PAT 1136 A Delayed Palindrome

Consider a positive integer N written in standard notation with k+1 digits a​i as ak ⋯a1a0​ with 0≤ai <10 for all i and a​k​​ >0. Then N is palindromic if and only if a​i =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.

#include<iostream> //水题
#include<algorithm>
using namespace std;
string getreverse(string a){
    string b;
    b.resize(a.size());
    copy(a.rbegin(), a.rend(), b.begin());
    return b;
}
string add(string a, string b){
    string s;
    int t=0;
    for(int i=a.size()-1; i>=0; i--){
        int p=a[i]-'0', q=b[i]-'0';
        s.insert(s.begin(), (p+q+t)%10+'0');
        t=(p+q+t)/10;
    }
    if(t!=0)
        s.insert(s.begin(), t+'0');
    return s;
}
bool judge(string a){
    int k=a.size()-1;
    for(int i=0; i<=k/2; i++)
        if(a[i]!=a[k-i])
            return false;
    return true;
}
int main(){
    string  a;
    cin>>a;
    if(judge(a)){
        cout<<a<<" is a palindromic number."<<endl;
        return 0;
    }   
    for(int i=0; i<10; i++){
        string b=getreverse(a);
        string t=add(a,b);
        cout<<a<<" + "<<b<<" = "<<t<<endl;
        if(judge(t)){
            cout<<t<<" is a palindromic number."<<endl;
            return 0;
        }   
        a=t;
    }
    cout<<"Not found in 10 iterations."<<endl;
    return 0;
}