FatMouse' Trade

题目连接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=40692#problem/A

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. 

样例输入样例输入

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

 样例输出

 13.33331.500 

 贪心思路：先交换性价比大的，在交换性价比

#include<stdio.h>#include<algorithm>using namespace std;struct mouse{    double j,f;    double cent;}a[1100];bool cmp(mouse x,mouse y){    return (x.cent>y.cent);}int main(){    int n,m,i;    double sum;    while(scanf("%d %d",&m,&n)!=EOF)    {        sum=0;        if(m==-1 && n==-1)        break;        else        {            for(i=0;i<n;i++)            {                scanf("%lf %lf",&a[i].j,&a[i].f);                a[i].cent=a[i].j/a[i].f;            }            sort(a,a+n,cmp);            for(i=0;i<n;i++)            {                if(a[i].f<=m)                {                    sum=sum+a[i].j;                    m=m-a[i].f;                }                else                {                    sum=sum+a[i].cent*m;                    break;                }            }        }        printf("%.3f\n",sum);    }    return 0;}