1024. Palindromic Number (25)

红太狼 2022-05-31 10:12 140阅读 0赞

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 (<= 1010) 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

--------------------

题目大意：

当一个数字顺这写和倒着写是同一个数字时，它被称为回文数。例如，1234321是一个回文数。所有的个位数字都是回文数。  
非回文数可以通过一系列的操作与回文数配对。首先，将非回文数反转，将结果加到原数中。如果结果不是一个回文数，这将被重复，直到找到一个回文数。例如，如果我们从67开始，我们可以在两步中获得一个回文数：67 + 76 = 143, and 143 + 341 = 484。  
给定任何正整数N，你的任务是找到与他配对的回文数和所需的步骤数。  
输入规格：  
每个测试用例包含一个输入文件。每个测试用例由两个正整数N和K组成，其中N(<=10的100次方)是初始数，K（<=100）是最大的步骤数。数字被一个空格隔开。  
输出规范：  
对于每个测试用例，输出两个数字，一个占一行。第一个数字始于N配对的回文数，第二个数字是用来找到回文数的步骤数。如果在K个步骤之后没有找到回文数，只需输出第K步和第K步得到的数字。

代码：

#include<stdio.h>
    #include<string.h>
    int main()
    {
        char str[1000],num[1000],result[1000];
        int i,j,n,m,k,t,l,p;
        scanf("%s %d",str,&k);
        strcpy(result,str);
        for(i=0;i<k;i++)
        {
            n=strlen(str);
            for(j=0;j<n/2;j++)
            {
                if(result[j]!=result[n-j-1])
                {
                    break;
                }
            }
            if(j>=n/2)
            {
               printf("%s\n",result);
               printf("%d",i);
               break;
            }
            else
            {
               for(j=0;j<n;j++)
               {
                   num[j]=result[n-j-1];
               }
               num[j]='\0';
               t=0;
               for(j=n-1,l=0;j>=0;j--)
               {
                   result[j]+=t;
                   t=((num[j]-'0')+(result[j]-'0'))/10;
                   str[l++]=((num[j]-'0')+(result[j]-'0'))%10+'0';
               }
               if(t!=0)
               {
                   str[l++]=t+'0';
               }
               str[l]='\0';
               p=strlen(str);
               for(j=0;j<p;j++)
               {
                   result[j]=str[p-j-1];
               }
               result[j]='\0';
            }
        }
        if(i>=k)
        {
            printf("%s\n",result);
            printf("%d",k);
        }
        return 0;
    }