习题:分别用成员函数和友元函数重载运算符,使对整型的运算符=、+、-、*、适用于分数运算。要求:(1)输出结果是最简分数(可以是带分数);(2)分母为1,只输出分子。

习题:分别用成员函数和友元函数重载运算符,使对整型的运算符=、+、-、*、适用于分数运算。要求:(1)输出结果是最简分数(可以是带分数);(2)分母为1,只输出分子。,第1张

习题:分别用成员函数和友元函数重载运算符,使对整型的运算符=、+、-、*、/适用于分数运算。要求:(1)输出结果是最简分数(可以是带分数);(2)分母为1,只输出分子。

个人答案:

#include 
using namespace std;
class fenshu {
public:
    fenshu(int n = 0, int d = 0): num(n), den(d) {}//分子n,分母d
    fenshu operator+(fenshu& c2);
    fenshu operator-(fenshu& c2);
    fenshu operator*(fenshu& c2);
    fenshu operator/(fenshu& c2);
    int gys(int a, int b);
    void display()const {
        if (den == 1)
            cout << num << endl;
        else
            cout << num << "/" << den << endl;
    }
private:
    int num, den;    //分子分母
};
int fenshu::gys(int a, int b) {
    int c;
    while (b != 0) {
        c = a % b;
        a = b;
        b = c;
    }
    return a > 0 ? a : -a;
}
fenshu fenshu::operator+(fenshu& c2) //加法的重载
{
    int d3, num3, g;
    d3 = den * c2.den;
    num3 = num * c2.den + c2.num * den;
    g = gys(d3, num3);
    d3 = d3 / g;
    num3 = num3 / g;
    return fenshu(num3, d3);
}
fenshu fenshu::operator-(fenshu& c2) //减法的重载
{
    int d3, num3, g;
    d3 = den * c2.den;
    num3 = num * c2.den - c2.num * den;
    g = gys(d3, num3);
    d3 = d3 / g;
    num3 = num3 / g;
    return fenshu(num3, d3);
}
fenshu fenshu::operator*(fenshu& c2) //乘法的重载
{
    int d3, num3, g;
    d3 = den * c2.den;
    num3 = num * c2.num;
    g = gys(d3, num3);
    d3 = d3 / g;
    num3 = num3 / g;
    return fenshu(num3, d3);
}
fenshu fenshu::operator/(fenshu& c2) //除法的重载
{
    int d3, num3, g;
    d3 = den * c2.num;
    num3 = num * c2.den;
    g = gys(d3, num3);
    d3 = d3 / g;
    num3 = num3 / g;
    return fenshu(num3, d3);
}
int main()
{
    int d1, n1, d2, n2;
    char c;
    cout << "Input x: ";
    cin >> d1 >> c >> n1;
    cout << "Input y: ";
    cin >> d2 >> c >> n2;
    fenshu c1(d1, n1), c2(d2, n2), c3;
    c3 = c1 + c2;
    cout << "x+y=";
    c3.display();
    c3 = c1 - c2;
    cout << "x-y=";
    c3.display();
    c3 = c1 * c2;
    cout << "x*y=";
    c3.display();
    c3 = c1 / c2;
    cout << "x/y=";
    c3.display();
    system("pause");
    return 0;
}

结果:

 

 

欢迎分享,转载请注明来源:内存溢出

原文地址:https://54852.com/zaji/5635563.html

(0)
打赏 微信扫一扫微信扫一扫 支付宝扫一扫支付宝扫一扫
上一篇 2022-12-16
下一篇2022-12-16

发表评论

登录后才能评论

评论列表(0条)

    保存