1136 A Delayed Palindrome （20 分）

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本文介绍了一种通过反复反转和累加非回文数直至得到回文数的算法，即所谓的延迟回文数。详细解释了算法流程，并通过示例展示了如何从任意正整数找到其配对的回文数，同时提供了实现该算法的C++代码。

1136 A Delayed Palindrome （20 分）

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=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:

Sample Output 1:

97152 + 25179 = 122331
122331 + 133221 = 255552
255552 is a palindromic number.

Sample Input 2:

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<bits/stdc++.h>
using namespace std;
string a,b,c;
void calc()
{
    int cat=0;
    int ba;
    c.clear();
    for(int i=0;i<a.size();i++)
    {
        ba=(a[i]-'0'+b[i]-'0'+cat)%10;
        cat=(a[i]-'0'+b[i]-'0'+cat)/10;
        c.push_back(ba+'0');
    }
    if(cat>0)
        c.push_back('1');
    reverse(c.begin(),c.end());
}

int main()
{
    cin>>a;
    b=a;
    reverse(b.begin(),b.end());
    int cnt=0;
    while(cnt<=10)
    {
        if(a==b)
        {
            cout<<a<<" is a palindromic number.";
             return 0;
        }
        else if(cnt==10)
        {
            cout<< "Not found in 10 iterations." ;
             return 0;

        }
        calc();
        cout<<a<<" + "<<b<<" = "<<c<<endl;
        a=b=c;
        reverse(b.begin(),b.end());
        cnt++;
    }

}