FatMouse' Trade

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. 

InputThe 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. 
OutputFor 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. 
Sample Input

5 3
7 2
4 3
5 2
20 3
25 18
24 15
15 10
-1 -1

Sample Output

13.333
31.500

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
struct kang
{
    double a, b;
};
bool cmp(const kang&a, const kang&b)
{
    return 1.0*a.a / a.b > 1.0*b.a / b.b;
}
int main()
{
    kang k[10000];
    int m, n;
    
    while (cin >> m >> n,n == -1 && m == -1)
    {
        double sum=0;
        for (int i = 0; i < n; i++)
            cin >> k[i].a >> k[i].b;
        sort(k, k + n , cmp);
        for (int i = 0; i < n&&m != 0; i++)
        {
            if (m - k[i].b >= 0)
            {
                sum += k[i].a;
                m -= k[i].b;    
            }
            else
            {
                sum += m * 1.0*k[i].a / k[i].b;
                m = 0;
            }
        }
        printf("%.3f\n", sum);
    }
    return 0;
}

简单的贪心题，自定义一个cmp就看解决，不过在用sort时注意区间为左闭右开；

也可以使用符号重载：

bool operator < (const kang & a, const kang & b)
{
return a.a / a.b > b.a / b.b;
}（已测试）