1136 A Delayed Palindrome

Consider a positive integer N written in standard notation with k+1 digits ai​ as ak​⋯a1​a0​ with 0≤ai​<10 for all i and ak​>0. Then N is palindromic if and only if ai​=ak−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.

高精加模拟： 

#include 
#include 
#include 
#include 
using namespace std;
 
bool check(string x) {
	int i = 0;
	while (x[i] == '0') {
		i++;
	}
	for (int j = 0; i < x.size(); i++, j++) {
		if (x[i] != x[x.size() - 1 - j]) {
			return false;
		}
	}
	return true;
}
 
int main() {
	int cnt = 0;
	string n;
	cin >> n;
	while (!check(n) && cnt < 10) {
		cnt++;
		string t = n, s = "";
		reverse(t.begin(), t.end());
		int x, y = 0;
		for (int i = n.size() - 1; i >= 0; i--) {
			x = (n[i] - '0' + t[i] - '0' + y) % 10;
			s += (x + '0');
			y = (n[i] - '0' + t[i] - '0' + y) / 10;
		}
		if (y) {
			s += (y + '0');
		}
		reverse(s.begin(), s.end());
		int i = 0;
		while (s[i] == '0') {
			i++;
		}
		if (i) {
			s.erase(0, i);
		}
		cout << n << " + " << t << " = " << s << endl;
		n = s;
	}
	if (cnt >= 10) {
		cout << "Not found in 10 iterations.";
	} else {
		cout << n << " is a palindromic number.";
	}
	return 0;
}