九度oj 1433

题目描述：

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. 

输入：

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.

输出：

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.

样例输入：

5 37 24 35 220 325 1824 1515 10-1 -1

样例输出：

13.333
31.500

#include<iostream>#include<algorithm>#include<stdio.h>using namespace std;struct good{       double j;       double f;       double s;       bool operator <(const good &a)const       {            return s>a.s;            }            }b[1000];            int main()            {                double m;                int n;                while(scanf("%lf%d",&m,&n)!=EOF)                {                                if(m==-1&&n==-1){                                break;}                                for(int i=0;i<n;i++)                                {                                        scanf("%lf%lf",&b[i].j,&b[i].f);                                        b[i].s=b[i].j/b[i].f;                                        }                                        sort(b,b+n);                                        int ins=0;                                        double ans=0;                                        while(m>0&&ins<n)                                        {                                                         if(m>b[ins].f)                                                         {                                                                       ans+=b[ins].j;                                                                       m-=b[ins].f;                                                                                                                                              }                                                                       else{                                                                       ans+=b[ins].j*m/b[ins].f;                                                                       m=0;}                                                                       ins++;                                                                       }                                                                       printf("%.3lf\n",ans);                                                                       }                                                                       }