軟件開發-這道編程題怎麼做？本人小白。

這道編程題怎麼做？本人小白。

重載全部6個關系運算符，運算符對pounds成員進行比較，並返回一個bool值，編程，它聲明一個包含6個Stonewt對象的數組，並在數組聲明中初始化前3個對象。然後使用循環來讀取用於設置剩余3個數組元素的值。接著報告最小的元素，最大的元素以及大於或等於11英石的元素的數量。

最佳回答：

 #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;
}