Armstrong Number Program

An integer number is called Armstrong number if sum of the cubes of its digits is equal to the number itself. For example:

370 is an armstrong number because:

370 = 3*3*3 + 7*7*7 + 0*0*0
    = 27 + 343 + 0
    = 370

Example: Check Armstrong Number using For loop

#include <iostream>  
using namespace std;    
int main() {
     int num, sum = 0, digit;
     cout<<"Enter a positive integer: ";
     cin>>num;
       for(int temp=num; temp!=0;){
        digit = temp % 10;
        sum = sum +(digit * digit * digit);
        temp = temp/10;
     }
       if(sum == num)
        cout<<num<<" is an Armstrong number.";
     else
        cout<<num<<" is not an Armstrong number.";
       return 0;
  }

Output

Enter a positive  integer: 370
370 is an Armstrong number.

You can also use while loop instead of for loop to check the Armstrong number:

Replace this part of the code:

for(int temp=num; temp!=0;){
      digit = temp % 10;
      sum = sum +(digit * digit * digit);
      temp = temp/10;
  }

With this

int temp = num;
  while(temp != 0)  {
     digit = temp % 10;
     sum = sum +(digit * digit * digit);
     temp = temp/10;
  }

Using While Loop

#include <iostream>  
using namespace std;    
int main() {
     int num, sum = 0, digit;
     cout<<"Enter a positive integer: ";
     cin>>num;
       int temp = num;
  while(temp != 0)  {
     digit = temp % 10;
     sum = sum +(digit * digit * digit);
     temp = temp/10;
  }
       if(sum == num)
        cout<<num<<" is an Armstrong number.";
     else
        cout<<num<<" is not an Armstrong number.";
       return 0;
  }

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