C program to Demonstrate the Recursive function.

Write a C program to demonstrate the recursive function.
Recursion  is the programming technique that a process invoking itself again and again. In this program, We reverse the given number and checks it is a palindrome or not. Read more about C Programming Language .

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#include<stdio.h>
#include<conio.h>

int main(){
clrscr();
int num,num1,rev;
printf("nEnter a number :n");
scanf("%d",&num);
num1=num;
//call recursive function
rev=reverse(num);

printf("nAfter reverse the number is :n%d",rev);
if(num1==rev){
printf("nnNumber %d is Palindromen",num1);
}else
{
printf("nnNumber %d is NOT a Palindromen",num1);
}

return 0;
}
int sum=0,r;
reverse(int num){
if(num){
r=num%10;
sum=sum*10+r;
reverse(num/10);
}
else
return sum;
return sum;
}

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C Program to find the strong number.

Write a C Program to find the given number is strong or not?
A number is called strong number if sum of the factorial of its digit is equal to number itself.
Example: 145 since 1! + 4! + 5! = 1 + 24 + 120 = 145.
Read more about C Programming Language .

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* You can use all the programs on www.c-program-example.com
* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
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*
* Happy Coding
***********************************************************/
#include<stdio.h>
#include<conio.h>

int main()
{
int num,i,f,r,sum=0,num1;
clrscr();
printf("Enter a number: ");
scanf("%d",&num);

num1=num;
while(num){
i=1,f=1;
r=num%10;

while(i<=r){
f=f*i;
i++;
}
sum=sum+f;
num=num/10;

}
if(sum==num1)
printf("%d is a strong number",num1);
else
printf("%d is not a strong number",num1);
getch();
return 0;

}
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C Program to add one to digits of a number

Write C Program to add one to digits of a number.
C Program that adds the 1 to each single digit of a number, i.e for Example 12345’s output is 23456.
If the digit is 9 it adds 1 and follows the carry system, 9 becomes 0 and 9’s left digit adds one more 1. I.e., 3491’s output is 4602. Read more about C Programming Language .

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* for personal and learning purposes. For permissions to use the
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* 
*                      Happy Coding
***********************************************************/

#include<stdio.h>
#include<conio.h>
int main()
{
 int num, sum = 0;
 int rem, check = 0;
 clrscr( );
 printf("Enter the required number:");
 scanf("%d",&num);
 printf("nGiven Number: %d",num);
 while(num>0) 
 {
  rem = num % 10;
  if(rem != 9)
  {
   if(check == 0) 
    sum = (10 * sum) + (rem + 1);
   else{
    sum = (10*sum) + (rem + 2);
    check = 0;
   }
  } 
  else{
   sum = (10 * sum) + 0;
   check = 1;
  }
  num = num/10;
 } 

 num = sum; sum=0;
 while(num > 0)
 {
  rem = num % 10;
  sum = (10*sum) + rem;
  num = num / 10;
 }
 printf("nAfter Adding one: %d",sum);
 getch( );
 return 0;
}
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C program to check whether a given 5 digit number is palindrome or not

Check for Palindromic number. C program to check whether a given 5 digit number is palindrome or not.

Problem Statement

Write a C program to check if a given 5 digit number is palindrome or not. Take input from the user, check if the given number is a palindromic number and display appropriate message in the end.

A number is said to be a palindromic number if it remains same when reversed. For example 12321 is a palindromic number, where as 12345 is not. You can read more about Palindromic number.

Solution

The program takes input number from the user. Then the digits of the given number are reversed. Reversed number is compared with the original number to check if they are equal. If they are equal, the original number is a palindrome. If they are not equal, then the number is not a palindrome.

The Program

/* C program to check whether a given 5 digit number is palindrome or not
* (c) www.c-program-example.com
*/
#include<stdio.h>
int main() {
int num, rem, i, rev = 0, num1, count = 0;
printf("Enter a five digit number:\n");
scanf("%d", &num);
num1 = num;
// reverse the given number
while(num > 0) {
rem = num % 10;
rev = rem + rev * 10;
num = num / 10;
count++;
}
if(count == 5) {
if(num1 == rev) { // if it's palindrome, reverse & original number will be same
printf("The Given Number %d is Palindrome", num1);
} else {
printf("The Given Number %d is NOT Palindrome",num1);
}
} else {
printf("The given %d number is not a five digit number!",num1);
}
return 0;
}
view raw palindrome.c hosted with ❤ by GitHub

Sample Output

Palindrome number or not

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C program to convert a Binary number into its equivalent Decimal, Octal and Hexadecimal numbers.

C Program to convert binary number into its equivalent Decimal, Octal, and Hexadecimal representations. In this program we convert a binary number to decimal. Then convert that decimal number to octal and Hexadecimal equivalent.

The program implements various functions to do the job.

  • binary_to_decimal : Convert given binary number to it’s decimal equivalent
  • decimal_to_octal : Convert from decimal to octal
  • decimal_to_hex : Convert from decimal to hexadecimal

The main function takes care of getting input from user, calling the above functions and displaying the results.

Read more here: What are binary, octal, and hexadecimal notation?

The Program

#include <stdio.h>
/*
* Function to convert a given decimal number to its
* hexadecimal equivalent.
*/
int decimal_to_hex(long number, char *hexstr, int pos) {
long i;
int new_position, n;
char hexchar;
new_position = pos;
if(number > 0) {
i = number % 16;
n = number / 16;
new_position = decimal_to_hex(n, hexstr, pos);
if(i >= 10) {
switch(i) {
case 10: hexchar = 'A'; break;
case 11: hexchar = 'B'; break;
case 12: hexchar = 'C'; break;
case 13: hexchar = 'D'; break;
case 14: hexchar = 'E'; break;
case 15: hexchar = 'F'; break;
}
} else {
hexchar = '0' + i;
}
hexstr[new_position++] = hexchar;
hexstr[new_position] = '\0';
}
return new_position;
}
/*
* Function to convert a given binary number to
* its decimal equivalent.
*/
int binary_to_decimal(int num) {
int rem, base = 1, decimal_number = 0;
while( num > 0) {
rem = num % 10;
if((rem == 0) || (rem == 1)) {
decimal_number = decimal_number + rem * base;
num = num / 10 ;
base = base * 2;
} else {
return -1; // Invalid binary number
}
}
return decimal_number;
}
/*
* Function to convert a given decimal number in to
* its octal equivalent.
*/
int decimal_to_octal(int number) {
int octal_number = 0, rem, base = 1;
while(number > 0) {
rem = number % 8;
octal_number = octal_number + rem * base;
base = base * 10;
number = number / 8;
}
return octal_number;
}
void main() {
int binary_numer, decimal_number, octal_number;
char hexstr[25] = "0";
printf("Enter a binary number: ");
scanf("%d", &binary_numer);
printf("Given Binary number is: %d\n", binary_numer);
decimal_number = binary_to_decimal(binary_numer);
if(decimal_number == -1) {
printf("\nInvalid binary number. Try again.");
return;
}
printf("Its Decimal equivalent is: %d", decimal_number);
octal_number = decimal_to_octal(decimal_number);
printf("\nIts octal equivalent is: %d", octal_number);
decimal_to_hex(decimal_number, hexstr, 0);
printf("\nIts Hexa Decimal equivalent is: %s", hexstr);
}

Sample Output

Binary to decimal, binary to octal, binary to hexadecimal

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Edit1 : 11th August 2017

  • Added screenshot of sample runs
  • Moved the code to it’s own Gist
  • Modified the program to make it more modular

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C Program to check the given number is Armstrong or not?

C Program to check the given number is Armstrong number or not?. Armstrong number is a number that is the sum of its own digits each raised to the power of the number of digits. Example: 153 = 1^3 + 5^3 + 3^3. Read more about C Programming Language .

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* programs for commercial purposes,
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* and browse!
*
* Happy Coding
***********************************************************/


#include<stdio.h>
void main()
{
int num,num1,arms=0,rem;

printf("Enter the number:n");
scanf("%d",&num);
num1=num;
while(num>0)
{
rem=num%10;
arms=arms+rem*rem*rem;
num=num/10;
}
if(num1==arms)
{
printf(" n%d is an Armstrong number",num1);
}
else
{
printf("n%d is NOT an Armstrong number",num1);
}

}
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C Program to convert a Roman numeral to its decimal equivalent.

C Program to convert a Roman numeral to its decimal equivalent. Roman numbers are the oldest number system used in ancient Rome. They use the combination of letters from Latin alphabet to represent the system. We used the if-else and for statements to solve this problem. Read more about C Programming Language .

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* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
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* and browse!
*
* Happy Coding
***********************************************************/



#include<stdio.h>
#include<conio.h>
#include<string.h>
#include<stdlib.h>

void main()
{

int *a,len,i,j,k;
char *rom;

clrscr();

printf("Enter the Roman Numeral:");
scanf("%s",rom);

len=strlen(rom);

for(i=0;i<len;i++)
{
if(rom[i]=='I')
a[i]=1;
else if(rom[i]=='V')
a[i]=5;
else if(rom[i]=='X')
a[i]=10;
else if(rom[i]=='L')
a[i]=50;
else if(rom[i]=='C')
a[i]=100;
else if(rom[i]=='D')
a[i]=500;
else if(rom[i]=='M')
a[i]=1000;
else
{
printf("nInvalid Value");
getch();
exit(0);
}
}
k=a[len-1];
for(i=len-1;i>0;i--)
{
if(a[i]>a[i-1])
k=k-a[i-1];
else if(a[i]==a[i-1] || a[i]<a[i-1])
k=k+a[i-1];
}
printf("nIts Decimal Equivalent is:");
printf("%d",k);
getch();
}
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C Program to convert a given decimal number into its binary equivalent

C Program to convert a given decimal number into its binary equivalent. In this program we convert the given decimal based number system(0, 1, 2, 3, —–,9) to Binary number system(0 and 1). Read more about C Programming Language .

/***********************************************************
* You can use all the programs on www.c-program-example.com
* for personal and learning purposes. For permissions to use the
* programs for commercial purposes,
* contact [email protected]
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* and browse!
*
* This program was originally published at
* http://www.c-program-example.com/2011/09/you-can-use-all-programs-on-www_8419.html
* Happy Coding
***********************************************************/
/* Write a program to convert a given decimal numbers into its binary equivalent . */
#include<stdio.h>
int main()
{
int num,i=0,binary=0,num1, biny[30],counter=0;
printf("Enter the Decimal number:\n");
scanf("%d",&num);
num1=num;
do
{
binary=num%2;
biny[i]=binary;
num=num/2;
i++;
counter++;
} while(num>0);
printf("\nThe given decimal %d number binary equalant is :\n",num1);
counter--;
while(counter>=0){
printf("%d" ,biny[counter]);
counter--;
}
return 0;
}

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