QUESTION

A positive integer is called a palindrome if its representation in the decimal system is the same when read from left to right and from right to left. For a given positive integer K, write the value of the smallest palindrome larger than K to output.\n\nInput\nThe first line contains an integer, which corresponds to K. Assume that K is less than200000.\nOutput\nOutput consists of a single integer, which corresponds to the smallest palindrome larger than K.\n\n.

“TESTCASE_1”: “808\n###—###SEPERATOR—###—\n818”, “TESTCASE_2”: “2133\n###—###SEPERATOR—###—\n2222”, “TESTCASE_3”: “0\n###—###SEPERATOR—###—\n0”, “TESTCASE_4”: “0\n###—###SEPERATOR—###—\n0”, “TESTCASE_5”: “0\n###—###SEPERATOR—###—\n0

ANSWER

#include<stdio.h>
        #include<string.h>
        #include<stdlib.h>
        int main()
        {
            char a[1000002];
            int i,flag,check,l,m,k;
           
            
                flag=1;
                  scanf("%[^\n]%*c", a);

                l=strlen(a);
                //printf
                k=l>>1;
                check=0;
                for(i=0;i<l;i++)
                {
                    if(a[i]!='9')
                    {
                        flag=0;
                        break;
                    }
                }
                if(flag==1)
                {
                    a[0]='1';
                    for(i=1;i<l;i++)
                    a[i]='0';
                    a[l]='1';
                    a[l+1]='\0';
                    check=1;
                }
                flag=0;
                if(check!=1)
                {
                    for(i=0;i<k;i++)
                    {
                        if(a[i]<a[l-1-i])
                        flag=-1;
                        else if(a[i]>a[l-1-i])
                        flag=1;
                        a[l-1-i]=a[i];
                    }
         
                    if(!(l&1))
                    m=k-1;
                    else m=k;
                    if(flag==0||flag==-1)
                    {
                        i=0;
                        while(a[m-i]=='9')
                        {
                            a[m-i]='0';
                            a[l-1-m+i]='0';
                            i++;
                        }
         
                        a[m-i]++;
                        a[l-1-m+i]=a[m-i];
                }
            
            printf("%s\n",a);
            }
            return 0;
        }