这道编程题怎么做？本人小白。

Question

重载全部6个关系运算符，运算符对pounds成员进行比较，并返回一个bool值，编程，它声明一个包含6个Stonewt对象的数组，并在数组声明中初始化前3个对象。然后使用循环来读取用于设置剩余3个数组元素的值。接着报告最小的元素，最大的元素以及大于或等于11英石的元素的数量。

Answer 1

 #include <iostream>
using namespace std;

class Stonewt
{
private:
    int stone;
    double pounds;
public:
    Stonewt()
    {
        stone = 0;
        pounds = 0.0;
    }
    Stonewt(double lbs)
    {
        stone = (int)lbs / 14;
        pounds = lbs - ((int)lbs / 14 * 14);
    }
    Stonewt(int stn, double lbs)
    {
        stone = stn + (int)lbs / 14;
        pounds = lbs - ((int)lbs / 14 * 14);
    }
    void show()
    {
        cout << stone << "stn " << pounds << "lbs." << endl;
    }
    bool operator > (const Stonewt &s)
    {
        if (stone == s.stone)
            return pounds > s.pounds;
        else
            return stone > s.stone;
    }
    bool operator < (const Stonewt &s)
    {
        if (stone == s.stone)
            return pounds < s.pounds;
        else
            return stone < s.stone;
    }
    bool operator == (const Stonewt &s)
    {
        if (stone == s.stone)
            return pounds == s.pounds;
        else
            return false;
    }
    bool operator >= (const Stonewt &s)
    {
        return *this > s || *this == s;
    }
    bool operator <= (const Stonewt &s)
    {
        return *this < s || *this == s;
    }
    bool operator != (const Stonewt &s)
    {
        return !(*this == s);
    }
};

int main()
{
    Stonewt arr[6] = { Stonewt(28.5), Stonewt(50.1), Stonewt(1, 2.1) };
    arr[3] = Stonewt(1, 20.0);
    arr[4] = Stonewt(12, 0.4);
    arr[5] = Stonewt(12, 0.6);
    Stonewt max = arr[0];
    Stonewt min = arr[0];
    int n = 0;
    Stonewt c = Stonewt(11, 0.0);
    for (int i = 0; i < 6; i++)
    {
        arr[i].show();
        if (arr[i] > max) max = arr[i];
        if (arr[i] < min) min = arr[i];
        if (arr[i] >= c) n++;
    }
    cout << "max "; max.show();
    cout << "mix "; min.show();
    cout << "> 11 stone: " << n << endl;
    return 0;
}