PAT A1034 有理数四则运算 (20分)

题目链接：https://pintia.cn/problem-sets/994805260223102976/problems/994805287624491008

题目描述
本题要求编写程序，计算 2 个有理数的和、差、积、商。

输入
输入在一行中按照 a1/b1 a2/b2 的格式给出两个分数形式的有理数，其中分子和分母全是整型范围内的整数，负号只可能出现在分子前，分母不为 0。

输出
分别在 4 行中按照 有理数1 运算符 有理数2 = 结果 的格式顺序输出 2 个有理数的和、差、积、商。注意输出的每个有理数必须是该有理数的最简形式 k a/b，其中 k 是整数部分，a/b 是最简分数部分；若为负数，则须加括号；若除法分母为 0，则输出 Inf。题目保证正确的输出中没有超过整型范围的整数。

样例输入
2/3 -4/2

样例输出
2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)

代码

#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;
}