【PAT (Advanced Level) Practice】1024 Palindromic Number (25 分)

A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, $1234321$ is a palindromic number. All single digit numbers are palindromic numbers.

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. For example, if we start from 67, we can obtain a palindromic number in $2$ steps: $67 + 76 = 143$, and $143 + 341 = 484$.

Given any positive integer $N$, you are supposed to find its paired palindromic number and the number of steps taken to find it.

Input Specification:
Each input file contains one test case. Each case consists of two positive numbers $N$ and $K$, where $N (≤10^{10})$ is the initial numer and $K (≤100)$ is the maximum number of steps. The numbers are separated by a space.

Output Specification:
For each test case, output two numbers, one in each line. The first number is the paired palindromic number of $N$, and the second number is the number of steps taken to find the palindromic number. If the palindromic number is not found after $K$ steps, just output the number obtained at the $K$th step and $K$ instead.

Sample Input 1:

67 3

Sample Output 1:

484
2

Sample Input 2:

69 3

Sample Output 2:

1353
3

---

#include<iostream>
#include<vector>

using namespace std;

int k;
string s;

bool check(vector<int> & a){
    
    for(int i = 0, j = a.size() - 1; i < j; i++, j--)
        if(a[i] != a[j])
            return false;
    return true;
}

vector<int> add(vector<int>& a, vector<int>& b){
    
    vector<int> c;
    
    int t = 0;
    for(int i = 0; i < a.size() || i < b.size() || t; i++){
        
        if(i < a.size()) t += a[i];
        if(i < b.size()) t += b[i];
        c.push_back(t % 10);
        t /= 10;
    }
    
    return c;
}


int main(){
    
    cin >> s >> k;
    
    vector<int> a;
    for(int i = s.size() - 1; i >= 0; i--) a.push_back(s[i] - '0');
    
    int cnt = 0;
    while(!check(a) && cnt < k){
        cnt++;
        vector<int> b(a.rbegin(), a.rend());
        a = add(a, b);
    }
    
    for(int i = a.size() - 1; i >= 0; i--)
        cout << a[i];
    
    cout << '\n' << cnt;
    
    return 0;
}