北航：2018年计算机学院研究生推免机试第一题

同时也是PAT甲级1024

1024 Palindromic Number (25)（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 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

/*北航计算机学院2018年夏令营推免上机考试第一题*/
/*同时也是PAT甲级考试1024*/
/*真的很难，对字符串、模拟问题需要特别熟练，包括大整数运算等*/
#include 
#include 
#include 
#include 
using namespace std;

static char s1[100];//数的范围在10^9以内，100应该是够了
static char s2[100];

int is_huiwen(char s[],int n)//判断该串是否是回文串
{
    int f=1;
    for(int i=0;i=0;u++,j--)
                s2[u]=s1[j];
            s2[u]='\0';
            /*大整数加法，s1+s2*/
            ck=0;//ck是进位
            for(i=0;i=k)//注意哦，这里的循环变量是ii不是i，再次强调！！！ii>=k说明k步之后仍然不是回文，那么就直接输出吧
        {
            cout<