1136 A Delayed Palindrome （20 分）

Consider a positive integer N written in standard notation with k+1 digits a​i​​ as a​k​​⋯a​1​​a​0​​ with 0 for all i and a​k​​>0. Then N is palindromic if and only if a​i​​=a​k−i​​ 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.

#include<iostream>
#include<algorithm>
using namespace std;bool isPalindromic(string &str){int len = str.size();for(int i = 0; i < len/2; i++){if(str[i] != str[len - i - 1]) return false;}return true;
}string add(const string &A,const string &B){string C;int len = A.size();int carry = 0;for(int i = len - 1; i >= 0; i--){int temp = A[i] - '0' + B[i] - '0' + carry;C += temp % 10 +'0';carry = temp / 10;}if(carry != 0) C += carry + '0';reverse(C.begin(),C.end());return C;
}int main(){string A,B,C;cin >> A;int cnt = 10;if(isPalindromic(A)){cout << A << " is a palindromic number.";return 0;}while(cnt--){B = A;reverse(A.begin(),A.end());C = add(A,B);cout << B << " + " << A << " = " << C << endl;if(isPalindromic(C)){cout << C << " is a palindromic number.";return 0;}A = C;}cout <<"Not found in 10 iterations.";return 0;
}