hdoj1009解题报告

Problem Description

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.

 

Input

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.

 

Output

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.

 

题意:很久之前做的,忘了,自己看吧

#include <iostream>
#include <iomanip>
#include<cstdio>
using namespace std;

int main()
{
    int m, n;
    double a[10001];
    int j[10001], f[10001];
    while (true)
    {
        cin >> m >> n;
        if (m == -1 && n == -1)break;
        for (int i = 0; i < n; i++)
        {
            cin >> j[i] >> f[i];
                a[i] = double(j[i]) / f[i];
        }
        double res = 0;
        while(m != 0 && n != 0)
        {
            double max = -1;
            int pos = 0;
            for (int i = 0; i < n; i++)
                if (max < a[i])
                {
                    max = a[i];
                    pos = i;
                }
            if (m <= f[pos])
            {
                res = res + m * a[pos];
                m = 0;
                break;
            }
            else
            {
                res = res + j[pos];
                m = m - f[pos];
                a[pos] = 0;
            }
        }
        printf("%.3lf\n",res);
    }
    return 0;
}