PAT 1136. A Delayed Palindrome (20)
Consider a positive integer N written in standard notation with k+1 digits ai as ak…a1a0 with 0 <= ai < 10 for all i and ak > 0. Then N is palindromic if and only if ai = aki for all i. Zero is written 0 and is also palindromic by definition. Nonpalindromic numbers can be paired with palindromic ones via a series of operations. First, the nonpalindromic 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.
验证一个数加上逆转它之后得到的数是否为一个回文数。 输入的数值不超过1000位，所以不可以用int或者long存储，而必须用string存储。 算法的思路很常规，将输入的数值reverse并加上数值本身，并判断它是否是回文数即可。见代码
1 
