FatMouse's Trade

问题 E: FatMouse's Trade

时间限制: 1 Sec  内存限制: 32 MB

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

4 2

4 7

1 3

5 5

4 8

3 8

1 2

2 5

2 4

-1 -1

样例输出

2.286

2.500

#include<stdio.h>#include<algorithm>using namespace std;struct list{    double value;    double j;    double f;};bool cmp(struct list a,struct list b){    return a.value>b.value;}int main(){    int n;    double m;    while(scanf("%lf%d",&m,&n),m!=-1||n!=-1)    {        struct list an[1005];        double sum=0;        for(int i=0;i<n;i++)        {            double j,f;            scanf("%lf%lf",&an[i].j,&an[i].f);            an[i].value=an[i].j/an[i].f;        }        sort(an,an+n,cmp);        for(int i=0;i<n;i++)        {            if(m>=an[i].f)            {                m-=an[i].f;                sum+=an[i].j;            }            else            {                sum+=m*an[i].value;                m=0;            }            if(m==0)                break;        }        printf("%.3f\n",sum);    }    return 0;}