回文数（python中函数的递归）

解题思路:

注意事项:

参考代码:

while True:
    try:
        n = int(input())
        count = 0
        for i in range(n):
            def func1(s):  # 判断是否为回文数
                t = s[:: -1]
                if s == t:
                    g = True
                else:
                    g = False
                return g

            def func2(b, count1):  # 记步数
                c = b[:: -1]
                f = func1(str(b)) # 如果是回文数则为True反之为False
                if not f:
                    b = int(c) + int(b)
                    count1 += 1
                    count2 = count1
                    func2(str(b), count2)
                if count1 > 8 and f:
                    print(0)
                if count1 <= 8 and f:
                    print(count1)
                    
            a = input()
            func2(a, count)
    except:
        break