SDJZU-FatMouse' Trade

http://sdjzu.acmclub.com/index.php?app=problem_title&assignment_id=1016&problem_id=2134

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. 

输入格式

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

分析：

#include <iostream>#include <algorithm>#include <iomanip>using namespace std;struct cat{    double food;    double java;    double avg;};bool cmp(cat a,cat b){    return a.avg>b.avg;}int main(){    int n,m,i;    struct cat ti[1005];    while (cin>>m>>n)    {        if(m==-1&&n==-1)break;        for(i=0; i<n; i++)        {            cin>>ti[i].java>>ti[i].food;            ti[i].avg=ti[i].java/ti[i].food;        }        sort(ti,ti+n,cmp);        double count=0;        double total=m;        for(i=0; i<n; i++)        {            if(total-ti[i].food>=0)//如果全部能装下            {                count+=ti[i].java;                total-=ti[i].food;            }            else            {                count+=total*ti[i].avg;                break;            }        }        cout<<fixed<<setprecision(3)<<count<<endl;    }    return 0;}