PAT 1136 A Delayed Palindrome python解法

1136 A Delayed Palindrome （20 分）
Consider a positive integer N written in standard notation with k+1 digits a_​i​​ as a_ki​​ ⋯a​₁​​ a​₀​​ with 0 ≤ a_​i​​ < 10 for all i and a_k​​ > 0. Then N is palindromic if and only if a_​i​​ = a_k-1​​ for all i. Zero is written 0 and is also palindromic by definition.

Non-palindromic numbers can be paired with palindromic ones via a series of operations. First, the non-palindromic number is reversed and the result is added to the original number. If the result is not a palindromic number, this is repeated until it gives a palindromic number. Such number is called a delayed palindrome. (Quoted from https://en.wikipedia.org/wiki/Palindromic_number )

Given any positive integer, you are supposed to find its paired palindromic number.

Input Specification:
Each input file contains one test case which gives a positive integer no more than 1000 digits.

Output Specification:
For each test case, print line by line the process of finding the palindromic number. The format of each line is the following:

A + B = C
where A is the original number, B is the reversed A, and C is their sum. A starts being the input number, and this process ends until C becomes a palindromic number – in this case we print in the last line C is a palindromic number.; or if a palindromic number cannot be found in 10 iterations, print Not found in 10 iterations. instead.

Sample Input 1:
97152
Sample Output 1:
97152 + 25179 = 122331
122331 + 133221 = 255552
255552 is a palindromic number.
Sample Input 2:
196
Sample Output 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.

题意：给出一个数x，求这个数加上它的反转数，判断相加结果是否为回文数，是则输出x is a palindromic number.否则继续将结果加上结果的反转进行判断，如果10次之后还没有获得回文数，则输出Not found in 10 iterations.

解题思路：定义两个函数，一个函数palindromic(n)判断n是否是回文数，另一个函数f_sum(a)按照题意进行求和操作，循环10次，每次判断n是否为回文数，是则输出相应结果并跳出循环，不是则继续进行求和，10次循环完成后输出相应结果。

n = input()
def palindromic(n):
    if n[::-1] == n:
        print(n +' is a palindromic number.')
        return True
    else:
        return False
def f_sum(a):
    b = int(a[::-1])
    a = int(a)
    print('%d + %d = %d'%(a,b,a+b))
    return str(a+b)
for i in range(10):
    if palindromic(n):
        break
    n = f_sum(n)
else:
    print('Not found in 10 iterations.')