1136 A Delayed Palindrome （20 分）

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本文介绍了一种算法，用于找到任意正整数的配对回文数。通过将非回文数与其反转数相加，重复此过程直到得到回文数。文章详细解释了这一过程，并提供了一个C++实现示例，展示了如何判断一个数是否为回文数，以及如何进行配对。

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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 Cbecomes 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.

代码如下：

#include<bits/stdc++.h>
using namespace std;
string  add (string a )
{
   string b=a,ans;
   reverse(b.begin(),b.end());
   int len=a.length(),carry=0;
   for (int i=0;i<len;i++)
   {
       int num=(a[i]-'0')+(b[i]-'0')+carry;
       carry=0;
       if (num>=10)
       {
           carry=1;
           num=num-10;
       }
       ans+=char(num+'0');
   }
   if (carry==1)  ans+='1';
   reverse(ans.begin(),ans.end());
   return ans;
}

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