)a).r > (*(struct node *)b).r?1:-1;
}
int main()
{
    int m,n,i;
    double ans;
    while(scanf("%d%d",&m,&n)!=EOF)
    {
        memset(a,0,sizeof(a));
        if(m==-1&&n==-1)
        {
            break;
        }
        ans=0;
        for(i=0;i<n;i++)
        {
            scanf("%lf%lf",&a[i].f,&a[i].j);
            a[i].r=a[i].j/a[i].f;
        }
        qsort(a,n,sizeof(a[0]),cmp);
        for(i=0;i<n;i++)
        {
            if(m>=a[i].j)
            {
                ans+=a[i].f;
                m-=a[i].j;
            }
            else
            {
                ans+=(a[i].f/a[i].j)*m;
                break;
            }
        }
        printf("%.3lf\n",ans);
    }
    return 0;
}