The generation of palindromic numbers within a given range is best done with a computer program. Space it limited so an example of program code cannot be shown here, but the Codecast website gives full guidance.
#include <stdio.h> #include<conio.h> void main () { int a,r,n,sum=0; clrscr(); printf("enter the number"); scanf("%d",& n); c=n while (n!=0) { a=n%10; n=n/10; sum=sum*a+a } printf("the reverse is %d \n",sum); if(c==sum) printf("\n the given number is palindrome"); else printf("\n the given number is not a palindrome"); getch(); }
This program only suits PHP. If you want a proper one try C program for it available on web <body> <?php if(isset($_POST['submit'])) { $text = $_POST['text']; $string = mysql_real_escape_string($text); $invert = strrev($string); if($string == $invert) { echo "<br>Given number is a palindrome!!"; } else { echo "<br>Given number is not a palindrome!!"; } } ?> <form method="post"> <input type="text" name="text" id="text" /> <input type="submit" name="submit" value="Submit" id="submit" onclick="<?php $_SERVER['PHP_SELF']; ?>" /> </form> </body>
#include <stdio.h> #include <conio.h> void main() { int num,rev=0,m,r; clrscr(); printf("enter any number"); scanf("%d",&num); m=num; while(num>0) { r=num%10; rev=rev*10+r; num=num/10; } if(rev==m) printf("given number is palindrome"); else printf("given number is not palindrome"); getch(); } this is the answer.
/*To check whether a string is palindrome*/includeincludevoid main () { int i,j,f=0; char a[10]; clrscr (); gets(a); for (i=0;a[i]!='\0';i++) { } i--; for (j=0;a[j]!='\0';j++,i--) { if (a[i]!=a[j]) f=1; } if (f==0) printf("string is palindrome"); else printf("string is not palindrome"); getch (); }
#include <stdio.h> #include <string.h> #define N 100 #define PALINDROME 0 #define NONPALINDROME 1 /*Program that tells you whether what you enter is a palindrome or not*/ char pal[N]; //input line int i; //counter int k; //counter int tag; int flag = PALINDROME; /*flag=0 ->palindrome, flag =1 ->nonpalindrome*/ int main() { printf("Enter something: \n"); scanf("%s", pal); tag = strlen(pal); /*determine length of string*/ /* pointer running from the beginning and the end simultaneously looking for two characters that don't match */ /* initially assumed that string IS a palindrome */ /* the following for loop looks for inequality and flags the string as a non palindrome the moment two characters don't match */ for (i=0,k=tag-1; i=0; i++,k--) { if(pal[i] != pal[k]) { flag=NONPALINDROME; break; } } if(flag == PALINDROME) { printf("This is a palindrome\n"); } else { printf("This is NOT a palindrome\n"); } return 0; } #include <stdio.h> #include <string.h> #define N 100 #define PALINDROME 0 #define NONPALINDROME 1 /*Program that tells you whether what you enter is a palindrome or not*/ char pal[N]; //input line int i; //counter int k; //counter int tag; int flag = PALINDROME; /*flag=0 ->palindrome, flag =1 ->nonpalindrome*/ int main() { printf("Enter something: \n"); scanf("%s", pal); tag = strlen(pal); /*determine length of string*/ /* pointer running from the beginning and the end simultaneously looking for two characters that don't match */ /* initially assumed that string IS a palindrome */ /* the following for loop looks for inequality and flags the string as a non palindrome the moment two characters don't match */ for (i=0,k=tag-1; i=0; i++,k--) { if(pal[i] != pal[k]) { flag=NONPALINDROME; break; } } if(flag == PALINDROME) { printf("This is a palindrome\n"); } else { printf("This is NOT a palindrome\n"); } return 0; }
You write the number and then follow it with the digits in reverse order.
Reverse the string and compare it to the original. If they match, then it is a palindrome.
civic
civic
Civic
civic
Yes, you can use regex to determine if a given string is a palindrome by reversing the string and then comparing it to the original string using regex.
To find the next numerical palindrome, start with the given number and increment it by one. Check if the new number reads the same forwards and backwards. If it does, you've found your next palindrome; if not, repeat the process until you find one. For efficiency, you can skip directly to the next candidate by considering the structure of palindromes, focusing on the digits that mirror around the center.
A civic reception.
50 80
depends on what the given set is
#include <stdio.h> #include<conio.h> void main () { int a,r,n,sum=0; clrscr(); printf("enter the number"); scanf("%d",& n); c=n while (n!=0) { a=n%10; n=n/10; sum=sum*a+a } printf("the reverse is %d \n",sum); if(c==sum) printf("\n the given number is palindrome"); else printf("\n the given number is not a palindrome"); getch(); }