2023.5.16

相信同学们对复数运算符重载已经相当熟悉啦，那今天那我们来看看分数又该如何处理呢?定义一个分数类FS，有私有成员分子fz，分母fm。另有公有成员函数FS operator + (const FS &f)对运算符“+”进行重载，实现两个分数相加。题目首先给出一个整型数n，紧跟着2n行输入，输入形如3z4m，代表分子为3，分母为4。其中分母不为0，输入时分母可以为负数，但输出时分母必须为正数。
要求对分数进行两两求和，并化简。(备注说明：分数为0时，表示成0z1m，如果结果为负数，那么分子取负数，分母为正数)

#include <iostream>
using namespace std;
void HJ(int &fz, int &fm)
{
    int gy = 1;
    while ((gy < fz) || (gy < fm))
    {
        gy++;
        if (((fz % gy) == 0) && ((fm % gy) == 0))
        {
            fz /= gy;
            fm /= gy;
        }
    }
    
}
class FS
{
    friend void operator >>(istream& cin, FS& fs);
public:
    FS operator+(FS& f)
    {
        if (this->fm < 0)
        {
            this->fm = -this->fm;
            this->fz = -this->fz;
        }
        if (f.fm < 0)
        {
            f.fm = -f.fm;
            f.fz = -f.fz;
        }
        if (this->fm == f.fm)
        {
            int fz = this->fz + f.fz;
            int fm = this->fm;
            HJ(fz,fm);
            cout << fz << "z" << fm << "m" << endl;
        }
        else if (this->fz == 0 || f.fz == 0)
        {
            cout << "0z1m" << endl;
        }
        else
        {
            int fz = (this->fz * f.fm) + (this->fm * f.fz);
            int fm = this->fm * f.fm;
            HJ(fz,fm);
            cout << fz << "z" << fm << "m" << endl;
        }
        return f;
    }
private:
    int fz;
    int fm;
};
void operator >>(istream& cin, FS& fs)
{
    char Z;
    char M;
    cin >> fs.fz >> Z >> fs.fm >> M;
}
void test01()
{
    int n = 0;
    cin >> n;
    FS fs[6];
    for (int i = 0; i < 2 * n; i++)
    {
        cin >> fs[i];
    }
    for (int j = 0; j < 2 * n; j += 2)
    {
        fs[j] + fs[j + 1];
    }
}
int main()
{
    test01();
    return 0;
}