C Program to check Armstrong Number

Given a number x, determine whether the given number is Armstrong number or not.

A positive integer of n digits is called an Armstrong number of order n (order is number of digits) if.

abcd... = pow(a,n) + pow(b,n) + pow(c,n) + pow(d,n) + ....

Example:

Input : 153
Output : Yes
153 is an Armstrong number.
1*1*1 + 5*5*5 + 3*3*3 = 153

Input : 120
Output : No
120 is not a Armstrong number.
1*1*1 + 2*2*2 + 0*0*0 = 9

Input : 1253
Output : No
1253 is not a Armstrong Number
1*1*1*1 + 2*2*2*2 + 5*5*5*5 + 3*3*3*3 = 723

Input : 1634
Output : Yes
1*1*1*1 + 6*6*6*6 + 3*3*3*3 + 4*4*4*4 = 1634

C++

// C++ program to determine whether the number is 
// Armstrong number or not 
#include<bits/stdc++.h> 
using namespace std; 
  
/* Function to calculate x raised to the power y */
int power(int x, unsigned int y) 
{ 
    if( y == 0) 
        return 1; 
    if (y%2 == 0) 
        return power(x, y/2)*power(x, y/2); 
    return x*power(x, y/2)*power(x, y/2); 
} 
  
/* Function to calculate order of the number */
int order(int x) 
{ 
    int n = 0; 
    while (x) 
    { 
        n++; 
        x = x/10; 
    } 
    return n; 
} 
  
// Function to check whether the given number is 
// Armstrong number or not 
bool isArmstrong(int x) 
{ 
    // Calling order function 
    int n = order(x); 
    int temp = x, sum = 0; 
    while (temp) 
    { 
        int r = temp%10; 
        sum += power(r, n); 
        temp = temp/10; 
    } 
  
    // If satisfies Armstrong condition 
    return (sum == x); 
} 
  
// Driver Program 
int main() 
{ 
    int x = 153; 
    cout << isArmstrong(x) << endl; 
    x = 1253; 
    cout << isArmstrong(x) << endl; 
    return 0; 
} 

Please refer complete article on Program for Armstrong Numbers for more details!

My Personal Notes

C++

Recommended Posts: