設計一個分數類。要求：
1.分數中包含的分數運算有：兩個分數的加、減、乘、除運算。
2.分數的輸出格式是：“分子/分母”。
3.編寫一個測試程式進行測試。

C++版本請複製連結檢視http://blog.csdn.net/bee0_0/article/details/78106442package Cal;

package Cal;

import java.util.Scanner;

public class Fraction {
    private int numerator1;
    private int denominator1;
    private int numerator2;
    private int 
 
denominator2;
    private int numerator3;
    private int denominator3;

    void set(int numerator1, int denominator1, int numerator2, int denominator2) {
        this.denominator1 = denominator1;
        this.denominator2 = denominator2;
        this.numerator1 = numerator1;
        this.numerator2 
 
= numerator2;
    }

    void multiply() {
        //乘法運算
numerator3 = numerator1 * numerator2;
        //結果分子 = 分子1 * 分子2
denominator3 = denominator1 * denominator2;
        //結果分母 = 分母1 * 分母2
output(numerator1, denominator1, numerator2,
                denominator2, numerator3, denominator3, '*');
    }

    void 
 
divided() {
        //除法運算
numerator3 = numerator1 * denominator2;
        denominator3 = denominator1 * numerator2;
        output(numerator1, denominator1, numerator2,
                denominator2, numerator3, denominator3, '/');
    }

    void plus() {
        int a, b, c;
        a = denominator1;
        b = denominator2;
        //使用相減法求最大公約數
if (a == b) {
            denominator3 = a;
            numerator3 = numerator1 + numerator2;
        } else {
            while (a != b) {
                if (a > b) a = a - b;
                else b = b - a;
            }
            int t;//最小公倍數 = t
t = (denominator1 * denominator2) / a;
            //最小公倍數 =（ 分母1 * 分母2 ）/最大公約數
denominator3 = t;//結果的分母 = 最小公倍數
numerator3 = ((t / denominator1)*denominator1) + ((t / denominator2)*denominator2);
            //結果的分子 = （最小公倍數/分母1*分母2+ （最小公倍數/分母2*分子2
}
        output(numerator1, denominator1, numerator2,
                denominator2, numerator3, denominator3, '+');
    }

    void minus() {
        int a, b, c;
        //使用相減法求最大公約數
double p, q;
        a = denominator1;
        b = denominator2;
        if (a == b) {
            a = denominator3;
            //當分母相同時分子相見應 大 - 小
if (numerator1 > numerator2) {
                numerator3 = numerator1 - numerator2;
            } else {
                numerator3 = numerator2 - numerator1;
            }
        } else {
            while (a != b) {
                if (a > b) a = a - b;
                else b = b - a;
            }
            int t;//令最小公倍數 = t
t = (denominator1 * denominator2) / a;
            //最小公倍數 =（ 分母1 * 分母2 ）/最大公約數
denominator3 = t;
            //結果的分母 = 最小公倍數
p = (double) numerator1 / (double) denominator1;
            q = (double) numerator2 / (double) denominator2;
            if ( p > q) {
                numerator3 = ((t / denominator1)*numerator1)- ((t / denominator2)*numerator2);
                output(numerator1, denominator1, numerator2,
                        denominator2, numerator3, denominator3, '-');
            } else {
                numerator3 = ((t / denominator2)*numerator2)- ((t / denominator1)*numerator1);
                output(numerator1, denominator1, numerator2,
                        denominator2, numerator3, -denominator3, '-');
            }
        }
    }

    public static void main(String[] arges) {
        Fraction a = new Fraction();//建立名為a的物件
Scanner in = new Scanner(System.in);
        int x, y;//x = numerator1; y = denominator1;
int m, n;//m = numerator2; n = denominator2;
System.out.println("請輸入第一個數的分子和分母 :");
        x = in.nextInt();
        y = in.nextInt();
        System.out.println("請輸入第二個數的分子和分母 :");
        m = in.nextInt();
        n = in.nextInt();
        a.set(x, y, m, n);
        //將x，y，m，n通過set函式分別賦給 :
        //numerator1 denominator1 numerator2 denominator2
a.multiply();
        a.divided();
        a.plus();
        a.minus();
    }

    void output(int a, int b, int c, int d, int e, int f, char name) {
        System.out.println(a + "/" + b + " " + name + " " + c + "/" + d + "=" + e + "/" + f);
        //    a   /   b     "+ - * /"    c   /   d    =    e   /   f
}
};