【每日一题1108】执着的单数

:woman_mage: 给定一个正整数n,请编写一个函数,将组成n的每个数字操作相乘,直到最终乘数结果是个位数为止。返回相乘操作的总次数。

示例:
输入:39,输出:3。因为39=27,27=14,1*4=4。

题目难度:简单
题目来源:codewars

def persistence(n: int) -> int:
    pass

assert persistence(39) == 3
assert persistence(999) == 4
assert persistence(8) == 1
def persistence(n):
    count=0
    result=n
    if result<10:
        return 1
    while result>=10:
        words=list(str(result))
        result=1
        for word in words:
            result*=int(word)
        count+=1
    return count
assert persistence(39) == 3
assert persistence(999) == 4
assert persistence(8) == 1
def persistence(n: int) -> int:
    times = 1
    while True:
        multiplier_sum = 1
        for i in [int(n) for n in str(n)]:
            multiplier_sum *= i
        if len(str(multiplier_sum)) == 1:
            return times
        else:
            n = multiplier_sum
        times += 1


assert persistence(39) == 3
assert persistence(999) == 4
assert persistence(8) == 1
def persistence(self, n: int) -> int:
    '''
    给定一个正整数n,请编写一个函数,将组成n的每个数字操作相乘,直到最终乘数结果是个位数为止。返回相乘操作的总次数。

    示例:
    输入:39,输出:3。因为39=27,27=14,1*4=4。
    :param n:
    :return:
    '''
    times = 0

    # 先得数每位相乘的结果
    cur_data = n
    list_data = []
    while cur_data > 9:
        data = cur_data
        list_data.append(data)
        result = 1
        while data != 0:
            result = result * (int(data%10))
            data =int(data / 10)
        times += 1
        cur_data = result
    print(list_data)
    return times
    pass

def test_persistence(self):
    assert self.persistence(0) == 0
    assert self.persistence(39) == 3
def persistence(n:int):
    m=str(n)
    if len(m)==1:
        return 1
    count=0
    while len(m)>1:
        res=1
        for i in m:
            res=res*int(i)
        count=count+1
        m=str(res)
    return count
# 思路:转化为字符串相乘计数

def persistence(n):
result = 1

if len(list(str(n))) == 1:
    return 1
elif len(list(str(n))) > 1:
    num = 0
    while True:
        for i in list(str(n)):
            # print(result)
            result *= int(i)
        # 将两位数相乘的结果赋值给n
        n = result

        # result重置为1,不然result值越来越大
        result = 1
        num += 1
        # print(f"第{num}次",n)
        if n < 10:
            break
    return num

print(persistence(39))
print(persistence(999))
print(persistence(9))
assert persistence(39) == 3
assert persistence(999) == 4
assert persistence(8) == 1

def persistence(n):
    if n<10:
        return 1
    count = 0
    while n//10!=0:
        each_number = []
        while n>0:
            each_number.append(n%10)
            n = n//10
        result = 1
        for each in each_number:
            result *= each
        n = result
        count += 1
    return count

当数字小于10时,直接返回1
当数字大于10时,从个位数开始将每个数字存入列表each_number中(求余+对10整除),然后遍历each_number相乘得出新的数字,计数器加一

Python参考题解:

from functools import reduce
from operator import mul

def persistence(n: int) -> int:
    
    if len(str(n)) == 1:
        return 1
    
    count = 0
    while len(str(n))>1:
        n = reduce(mul, [int(c) for c in str(n)])
        count += 1
        
    return count

persistence(39)

assert persistence(39) == 3
assert persistence(999) == 4
assert persistence(8) == 1

思路:判断输入的数字是多位的,则进行while循环计算因子的乘积,直到成为单位数。顺便统计操作的次数。

Python参考题解:

def persistence(n: int) -> int:
    
    def rec(n: int): 
        if n > 10:
            # 计算字符串的因子乘积
            total = 1
            for c in str(n):
                total *= int(c)
            # 将乘积结果继续调用自己 
            return 1 + rec(total)
        else:
            return 0

    return 1 if n < 10 else rec(n)


assert persistence(39) == 3
assert persistence(999) == 4
assert persistence(8) == 1

思路: 利用递归,实现乘积运算和统计。

使用%来获取十位与个位,然后相乘

def number(n):
    # 计数器
    count = 0

    # while循环:在满足条件的情况下执行
    while (n > 9):
        # 判断是否是两位数
        if (n > 9):
            # 每循环一次计数器加一
            count += 1
            # 分解出十位的数字,因为分解后等于小数,转换成int整数类型
            shiwei = int(n % 100 / 10)
            # 分解出个位的数字
            gewei = n % 10
            # 获得结果
            jieguo = shiwei * gewei
            # 输出结果
            print(f'{shiwei} * {gewei} = {jieguo}')
            # 将结果重新赋值给函数,再次循环、判断
            n = jieguo

    print(f'一共相乘了{count}次')

# 调用函数
number(98)
输出结果:
9 * 8 = 72
7 * 2 = 14
1 * 4 = 4
一共相乘了3次
def persistence(n: int) -> int:
    '''
     给定一个正整数n,请编写一个函数,将组成n的每个数字操作相乘,直到最终乘数结果是个位数为止。返回相乘操作的总次数。
     示例:
     输入:39,输出:3。因为39=27,27=14,1*4=4。
    :param n:
    :return:
    '''
    count = 0
    while True:
        # 将整数转换成字符串,再将每个数字拆开,组成列表;再对列表做遍历,转换成int类型相乘;将相乘后的结果再赋给n
        num_str = str(n)
        num = "_".join(num_str).split("_")
        multiply = 1
        for i in num:
            multiply *= int(i)
        n = multiply
        count += 1
        # 直到n为个数(<10)时,跳出循环
        if n < 10:
            break
    return count


assert persistence(39) == 3
assert persistence(999) == 4
assert persistence(8) == 1
def persistence(n: int) -> int:
    if len(str(n)) == 1:
        return 1
    count = 0
    while len(str(n)) > 1:
        result = 1
        for i in str(n):
            result *= int(i)
        n = result
        count += 1
    return count


assert persistence(39) == 3
assert persistence(999) == 4
assert persistence(8) == 1
def persistence(n: int) -> int:
    def mul(n):
        result = 1
        L = [int(i) for i in list(str(n))]
        for i in L:
            result *= i
        return result

    count =0
    while(1):
        n = mul(n)
        count += 1
        if n <10:
            break
    return count

assert persistence(39) == 3
assert persistence(999) == 4
assert persistence(8) == 1
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