解法一：回文数+高精度

模拟，注意输入范围，需要使用高精度。

import java.io.*;
import java.math.BigInteger;
public class Main {
    public static void main(String[] args) throws IOException {
        StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        in.nextToken();
        BigInteger N = new BigInteger(String.valueOf((long)in.nval));
        in.nextToken();
        int K = (int) in.nval;
        for (int i = 0; i < K; ++i) {
            if (isPalindromic(N)) {
                out.println(N);
                out.println(i);
                out.flush();
                return;
            }
            N = plus(N);
        }
        out.println(N);
        out.println(K);
        out.flush();
    }
    private static BigInteger plus(BigInteger n) {
        BigInteger s = new BigInteger("0");
        char[] chars = n.toString().toCharArray();
        for (int i = chars.length - 1; i >= 0; --i) {
            BigInteger bit = new BigInteger(String.valueOf(chars[i] - 48));
            s = s.multiply(BigInteger.TEN).add(bit);
        }
        return s.add(n);
    }
    private static boolean isPalindromic(BigInteger n) {
        boolean flag = true;
        char[] chars = n.toString().toCharArray();
        int len = chars.length;
        for (int i = 0; i < len / 2; ++i) {
            if (chars[i] != chars[len - i - 1]) {
                flag = false;
                break;
            }
        }
        return flag;
    }
}