Question

我想在python（3.6.5）中编写一个程序来说明例如1/7。对于此示例，输出应类似于：“长度：6，重复数字：142857”。到目前为止，我已经知道了：

n = int(input("numerator: "))
d = int(input("denominator: "))

def t(n, d):
    x = n * 9
    z = x
    k = 1
    while z % d:
        z = z * 10 + x
        k += 1
        print ("length:", k)
        print ("repeated numbers:", t)

    return k, z / d

t(n, d)

Answer 1

执行print ("repeated numbers:", t)会打印t函数本身的表示形式，而不是其输出。

这是代码的修复版本。我使用Python 3.6+ f字符串将重复的数字转换为字符串，并在前面添加零以使其长度正确。

def find_period(n, d):
    z = x = n * 9
    k = 1
    while z % d:
        z = z * 10 + x
        k += 1

    digits = f"{z // d:0{k}}"
    return k, digits

# Test

num, den = 1, 7
period, digits = find_period(num, den)
print('num:', num, 'den:', den, 'period:', period, 'digits:', digits)

num, den = 1, 17
period, digits = find_period(num, den)
print('num:', num, 'den:', den, 'period:', period, 'digits:', digits)

输出

num: 1 den: 7 period: 6 digits: 142857
num: 1 den: 17 period: 16 digits: 0588235294117647

这行可能有点神秘：

f"{z // d:0{k}}"

它说：找到小于或等于z除以d的最大整数，将其转换为字符串，并在左侧用零填充（如有必要），以得到一个整数。 k的长度。

正如Goyo在评论中指出的那样，该算法并不完美。如果小数点包含任何非重复部分，即分母具有2或5的因数，它将陷入循环。请查看是否可以找到一种处理方式。

Answer 2

这是我对https://www.geeksforgeeks.org/find-recurring-sequence-fraction/的Python实现

def fraction_to_decimal(numerator, denominator):
    """ This function returns repeating sequence of a fraction.
        If repeating sequence doesn't exits, then returns empty string """

    # Create a map to store already seen remainders
    # remainder is used as key and its position in
    # result is stored as value. Note that we need
    # position for cases like 1/6.  In this case,
    # the recurring sequence doesn't start from first
    # remainder.
    result = ""
    mapping = {}

    # Find first remainder
    remainder = numerator % denominator

    # Keep finding remainder until either remainder
    # becomes 0 or repeats
    while remainder != 0 and remainder not in mapping:

        # Store this remainder
        mapping[remainder] = len(result)

        # Multiply remainder with 10
        remainder = remainder * 10

        # Append remainder / denominator to result
        result_part = int(remainder / denominator)
        result += str(result_part)
        # print(f"Result: {result}")

        # Update remainder
        remainder = remainder % denominator
        # print(f"Map: {mapping}")

    return result

if __name__ == '__main__':
    result = fraction_to_decimal(1, 7)
    if result == "":
        print("No recurring sequence")
    else:
        print(f"\nLenght of recurring sequence: {len(result)}")
        print(f"\nRecurring sequence is {result}\n")

计算循环小数的周期长度

2 个答案: