C/C++_2019_8_10（有理数运算 ）

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题目描述

Rational Arithmetic(20) | 时间限制：1秒 | 内存限制：32768K
For two rational numbers, your task is to implement the basic arithmetics, that is, to calculate their sum, difference,product and quotient.

输入描述

Each input file contains one test case, which gives in one line the two rational numbers in the format “a1/b1 a2 / b2”.The numerators and the denominators are all in the range of long int.Ifthere is a negative sign, it mustappear only in front of the numerator.The denominators are guaranteed to be non - zero numbers.

输出描述

For each test case, print in 4 lines the sum, difference, product and quotient of the two rational numbers,respectively.The format of each line is “number1 operator number2 = result”.Notice that all the rational numbers must be in their simplest form “k a/b”, where k is the integer part, and a / b is the simplest fraction part.If the number is negative, it must be included in a pair of parentheses.If the denominator in the division is zero, output “Inf” as the result.It is guaranteed that all the output integers are in the range of long int.

示例

输入
5 / 3 0 / 6
输出
1 2 / 3 + 0 = 1 2 / 3
1 2 / 3 - 0 = 1 2 / 3
1 2 / 3 * 0 = 0
1 2 / 3 / 0 = Inf
Rational Arithmetic

题目翻译

24505 - 有理数运算
实现对两个有理数的基本运算，包括加、减、乘、除。

输入描述

每个输入文件只包含一个测试用例，测试用例会给出一行数据，格式为“a1 / b1 a2 / b2”分子分母的范围都在长整型的范围内，如果数字为负，则符号只会出现在分子的前面。分母一定是非零数。

输出描述

针对每个测试用例，都输出四行，分别是这两个有理数的和、差、积和商，格式为“数1 操作符 数2 = 结果”。注意，所有的有理数都将遵循一个简单形式“k a / b”，其中k是整数部分，a / b是最简分数形式，如果该数为负数，则必须用括号包起来。如果除法中的除数为0，则输出“Inf”。结果中所有的整数都在long int的范围内。

题目解析

本题看上去不难，但是存在几个问题：
1、除数为0，这个很好解决，做个判断即可。
2、负数的输出，这个只要一个标签即可。
3、题目中虽然没有明说，但是这个数字处理后其实是有可能不存在分数部分或者整数部分的。也就是说将数据处理完形成k a / b的格式后，有可能只有一个k，也可能只有一个a / b，也有可能两者皆有，所以要分别考虑这几种情况。

解题思路

可以尝试实现一个有理数类，将数据处理后重载一下加减乘除即可。处理数据的方法就是除一下mod一下的问题，加减乘除遵循基本的分数加减乘除原则，最后求一下最大公约数，做一下约分，再处理一下数据，就OK了。

示例代码

#include 
#include 
//long long是两个关键字拼起来的，用起来很不方便，重命名一下
typedef long long _sint64;
//有理数类的声明
class RationalNumber{
	bool m_infinate; //处理除数为零
	bool m_negative; //处理负数
	_sint64 m_numerator; //分子，方便输出
	_sint64 m_denominator; //分母
	_sint64 m_integer; //整数部分
	_sint64 m_numeratorAll; //记录无整数分数的分子，方便进行运算
	_sint64 calcGCD(_sint64 a, _sint64 b); //求最大公约数的函数
public:
	RationalNumber(_sint64 numerator, _sint64 denominator); //构造函数
	RationalNumber operator+(RationalNumber const& o) const; //四则运算重载
	RationalNumber operator-(RationalNumber const& o) const;
	RationalNumber operator*(RationalNumber const& o) const;
	RationalNumber operator/(RationalNumber const& o) const;
	//输出流运算符重载
	friend std::ostream &operator<<(std::ostream &os, RationalNumber const& o);
};
//有理数类每个方法的实现
_sint64 RationalNumber::calcGCD(_sint64 a, _sint64 b)
{
	if (b == 0)
	{
		return a;
	}
	//辗转相除法
	return calcGCD(b, a % b);
}
RationalNumber::RationalNumber(_sint64 numerator, _sint64 denominator)
{
	m_negative = false;
	m_infinate = false;
	//处理分母为零的情况
	if (denominator == 0)
	{
		m_infinate = true;
		return;
	}
	//这里这样写，是因为在通过计算结果进行构造过程中，有可能出现分子分母均为负的情况。
	if (numerator < 0)
	{
		m_negative = !m_negative;
	}
	if (denominator < 0)
	{
		m_negative = !m_negative;
	}
	//计算整数、分子、分母。其中分母要参与下面的运算，所以不能是负的，用abs取绝对值，分子要保留原值
	m_integer = numerator / denominator;
	m_numerator = numerator - m_integer * denominator;
	m_denominator = abs(denominator);
	//约分，注意传给子函数的分子必须是正的，分母上面处理过了
	if (m_numerator)
	{
		_sint64 maxtmp = calcGCD(abs(m_numerator), m_denominator);
		if (maxtmp)
		{
			m_numerator /= maxtmp;
			m_denominator /= maxtmp;
		}
	}
	//计算约分后假分数版的分子，因为后续运算是不需要整数部分的，所以必须用假分数的分子算。
	m_numeratorAll = m_numerator + m_integer * m_denominator;
}
//以下为分数的加减乘除，统统使用m_numeratorAll（假分数的分子）进行运算。
RationalNumber RationalNumber::operator+(RationalNumber const& o) const
{
	_sint64 numerator = (m_numeratorAll * o.m_denominator) +
		(o.m_numeratorAll * m_denominator);
	_sint64 denominator = m_denominator * o.m_denominator;
	return RationalNumber(numerator, denominator);
}
RationalNumber RationalNumber::operator-(RationalNumber const& o) const
{
	_sint64 numerator = (m_numeratorAll * o.m_denominator) -
		(o.m_numeratorAll * m_denominator);
	_sint64 denominator = m_denominator * o.m_denominator;
	return RationalNumber(numerator, denominator);
}
RationalNumber RationalNumber::operator*(RationalNumber const& o) const
{
	_sint64 numerator = m_numeratorAll * o.m_numeratorAll;
	_sint64 denominator = m_denominator * o.m_denominator;
	return RationalNumber(numerator, denominator);
}
RationalNumber RationalNumber::operator/(RationalNumber const& o) const
{
	_sint64 numerator = m_numeratorAll * o.m_denominator;
	_sint64 denominator = m_denominator * o.m_numeratorAll;
	return RationalNumber(numerator, denominator);
}
std::ostream &operator<<(std::ostream &os, RationalNumber const& o)
{
	//分母为0的情况就不用继续了
	if (o.m_infinate)
	{
		os << "Inf";
		return os;
	}
	//整数和分子为0那干脆就是0了
	if (o.m_numerator == 0 && o.m_integer == 0)
	{
		os << "0";
		return os;
	}
	//负数打印括号和负号
	if (o.m_negative)
	{
		os << "(-";
	}
	//有整数就打整数
	if (o.m_integer)
	{
		os << abs(o.m_integer);
		if (o.m_numerator) //整数小数都有就打个空格隔开
		{
			os << " ";
		}
	}
	//有分数就打分数，分母已经abs过了，这里可以不用
	if (o.m_numerator)
	{
		os << abs(o.m_numerator) << '/' << o.m_denominator;
	}
	//负数的后半边括号
	if (o.m_negative)
	{
		os << ")";
	}
	return os;
}
int main()
{
	_sint64 n1, d1, n2, d2;
	scanf("%lld/%lld %lld/%lld", &n1, &d1, &n2, &d2);
	RationalNumber rn1(n1, d1), rn2(n2, d2);
	//轻松+愉快的使用函数时间
	std::cout << rn1 << " + " << rn2 << " = " << rn1 + rn2 << '\n';
	std::cout << rn1 << " - " << rn2 << " = " << rn1 - rn2 << '\n';
	std::cout << rn1 << " * " << rn2 << " = " << rn1 * rn2 << '\n';
	std::cout << rn1 << " / " << rn2 << " = " << rn1 / rn2 << '\n';
	return 0;
}