PAT甲级——1088 Rational Arithmetic——分数 20

题目

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解法1

点击查看代码

#include<iostream>
#include<cmath>
using namespace std;

struct Fraction {
    long long num, den;
    Fraction(long long n = 0, long long d = 1) {
        if (d < 0) {
            n = -n;
            d = -d;
        }
        long long g = gcd(abs(n), d);
        num = n / g;
        den = d / g;
    }

    // 最大公约数
    long long gcd(long long a, long long b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    // 加减乘除，运算符重载
    Fraction operator+(const Fraction& f) {
        return Fraction(num * f.den + f.num * den, den * f.den);
    }
    Fraction operator-(const Fraction& f) {
        return Fraction(num * f.den - f.num * den, den * f.den);
    }
    Fraction operator*(const Fraction& f) {
        return Fraction(num * f.num, den * f.den);
    }
    Fraction operator/(const Fraction& f) {
        if (f.num == 0) return Fraction(1, 0); // 用 1/0 表示 Inf
        return Fraction(num * f.den, den * f.num);
    }

    // 转换成字符串
    string to_string() const {
        if (den == 0) return "Inf";
        long long n = num, d = den;
        bool neg = n < 0;
        n = abs(n);
    
        long long int_part = n / d;
        long long rem = n % d;
    
        string result;
        if (neg) result += "(-";
    
        if (int_part != 0) result += std::to_string(int_part);
        if (int_part != 0 && rem != 0) result += " ";
        if (rem != 0) result += std::to_string(rem) + "/" + std::to_string(d);
        if (int_part == 0 && rem == 0) result += "0";
    
        if (neg) result += ")";
        return result;
    }
};

int main() {
    long long a1, b1, a2, b2;
    scanf("%lld/%lld %lld/%lld", &a1, &b1, &a2, &b2);

    Fraction f1(a1, b1), f2(a2, b2);

    string s1 = f1.to_string();
    string s2 = f2.to_string();

    cout << s1 << " + " << s2 << " = " << (f1 + f2).to_string() << endl;
    cout << s1 << " - " << s2 << " = " << (f1 - f2).to_string() << endl;
    cout << s1 << " * " << s2 << " = " << (f1 * f2).to_string() << endl;
    cout << s1 << " / " << s2 << " = " << (f1 / f2).to_string() << endl;

    return 0;
}