Armstrong Number program in C and C++

In this tutorial, we are going to create a program in C and C++ to check whether a user-entered number is an Armstrong number or not but first let’s see what is an Armstrong number.

An Armstrong Number number is an n digit integer such that the sum of each digit to the power n is equal to the number itself, for example:

Ex-1 : 1634  n = 4  1^n + 6^n + 3^n + 4^n = 1634  Ex-2 : 371  n = 3  3^n + 7^n + 1^n = 371

Now let us see the code to check whether a number is Armstrong or not in C and C++.

C program to check whether a number is Armstrong or not.

#include<stdio.h> #include<math.h>  int main(){ int n,y,len = 0,x,sum = 0; printf("Enter a number: "); scanf("\%d",&n); y = n;//copy variable to preserve the value of the entered input  //below while loop will calculate the number of digits in the number. while(y!=0){ y = y/10; len++; //length of the number } y = n; while(n!=0){ x = n\%10;//this expression will give us each digit of the given number. sum += pow(x,len);//sum variable to calculate x1^n + x2^n +...+ xm^n n = n/10; } if(sum == y) printf("The number is an Armstrong Number."); else printf("Number is not an armstrong number.");  }

C++ program to check whether a number is Armstrong or not

#include<iostream> #include<math.h>  using namespace std;  int main(){ int n,y,len = 0,x,sum = 0; cout<<"Enter a number: "; cin>>n; y = n;//copy variable to preserve the value of the entered input  //below while loop will calculate the number of digits in the number. while(y!=0){ y = y/10; len++; //length of the number } y = n; while(n!=0){ x = n\%10;//this expression will give us each digit of the given number. sum += pow(x,len);//sum variable to calculate x1^n + x2^n +...+ xm^n n = n/10; } if(sum == y) cout<<"The number is an Armstrong Number."; else cout<<"Number is not an Armstrong Number.";  }

Output:

Enter a number: 371 The number is an Armstrong Number. --------------------------------

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