给定一个 k+1 位的正整数 N，写成 a_k⋯a₁a₀的形式，其中对所有 i 有 0≤a_i<10 且 a_k>0。N 被称为一个回文数，当且仅当对所有 i有 a_i=a_k−i。零也被定义为一个回文数。

非回文数也可以通过一系列操作变出回文数。首先将该数字逆转，再将逆转数与该数相加，如果和还不是一个回文数，就重复这个逆转再相加的操作，直到一个回文数出现。如果一个非回文数可以变出回文数，就称这个数为延迟的回文数。（定义翻译自 https://en.wikipedia.org/wiki/Palindromic_number ）

给定任意一个正整数，本题要求你找到其变出的那个回文数。

输入格式：

输入在一行中给出一个不超过1000位的正整数。

输出格式：

对给定的整数，一行一行输出其变出回文数的过程。每行格式如下

A + B = C

其中 A 是原始的数字，B 是 A 的逆转数，C 是它们的和。A 从输入的整数开始。重复操作直到 C 在 10 步以内变成回文数，这时在一行中输出 C is a palindromic number.；或者如果 10 步都没能得到回文数，最后就在一行中输出 Not found in 10 iterations.。

输入样例 1：

输出样例 1：

97152 + 25179 = 122331
122331 + 133221 = 255552
255552 is a palindromic number.

输入样例 2：

输出样例 2：

196 + 691 = 887
887 + 788 = 1675
1675 + 5761 = 7436
7436 + 6347 = 13783
13783 + 38731 = 52514
52514 + 41525 = 94039
94039 + 93049 = 187088
187088 + 880781 = 1067869
1067869 + 9687601 = 10755470
10755470 + 07455701 = 18211171
Not found in 10 iterations.

编程语言：Python

解题思想：

1.首先对特例情况进行判断

2.利用字符串拼接得到B

3.判断是否回文可以使用两边往中间逐位比较的方法，如165561，依次进行1==1、6==6、5==5的比较

时间复杂度：O(NlogN)

代码如下：

Number = input()  # 输入原始数字
Count = 0  # 操作的次数
A = Number
 
Palindrome = True  # 标记是否为回文数，默认为是
for i in range(len(A)):  # 特例：输入的数就是回文数
    if A[i] != A[len(A) - i - 1]:
        Palindrome = False
        break
 
if Palindrome:
    print(A + " is a palindromic number.")
else:
    while True:
        B = ""  # 逆转数
        for i in range(len(A) - 1, -1, -1):
            B += A[i]
        C = str(int(A) + int(B))
        Count += 1  # 完成一次操作
        print(A + " + " + B + " = " + C)  # 输出运算结果
        Palindrome = True  # 每次都将标记重置
        for i in range(len(C)):  # 遍历C，从左边、右边同时逐位往中间判断
            if C[i] != C[len(C) - i - 1]:  # 两字符不相同
                Palindrome = False
                break
        if Palindrome:  # C是回文数
            print(C + " is a palindromic number.")
            break
        else:  # C不是回文数，继续操作
            A = C
        if Count == 10:  # 操作次数超过规定
            print("Not found in 10 iterations.")
            break

< >

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PAT乙级:1079 延迟的回文数