PAT乙级 1034有理数四则运算 python

稍麻烦，讲道理觉得应该是一道甲级的25分题才对……

分类讨论思想，程序模块化才能不乱。

格式转换要趁早。

三元表达式让代码更易读。

def gcd(m, n):  # 欧几里得算法
    m, n = abs(m), abs(n)
    if m < n:
        m, n = n, m
    while m % n != 0:
        r = m % n
        m, n = n, r
    return int(n)  # 最大公约数

def deal(x):#由特殊到一般
    x1, x2 = list(map(int, x.split('/')))
    if int(x1) == 0:
        return '0'
    if x1 % x2 == 0:
        return '(' + str(int(x1 / x2)) + ')' if x[0] == '-' else int(x1 / x2)
    n = gcd(x1, x2)
    x1, x2 = int(x1 / n), int(x2 / n)  # 化简
    if abs(x1) <= abs(x2):
        return '(' + str(x1) + '/' + str(x2) + ')' if x[0] == '-' else str(x1) + '/' + str(x2)
    else:
        if x[0] == '-':
            return '(-' + str(int(abs(x1) // x2)) + ' ' + str(abs(x1) - int(abs(x1) // x2 * x2)) + '/' + str(x2) + ')'
        else:
            return str(int(x1 // x2)) + ' ' + str(x1 - int(x1 // x2 * x2)) + '/' + str(x2)

def jia(x1, x2, y1, y2):
    g = gcd(x2, y2)
    mu, zi = x2 * y2 / g, x1 * y2 / g + y1 * x2 / g
    return result(mu, zi)
def jian(x1, x2, y1, y2):
    y1 = -y1
    g = gcd(x2, y2)
    mu, zi = x2 * y2 / g, x1 * y2 / g + y1 * x2 / g
    return result(mu, zi)
def cheng(x1, x2, y1, y2):
    mu, zi = x2 * y2, x1 * y1
    return result(mu, zi)
def chu(x1, x2, y1, y2):
    mu, zi = x2 * y1, x1 * y2
    return result(mu, zi)

def result(mu, zi):
    return str(int(abs(zi))) + '/' + str(int(abs(mu))) if mu * zi >= 0 else '-' + str(int(abs(zi))) + '/' + str(int(abs(mu)))

a, b = input().split()
a1, a2 = list(map(int, a.split('/')))
b1, b2 = list(map(int, b.split('/')))
print(deal(a), '+', deal(b), '=', deal(jia(a1, a2, b1, b2)))
print(deal(a), '-', deal(b), '=', deal(jian(a1, a2, b1, b2)))
print(deal(a), '*', deal(b), '=', deal(cheng(a1, a2, b1, b2)))
print(deal(a), '/', deal(b), '=', deal(chu(a1, a2, b1, b2))) if b1 != 0 else print(deal(a), '/', deal(b), '=', 'Inf')