分數四則運算
Time Limit: 1000 ms Memory Limit: 65536 KiB

Problem Description
編寫程式，實現兩個分數的加減法

Input
輸入包含多行資料；
每行資料是一個字串，格式是"a/boc/d"，其中a, b, c, d為數字（每個數字保證為正數並且不存在正號）。o是運算子"+“或者”-","*",""。
資料以EOF結束，輸入資料保證合法。

Output
直接輸出結果，並且注意結果應符合書寫習慣，沒有多餘的符號、分子、分母，並且化簡至最簡分數形式。

Sample Input
1/100+3/100
1/4-1/2
1/3-1/3
1/2*2/1
1/2\1/2

Sample Output
1/25
-1/4
0
1
1

AC程式碼：

import java.util.Scanner;

class Fs {
	int fz, fm;

	public Fs(int x, int y) {
		fz = x;
		fm = y;
	}

	public Fs add(Fs fs) {
		int Fz = fz * fs.fm + fm * fs.fz;
		int Fm = fm * fs.fm;
		return new Fs(Fz, Fm);
	}

	public Fs sub(Fs fs) {
		int Fz = fz * fs.fm - fm * fs.fz;
		int Fm = fm * fs.fm;
		return new Fs(Fz, Fm);
	}

	public Fs nul(Fs fs) {
		int Fz = fz * fs.fz;
		int Fm = fm * fs.fm;
		return new Fs(Fz, Fm);
	}

	public Fs div(Fs fs) {
		int Fz = fz * fs.fm;
		int Fm = fm * fs.fz;
		return new Fs(Fz, Fm);
	}

	public int gcd(int n, int m) {  //求最小公倍數
		int r = n % m;
		while (r != 0) {
			n = m;
			m = r;
			r = n % m;
		}
		return m;
	}

	public String toString() {
		int d = gcd(fz, fm);
		fz /= d;
		fm /= d;
		String str = fz * fm < 0 ? "-" : "";
		if (fz % fm == 0) {
			str += Math.abs(fz / fm);  //絕對值
		} else {
			str += Math.abs(fz) + "/" + Math.abs(fm);
		}
		return str;
	}
}

public class Main {
	public static void main(String[] args) {
		Scanner mi = new Scanner(System.in);
		while (mi.hasNext()) {
			String str = mi.nextLine();
			String[] p = str.split("\\D+");
			int a = Integer.parseInt(p[0]);
			int b = Integer.parseInt(p[1]);
			int c = Integer.parseInt(p[2]);
			int d = Integer.parseInt(p[3]);
			Fs fs = new Fs(a, b);
			Fs fs1 = new Fs(c, d);
			String[] q = str.split("\\d+");
			char op = q[2].charAt(0);
			switch (op) {
			case '+':
				fs = fs.add(fs1);
				break;
			case '-':
				fs = fs.sub(fs1);
				break;
			case '*':
				fs = fs.nul(fs1);
				break;
			case '\\':
				fs = fs.div(fs1);
				break;
			}
			System.out.println(fs);
		}
		mi.close();
	}
}
 

——————
餘生還請多多指教！