PAT A1034 有理數四則運算 (20分)

題目鏈接：https://pintia.cn/problem-sets/994805260223102976/problems/994805287624491008

題目描述
本題要求編寫程序，計算 2 個有理數的和、差、積、商。

輸入
輸入在一行中按照 a1/b1 a2/b2 的格式給出兩個分數形式的有理數，其中分子和分母全是整型範圍內的整數，負號只可能出現在分子前，分母不爲 0。

輸出
分別在 4 行中按照 有理數1 運算符 有理數2 = 結果 的格式順序輸出 2 個有理數的和、差、積、商。注意輸出的每個有理數必須是該有理數的最簡形式 k a/b，其中 k 是整數部分，a/b 是最簡分數部分；若爲負數，則須加括號；若除法分母爲 0，則輸出 Inf。題目保證正確的輸出中沒有超過整型範圍的整數。

樣例輸入
2/3 -4/2

樣例輸出
2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)

代碼

#include <cstdio>
#include <algorithm>
using namespace std;

typedef long long ll;

struct fraction {
	ll up, down;
};

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a % b);
}


fraction reduction(fraction result) {
	if(result.down < 0) {
		result.up = -result.up;
		result.down = -result.down;
	}
	if(result.up == 0)
		result.down = 1;
	else {
		int d = gcd(abs(result.up), abs(result.down));
		result.up /= d;
		result.down /= d;
	}
	return result;
}

fraction add (fraction f1, fraction f2) {
	fraction result;
	result.up = f1.up * f2.down + f2.up * f1.down;
	result.down = f1.down * f2.down;
	return reduction(result);
}

fraction minu (fraction f1, fraction f2) {
	fraction result;
	result.up = f1.up * f2.down - f2.up * f1.down;
	result.down = f1.down * f2.down;
	return reduction(result);
}

fraction multi (fraction f1, fraction f2) {
	fraction result;
	result.up = f1.up *  f2.up;
	result.down = f1.down * f2.down;
	return reduction(result);
}

fraction divide (fraction f1, fraction f2) {
	fraction result;
	result.up = f1.up * f2.down;
	result.down = f1.down * f2.up;
	return reduction(result);
}

void showResult(fraction r) {
	r = reduction(r);	
	if(r.up < 0)
		printf("(");				
	if(r.down == 1)
		printf("%lld", r.up);
	else if(abs(r.up) % r.down == 0)			
		printf("%lld", r.up /  r.down);
	else if(abs(r.up) > r.down)
		printf("%lld %lld/%lld", r.up / r.down, abs(r.up) % r.down, r.down);
	else
		printf("%lld/%lld", r.up, r.down);
	if(r.up < 0)
		printf(")");
}

int main() {
	fraction a, b;
	scanf("%lld/%lld %lld/%lld", &a.up, &a.down, &b.up, &b.down);

	showResult(a);
	printf(" + ");
	showResult(b);
	printf(" = ");
	showResult(add(a, b));
	printf("\n");

	showResult(a);
	printf(" - ");
	showResult(b);
	printf(" = ");
	showResult(minu(a, b));
	printf("\n");

	showResult(a);
	printf(" * ");
	showResult(b);
	printf(" = ");
	showResult(multi(a, b));
	printf("\n");

	showResult(a);
	printf(" / ");
	showResult(b);
	printf(" = ");
	if(b.up == 0)
		printf("Inf");
	else
		showResult(divide(a, b));
	printf("\n");
	
	return 0;
}