HDU 1009 FatMouse' Trade

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 37 24 35 220 325 1824 1515 10-1 -1

 

Sample Output

13.33331.500
贪心   可以买一部份
#include <stdio.h>#include <algorithm>using namespace std;struct p{    double a,b;}bao[10000];bool cmp(p a,p b){    return a.a/a.b>b.a/b.b;}int main(){    int m,n,i;    double sum;    while(~scanf("%d %d",&m,&n))    {        if(m<0&&n<0)            break;        for(i=0;i<n;i++)            scanf("%lf %lf",&bao[i].a,&bao[i].b);        sort(bao,bao+n,cmp);//            for(i=0;i<n;i++)//                printf("%lf %lf\n",bao[i].a,bao[i].b);        sum=0;        for(i=0;i<n;i++)        {            if(m>bao[i].b)            {                m-=bao[i].b;                sum+=bao[i].a;            }            else            {                sum+=m/bao[i].b*bao[i].a;break;            }        }        printf("%.3lf\n",sum);    }    return 33;}