Armstrong Number using Do-While Loop in C++
What is Do-While Loop?
An iterative loop that checks the condition at the end.
The Do-While loop can be used whenever a test condition is specific, as the control enters the loop at a minimum once before the condition is evaluated. This loop checks the given condition after the execution of looping statements.
The loop actions or statements will be repeated infinitely as long as the test requirements are met.
Note: DO-WHILE LOOP means “first do it, then check”.
Syntax:
The Syntax of the DO-WHILE loop is:
Do
{
statement(s);
}
while(expression);
What are Armstrong Numbers?
Armstrong numbers are those natural positive numbers which are equal to the sum of cubes of each digit.
For example
a b c d... = a*n + b*n + c*n + d*
Following is Do-While loop program which test a number entered by the user as an input is Armstrong or not:
#include<iostream>
Using namespace std;
int main ()
{
int n, c, b, a = 0;
/* declaring various variables and initialization the value of 'a' as 0 */
cout<<"Input any natural number that you want to check whether it is an Armstrong number or not: "<<;
/* here we let user to enter a number via this program line */
Cin>>c;
/* assigning the value entered by the user to variable 'n' */
n=c;
/* initializing assignment operator and assigning the value ‘c’ to ‘n’ */
/* Loop to calculate the sum of the cubes of its digits */
do{
b=n % 10;
/* initializing assignment operator and assigning the value of ‘n %10’ to ‘b’ */
a=a +b*b*b;
/* initializing assignment operator and assigning the value of ‘a +b*b*b’ to ‘a’ */
n=n/10;
} while(n>0); /* while (test expression or termination condition) */
/* if-else condition to print Armstrong or Not */
if (a ==c)
/* initialising an ‘equal to’ operator and putting the condition to check if the value of a is equal to b or not */
cout<<"The number %d you have entered is an Armstrong number.\n"<<c;
else // else statement
cout<<"The number %d, that you have entered is not an Armstrong number.\n"<<c;
/* printing the value of the result as an output */
getch ();
}Output:
Input any natural number that you want to check whether it is an Armstrong number or not: 407
The number 407 that you have entered is an Armstrong number.
Explanation of above code:
Initially, we have declared four variables – n, a, b, c. Then, we initialized the value of variable ‘a’ as 0, it means the initial value of variable ‘a’ will be 0 so that we can avoid the value of variable ‘a’ from going into negative.
After that, we wrote a statement which will allow the user to enter the value for the variable ‘c’ with the ‘scanf’ command. The value in variable c holds that value which we want to test.
Now, we initialized a ‘do-while’ loop with the test expression or termination condition (n>0) which means the value of the variable ‘n’ must always be more than 0.
After the do-while loop, we initialized an if else statement and put the condition (n=c). It means that, if the value of ‘n’ is equal to ‘c’, the statement of the IF part will print. And, if the condition false, then the statement of the else part will print.
Here, you can see how the above code is execute:
1. n>0 (407>0) here you can see the ‘do-while’ loop test expression is true,
b=n%10 (b=407%10) So, the remainder of n%10 is 7.
a=a +b*b*b (a = 0 +7*7*7) So, the value of a is 343.
n=n/10 (n=407/10) So, the value of n is 40.
2. n>0 (40>0) here you can see the do-while loop test expression is true,
b=n%10 (b=40%10) So, the remainder of n%10 is 0.
a=a +b*b*b (a = 343 +0*0*0) So the value of a is now 343.
n=n/10 (n=40/10) So the value of n is now 4.
3. n>0 (4>0) here you can see the do-while loop test expression is true,
b=n%10 (b=4%10) So, the remainder of n%10 is 4.
a=a +b*b*b (a = 343 +4*4*4) So, the value of a is now 407.
n=n/10 (n=4/10) So, the value of n is now 0.
4. n>0 (0>0) here you can see the do-while loop test expression becomes false. And, the program execution ends.