PTA 乙级 1034 有理数四则运算 (20分)

1034 有理数四则运算 (20分)

本题要求编写程序，计算 2 个有理数的和、差、积、商。

输入格式：

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。

输出格式：

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。

输入样例 1：

2/3 -4/2

输出样例 1：

2/3 + (-2) = (-1 1/3)
2/3 - (-2) = 2 2/3
2/3 * (-2) = (-1 1/3)
2/3 / (-2) = (-1/3)

输入样例 2：

5/3 0/6

输出样例 2：

1 2/3 + 0 = 1 2/3
1 2/3 - 0 = 1 2/3
1 2/3 * 0 = 0
1 2/3 / 0 = Inf

思路：

相加相减其实就是通分的步骤，然后通分完还要化简，化完后负数的时候要多括号， $0$为底的时候输出 $inf$。

#include<iostream>
#include<cmath>
using namespace std;
struct fraction{
	long long up;
	long long down;
};
long long gcd(long long a, long long b);
fraction add(fraction f1, fraction f2);
fraction minu(fraction f1, fraction f2);
fraction multi(fraction f1, fraction f2);
fraction divide(fraction f1, fraction f2);
fraction reduction(fraction result);
void showresult(fraction result);
int main()
{
	struct fraction f1, f2;
	scanf("%lld/%lld %lld/%lld", &f1.up, &f1.down, &f2.up, &f2.down);
	showresult(f1);
	cout << " + ";
	showresult(f2);
	cout << " = ";
	showresult(add(f1, f2));
	cout << "\n";
	showresult(f1);
	cout << " - ";
	showresult(f2);
	cout << " = ";
	showresult(minu(f1, f2));
	cout << "\n";
	showresult(f1);
	cout << " * ";
	showresult(f2);
	cout << " = ";
	showresult(multi(f1, f2));
	cout << "\n";
	showresult(f1);
	cout << " / ";
	showresult(f2);
	cout << " = ";
	if(f2.up == 0) cout << "Inf";
	else showresult(divide(f1, f2));
	cout << "\n";
	
} 
long long gcd(long long a, long long b)
{
	if(b == 0) return a;
	else return gcd(b, a % b);
}
fraction add(fraction f1, fraction f2)
{
	fraction result;
	result.up = f1.up * f2.down + f1.down * f2.up;
	result.down = f1.down * f2.down;
	return reduction(result);
}
fraction minu(fraction f1, fraction f2)
{
	fraction result;
	result.up = f1.up * f2.down - f1.down * f2.up;
	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);
}
fraction reduction(fraction result)
{
	if(result.down < 0){
		result.up = -result.up;
		result.down = -result.down;
	}
	if(result.up == 0) result.down = 1;
	else{
		long long d = gcd(abs(result.up), abs(result.down));
		result.up /= d;
		result.down /= d;
	}
	return result;
}
void showresult(fraction result)
{
	result = reduction(result);
	if(result.up < 0) cout << "(";
	if(result.down == 1) cout << result.up;
	else if(abs(result.up) > result.down) printf("%lld %lld/%lld", result.up / result.down, abs(result.up % result.down), result.down);
	else cout << result.up << "/" << result.down;
	if(result.up < 0) cout << ")";
	return ;
}