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.
## 解答
```cpp
#include<iostream>
#include<string>
using namespace std;
void output(string str, int depth)
{
bool flag = true;
int slength = str.length(), c = 0, tmp;
string res(slength, '\0'), strReverse(slength, '\0');
for (int i = slength-1; i >=0; i--) {
if (str[i] != str[slength - i - 1]) flag = false;
tmp = str[i] + str[slength - i - 1] - 2 * '0' + c;
if (tmp >= 10) {
c = 1;
tmp %= 10;
}
else c = 0;
res[i] = tmp + '0';
strReverse[slength - i - 1] = str[i];
}
if (flag == true) {
printf("%s is a palindromic number.\n", str.c_str());
return;
}
if (c == 1) res = string(1, '1') + res;
printf("%s + %s = %s\n", str.c_str(), strReverse.c_str(), res.c_str());
if (depth == 10) printf("Not found in 10 iterations.");
else output(res, depth + 1);
}
int main()
{
string num;
cin >> num;
output(num, 1);
return 0;
}
```

PAT A1136 A Delayed Palindrome (20point(s))