C++ Program to Check Armstrong Number

In this article, you will learn and get code to check whether an input number is an armstrong number or not in C++. The program is created in these ways:

• Check Armstrong Number using while loop
• Check Armstrong Number using for loop

But before starting these programs, let's first understand about Armstrong number.

What is an Armstrong Number ?

A number that equals to the sum of its own digits, where each digit raised to the power of number of digits. For example, 1634 is an Armstrong number, because:

1634 = 14 + 64 + 34 + 44
     = 1 + 1296 + 81 + 256
     = 1297 + 337
     = 1634

The result is equal to the number itself. So it is an Armstrong number.

Note - Because the total number of digit in 1634 is 4, so each of its digit raised to the power of 4.

Check Armstrong Number using while Loop

This is the first program of this article. It asks from user to enter a number and checks whether it is an Armstrong number or not.

The question is, write a program in C++ that checks whether a given number by user at run-time, is an Armstrong number or not. Here is its answer:

#include<iostream>
using namespace std;
int main()
{
    intnum, temp, noOfDigit=0, res=0, rem, pow, i;
    cout<<"Enter the Number: ";
    cin>>num;
    temp = num;
    while(num>0)
    {
        num = num/10;
        noOfDigit++;
    }
    num = temp;
    while(num>0)
    {
        rem = num%10;
        pow = 1;
        i = 0;
        while(i<noOfDigit)
        {
            pow = pow*rem;
            i++;
        }
        res = res + pow;
        num = num/10;
    }
    if(res==temp)
        cout<<"\nIt is an Armstrong Number";
    else
        cout<<"\nIt is not an Armstrong Number";
    cout<<endl;
    return 0;
}

This program was build and run under Code::Blocks IDE. Here is its sample run:

Now supply any number say 1634 and press ENTER key to check whether it is an Armstrong number or not as shown in the snapshot given below:

And here is another sample run, with user input, 153:

Note - Because 153 is a 3-digit number, therefore each digit will raised to the power of 3

The following block of code:

while(num>0)
{
    num = num/10;
    noOfDigit++;
}

is used to count total number of digits available in the given number.

The dry run of above program with user input, 153 goes like:

• Initial values, noOfDigit=0, res=0
• After entering the number at run-time, say 153. It gets stored in num. So num=153
• Using temp=num;, the value of num gets initialized to temp. So temp=153
• Now the condition of while loop gets evaluated
• That is, the condition, num>0 or 153>0 evaluates to be true, therefore program flow goes inside the loop And num/10 or 153/10 or 15 gets initialized to num. So num=15
• The value of noOfDigit gets incremented. So noOfDigit=1 (because its initial value is 0)
• Program flow goes back and evaluates the condition of while loop again
• The process continues until its condition evaluates to be false. Before its condition evaluated to be false, the value of num and noOfDigit will be 0 and 3. So noOfDigit=3
• Now the statement num = temp; initializes the value of temp to num. So num=153
• The condition of second while loop gets evaluated. That is, the condition num>0 or 153>0 evaluates to be true, therefore program flow goes inside the loop
• There num%10 or 153%10 or 3 gets initialized to rem. So rem=3
• And pow=1, i=0
• Now the condition of third while loop gets evaluated. That is, the condition, i<noOfDigit or 0<3 evaluates to be true, therefore program flow goes inside the loop
• And pow*rem or 1*3 or 3 gets initialized to pow
• The value of i gets incremented. So i=1
• Program flow goes back and evaluates the condition, i<noOfDigit again
• The process continues until its condition evaluates to be false
• Before its condition evaluated to be false, the value of pow will be 27. That is the result of 3³
• So the statement, res = res + pow; initializes the value of res + pow (or 0+27 or 27) to res. So res=27
• Finally num/10 or 153/10 or 15 gets initialized to num. So num=15
• Program flow again goes back and evaluates the condition, num>0 (of while loop) again
• The process continues, until its condition evaluates to be false
• After its condition evaluated to be false, the value of res will be 153
• And after exiting from the loop, do compare res with original value entered by user
• If both is equal, then it is an Armstrong number, otherwise it is not an Armstrong number

Check Armstrong Number using for Loop

Now let's create the same program as of previous one, using for loop. That is, this program does the same job as of previous program. The only difference is, in place of while loop, we've used for loop here:

#include<iostream>
using namespace std;
int main()
{
    intnum, temp, noOfDigit=0, res=0, rem, pow, i;
    cout<<"Enter the Number: ";
    cin>>num;
    for(temp=num; temp>0; temp=temp/10)
        noOfDigit++;
    for(temp=num; temp>0; temp=temp/10)
    {
        rem = temp%10;
        pow = 1;
        for(i=0; i<noOfDigit; i++)
            pow = pow*rem;
        res = res + pow;
    }
    if(res==num)
        cout<<"\nIt is an Armstrong Number";
    else
        cout<<"\nIt is not an Armstrong Number";
    cout<<endl;
    return 0;
}

This program produces exactly the same output as of previous program. Here is its sample run with user input, 371:

And here is the last sample run, with user input, 567:

Same Program in Other Languages

• C Check Armstrong or Not
• Java Check Armstrong or Not
• Python Check Armstrong or Not

C++ Online Test