1136 A Delayed Palindrome

回文数与延迟回文数

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1136 A Delayed Palindrome
题目链接
Consider a positive integer N written in standard notation with k+1 digits a
i
as a
k
⋯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−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.
回文数字的判定与变换。
参考代码：

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
bool is_palin(string s) {
	int k = s.size() - 1, i, n;
	n = k / 2;
	for (i = 0; i <= n; i++) {
		if (s[i] != s[k - i])break;
	}
	if (i == n+1)return true;
    return false;
}
string sum(string s, string s1) {
	int n = s.size() - 1, jinwei = 0, m, p;
	string res;
	for (int i = n; i >= 0; i--) {
		p = s[i] - '0' + s1[i] - '0' + jinwei;
		m = p % 10;
		jinwei = p / 10;
		res = to_string(m) + res;
	}
	if (jinwei != 0)res = to_string(jinwei) + res;
	return res;
}
int main() {
	string s, s1, res;
	int cnt = 0;
	cin >> s;
	while (!is_palin(s)){
		cnt++;
		if (cnt > 10) { cout << "Not found in 10 iterations.";return 0; }
		s1 = s;
		reverse(s1.begin(), s1.end());
		res = sum(s, s1);
		cout << s << " + " << s1 << " = " << res << endl;
		s = res;
	}
	cout << s << " is a palindromic number.";

	return 0;
}