How do you write a C program to check a Given Number Whether is it Palindrome or Not?

Answer 1

#include void main() { long int n; printf("ENTER A NUMBER: "); scanf("%ld",&n); int n1,mod; n1=n; int rev=0; while(n>0) { mod = n%10; rev = rev * 10 + mod; n = n / 10; } if (n1 == rev) printf("ENTERED NUMBER IS A PALINDROME\n"); else printf("ENTERED NUMBER NOT A PALINDROME\n"); }

Answer 2

#include<stdio.h>

unsigned reverse (unsigned num) {

unsigned dgt, acc = 0;

while (num>0) { dgt = num%10; // determine last digit of num (remainder after division by 10)

rev = rev * 10; // left-shift the accumulator by 1 decimal digit

rev = rev + dgt; // add the digit

num = num / 10; // right-shift the number by 1 decimal digit

}

return acc; // the accumulator is the reverse of original number

}

bool is_palindrome (unsigned num) {

return num == reverse (num);

}

int main (void) {

unsigned num;

printf ("Enter a number greater than zero: ");

scanf ("%u\n", &num);

if (is_palindrome (num))

printf ("The number is a palindrome\n");

else

printf ("The number is not a palindrome\n");

return 0;

}

Answer 3

//assume all files r included.

void main()

{

int n,n1,s=0,r;

cout<<"enter the number";cin>>n;

n1=n;

while(n)

{r=n%10;

s+=r*10;

n=n%10;

}

if (n1==s)

cout<<"no. is palindrome";

else cout<<endl<<"no. is not palindrome";

}

Answer 4

The easiest way to check if a number is a palindrome or not is to reverse the number and see if the value is the same as the original. To reverse a number in any base, use the following function:

unsigned int reverse(unsigned int num, unsigned int base=10)

{

unsigned int rev=0;

while( num )

{

rev*=base;

rev+=num%base;

num/=base;

}

return( rev );

}

Answer 5

3

#include

#include

main()

{

int n, num, digit, sum=0, rev=0;

printf("Input the number");

n=num;

do

{

digit=num%10;

sum+=digit;

rev=rev*10+digit;

num!=10;

}

while (num!=0);

printf("sum of the digits of the number=%d%d%d%d\n",sum);

printf("reverse of the number=%d%d%d%d%d%d%d\n", rev);

if(n==rev)

printf("the number is a palindrome\n");

else

print("the number is not a palindrome\n");

getch();

}

OUTPUT:

Run1:- Input the number 1234

sum of the digits of the number= 10

reverse of the number= 4321

the number is not a palindrome

Run2:- Input the number 1221

sum of the digits of the number= 6

reverse of the number= 1221

the number is a palindrome

Answer 6

/* Palindrome program by Shekhar */

#include<stdio.h>

#include<conio.h>

void main() {

int n, i = 0, k = 0;

clrscr();

printf( "Enter number: " );

scanf( "%d", &n );

int p = n;

while( n != 0 ) {

k = k * 10 + ( n % 10 );

n = n / 10;

i++;

}

if ( k==p ) {

printf( "number is palindrome\n" );

} else {

printf( "number is not palindrome\n" );

}

getch();

}

Answer 7

Reverse the digits then test to see if the new number is the same as the original number. If it is, the number is a palindrome, otherwise it is not.

To reverse an unsigned integer (any base, default is 10):

unsigned rev (unsigned num, unsigned base=10)

{

unsigned r=0;

while (num)

{

r*=base;

r+=num%base;

num/=base;

}

return r;

}

Answer 8

Checking a number to see if it's palindromic in nature is a fun task. If the data were in arrays, all you would need to do is:

- calculate the halfway point in the array

- loop "count" from 0 to halfway-1 {

-- ch1 = arr[count]

-- ch2 = arr[array length - count - 1]

-- if ch1 and ch2 are not equal, return false

- }

- return true

If you're using an int, however, then to keep things simple (and probably faster) you may wish to break the integer down into each of its digits using the "%" modulus and "/" divide operators) while storing each of those digits in an array. Then pass that array into the function above. The bonus is that you know the maximum size of the array depends upon the size of the integer. If you're using a regular ol' "int", the maximum value's going to be about 2 billion, so 10 digits are going to be the max. Negative integers won't be palindromic because of the prefixed "-".

The author will leave it to you, the intrepid programmer, to write the code. Enjoy!

Answer 9

#include

#include

main()

{

int n, num, digit, sum=0, rev=0;

printf("Input the number");

n=num;

do

{

digit=num%10;

sum+=digit;

rev=rev*10+digit;

num!=10;

}

while (num!=0);

printf("sum of the digits of the number=%d%d%d%d\n",sum);

printf("reverse of the number=%d%d%d%d%d%d%d\n", rev);

if(n==rev)

printf("the number is a palindrome\n");

else

print("the number is not a palindrome\n");

getch();

}

Answer 10

just use the modulo (%) fcn to get the palindrome of the number.

the algorithm is for you to think so you'll know.

it is just easy. try it!