HDU 1009 FatMouse' Trade

                     FatMouse' Trade

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
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<stdio.h>#include<stdlib.h>#include<algorithm>using namespace std;const int MAXN = 1010;struct node{    double j,f;    double r;} a[MAXN];bool cmp(node a,node b){    return a.r  >  b.r;}int main(){    int N;    double M;    double ans;    while(scanf("%lf%d",&M,&N))    {        if(M==-1&&N==-1) break;        for(int i=0; i<N; i++)        {            scanf("%lf%lf",&a[i].j,&a[i].f);            a[i].r=(double)a[i].j/a[i].f;        }        sort(a,a+N,cmp);        ans=0;        for(int i=0; i<N; i++)        {            if(M>=a[i].f)            {                ans+=a[i].j;                M-=a[i].f;            }            else            {                ans+=(a[i].j/a[i].f)*M;                break;            }        }        printf("%.3lf\n",ans);    }    return 0;}