C and C++ Programs to Reverse a Number

This post includes a C program and a C++ program to reverse a given number. That is, to reverse the order of digits. If 543 is entered, it will be displayed as 345.

C Program:

#include<stdio.h>
main()
{

C and C++ Programs to Check Whether a Number is Strong Number or Not

This post contains C and C++ program to check whether a given number is Strong number or not. A strong number is a number for which the sum of factorials of its digits is equal to the number itself. The first one is C program to check whether the input number is a strong number or not. An example is 145. 1!+4!+5!=1+24+120=145 Therefore 145 is a strong number.

C Program:

#include<stdio.h>
long int factorial(int n)

{

Binary to Gray Code Converter

This post is to teach you how to convert binary number to corresponding Gray code. The conversion is so simple. You can see it..

If an n bit binary number is represented by B_n B_n-1 ...B₁ and its Gray code equivalent by G_n G_n-1...G₁ where B_n and G_n are the most significant bits (MSBs), then the Gray code bits are obtained from the binary code as follows. The symbol ⊕ stands for the Exclusive OR (XOR) operation explained below.

The conversion procedure is as follows:

C and C++ Programs to Find Sum of Digits of a Given Number

The following are C and C++ programs to find sum of digits of any given integer.

C Program

#include<stdio.h>
#include<conio.h>
void main()
{

C and C++ Programs to Check Whether a Given Number is Armstrong Number or Not

This post contains two programs, a C++ program and a C program. These are programs to check whether a given number is an Armstrong number or not. An Armstrong number is a positive integer for which the sum of 'n'th power of all of its digits is equal to the number itself. For example, 153 is a 3 digit Armstrong number because 1³+5³+3³=153.

C ++ program:

# include <iostream.h>

# include <conio.h>

# include <math.h>

void main ()

{

clrscr();

int a,b=0,sum=0;

long int n;

cout<<"Enter the number to be checked\n ";

cin>>n;

if(n<1)
{
cout<<"\nThe number should be greater than 0";
}
else
{
a=n;
//counting the digits
while (a>0) 
{
a=a/10; 
b++;
}
a=n;
//adding up bth power of digits
while(a>0)
{
sum=sum+pow(( a%10) ,b);
a=a/10;
}
if(sum==n)
cout<<"\nThe number is an ARMSTRONG number";
else
cout<<"\nThe number is NOT an ARMSTRONG number";
}
getch();
}

C Program:

# include <stdio.h>

# include <conio.h>

# include <math.h>

void main ()

{

clrscr();

int a,b=0,sum=0;

long int n;

printf("Enter the number to check\n ");

scanf("%i",&n);
if(n<1)
{
printf ("\nThe number should be greater than 0");
}
else
{
a=n;
//counting the digits
while (a>0) 
{
a=a/10; 
b++;
}
a=n;
//adding up bth power of digits
while(a>0)
{
sum=sum+pow(( a%10) ,b);
a=a/10;
}
if(sum==n)
printf ("\nThe number is an ARMSTRONG number");
else
printf ("\nThe number is NOT an ARMSTRONG number");
}
getch();
}