FatMouse' Trade(Hdu 1009)

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
题解：贪心求解，每一次选择性价比最高的那一个；

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct app
{
    double a;
    double b;
    double p;
};
bool cmp(struct app a,struct app b)
{
    return a.p>b.p;
}
int main()
{
    int p,i,j,m,u;
    double a,k,pp;
    struct app b[1111];
    for(;scanf("%d%d",&p,&m)!=-1;)
    {
        if(p==-1&&m==-1)
            break;
        for(i=0;i<m;i++)
        {
            scanf("%lf%lf",&b[i].a,&b[i].b);
            b[i].p=b[i].a/b[i].b;
        }
        sort(b,b+m,cmp);
        u=0;pp=0.0;
        for(i=0;i<m;i++)
        {
            pp+=b[i].a;
            u+=b[i].b;
            if(u>p)
            {
            pp=(pp-b[i].a)+(p-u+b[i].b)*b[i].p;
            break;
            }
        }
        printf("%.3lf\n",pp);
    }
    return 0;
}