pta 用python写的程序测评说 非零返回

Question

发现python处理除法挺快的，就直接偷懒了。。
求大佬解惑，为什么会有非零返回

from functools import reduce
class judge():
    def is_divisible_by_2(self,s):
        if int(s[-1]) == 0 or int(s[-1])%2 ==0:
            return True
        else:
            return False


    def is_divisible_by_3(self,s):
        #计算各个数位的数字之和
        b = reduce((lambda x, y: int(x) + int(y)), a)
        if b%3 == 0:
            return True
        else:
            return False


    def is_divisible_by_5(self,s):
        #如果末尾是0，则一定能被2整除。故不用再次判断
        if s[-1] =='5':
            return True
        else:
            return False


    def is_divisible_by_7(self,s):
        while int(s) > 7:
            s = str(int(s[:-1])-2*int(s[-1]))
        if int(s) != 7:
            return False
        else:
            return True



    def is_divisible_by_11(self, s):
        #分别取出奇数和偶数位的数字
        tmp1 = s[::2]
        tmp2 = s[1::2]
        sum_tmp1 = reduce((lambda x, y: int(x) + int(y)), tmp1)
        sum_tmp2 = reduce((lambda x, y: int(x) + int(y)), tmp2)
        if int(sum_tmp2-sum_tmp1) ==0 or int(sum_tmp2-sum_tmp1) %11 == 0:
            return True
        else:
            return False



    def is_divisible_by_13(self, s):
        while len(s) > 2:
            s = str(int(s[:-1])+int(s[-1])*4)
        if int(s) %13 ==0:
            return True
        else:
            return False



    def is_divisible_by_17(self, s):
        while len(s) > 2:
            s = str(int(s[:-1]) + int(s[-1]) * 5)
        if int(s) % 17 == 0:
            return True
        else:
            return False

    def is_divisible_by_19(self, s):
        while len(s) > 2:
            s = str(int(s[:-1]) + int(s[-1]) * 2)
        if int(s) % 19 == 0:
            return True
        else:
            return False



    def judge_prime(self,num,prime):
        if prime < 20:
            p = getattr(self,'is_divisible_by_'+str(prime))
            return p(num)
        else:
            if (int(num)%prime) == 0:
                return True
            else:
                return False
            






if __name__ == '__main__':
    l,k = map(int, input().split(" "))
    n = input()
    # l,k = 20,5
    # n = '23654987725541023819'
    m = [n[i:i+k] for i in range(l-k+1)]
    is_prime = [True] * (len(m))
    prime_initial = [3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71]
    #原始
    for i in range(len(m)):
        tmp = judge()
        if tmp.is_divisible_by_2(m[i]):
            is_prime[i] = False
            # print(is_prime)
        else:
            a = list(filter(None, list(m[i])))
            for j in prime_initial:
                if tmp.judge_prime(m[i],j) and is_prime[i]:
                    is_prime[i] = False


    if any(is_prime):

        for i in range(len(is_prime)):
            if is_prime[i]:
                print(m[i],end='')
                break

    else:
        print('404')

Answer 1

不知道你这个问题是否已经解决, 如果还没有解决的话:

• 建议你看下这篇博客👉 ：【PTA-Python题库】详解 统计字符

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