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 Kth step and K instead.

Sample Input 1:

67 3

Sample Output 1:

484
2

Sample Input 2:

69 3

Sample Output 2:

1353
3
題目大意就是：給出一個大數，並將其反轉，在有限次k次相加內，若結果是回文數，那麽就輸出並且輸出計算次數；否則就直接輸出計算k次的結果。
註意到n可能會非常大，10的10次方，而且要進行k次（最多100次），所以用long long不可以了，用string來做大數運算，並且string有reverse函數。
還有主函數中的邏輯判斷問題，因為一個數如果是回文數，那麽運算結果肯定也是回文數，只要判斷一個數反轉後是否等於自身，如果相等，那麽
就已經找到回文數，就break即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
using namespace std;
//
string add(string s1,string s2){
    long long  len=s1.size();
    int a,b=0;
    for(int i=0;i<len;i++){
        if(b==0)
            a=(s1[i]-‘0‘)+(s2[i]-‘0‘);
        else{
            a=(s1[i]-‘0‘)+(s2[i]-‘0‘)+1;
            b=0;
        }
        if(a>=10){
            b=1;
            a%=10;
        }
        s2[i]=a+‘0‘;
    }
    if(b==1){
        s2=s2+‘1‘;
    }
    reverse(s2.begin(),s2.end());
    return s2;
}
bool isH(string s){
    long long  len=s.size();
    long long  f=len/2;
    bool flag=true;
    for(int i=0;i<f;i++)
        if(s[i]!=s[len-i-1]){
            flag=false;break;
        }
    return flag;
}
int main()
{
   string s;
   int k;
   cin>>s>>k;
   string s2;
   int n=0;
   bool flag=false;
   while(n<k){
        s2=s;
        reverse(s.begin(),s.end());
        if(s2==s){
            break;
        }else{
            s=add(s,s2);
            n++;
        }
   }
   cout<<s<<‘\n‘<<n;
   return 0;
}

PAT 1024 Palindromic Number