HDU 1009: fatmouse 'trade (simple greedy)

Fatmouse 'trade

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 41982 accepted submission (s): 13962

Problem descriptionfatmouse 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.
 
Inputthe 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.
 
Outputfor 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

The question is that mice use cat food for mouse food... (Orz ).. Calculate the maximum number of replicas .. A greedy question ..

#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<iostream>#include<vector>#include<queue>#include<cmath>#define f1(i, n) for(int i=0; i<n; i++)#define f2(i, n) for(int i=1; i<=n; i++)using namespace std;const int M = 1005;int n, m;double ans;double t;struct node{    double J;    double F;    double c;}Q[M];int cmp (node a, node b){    return a.c > b.c;}int main(){    while(scanf("%d%d", &n, &m)!=EOF)    {        ans = 0;        t = (double) n;        memset(Q, 0, sizeof(Q));        if(n==-1 && m==-1)            break;        f1(i, m)           {               scanf("%lf%lf", &Q[i].J, &Q[i].F);               Q[i].c = Q[i].J / Q[i].F;           }        sort(Q, Q+m, cmp);        f1(i, m)        {            if( Q[i].F<=t )                {                    ans += Q[i].J;                     t -= Q[i].F;                }            else            {                ans += t * Q[i].c;                break;            }        }        printf("%.3lf\n", ans);    }    return 0;}