C++ Program to Check the Number is Armstrong Number or Not
This is C++ Program to Check the Number is Armstrong Number or Not || C++ Programming Tutorial #8(beginner) . you will learn to check whether a number entered by the user is an Armstrong number or not. To understand this example, you should have the knowledge of the following C++ programming topics:
C++ if, if...else and Nested if...else
C++ while and do...while Loop
SOME PRACTICE QUESTION :
Q.1) Write a program in ‘C’ to print your name 5 times using while loop.
Q.2) Write a program in ‘C’ to input any integer and print your name that many times.
Q.3) Write a program in ‘C’ to print the series as 1 2 3 4 5 6 7 ..........n, where ‘n’ is user input.
Q.4) Write a program in ‘C’ to print the series as 1 3 7 15 31 ..........n, where n is given by user.
Q.5) Write a program in ‘C’ to find out sum of the following series:
1+2+3+4+ ……………. +n, where n is given by user.
Q.6) Write a program in ‘C’ to calculate the factorial of a given number.
Q.7) Write a program in ‘C’ to calculate the sum of digits of a given number.
source code :
/*
ARMSTRONG NUMBER : a number that is the sum of its own digits each raised to the power of the number of digits.
example : 153,1634
153 = 1^3 + 5^3 + 3^3 = 153
1634 = 1^4 + 6^4 + 3^4 + 4^4 = 1634
*/
#include<iostream>
#include<math.h>
using namespace std;
int main(){
int n,count=0;
cout<<"enter the number : ";
cin>>n;
int n1=n,rem,sum=0;
int n2=n;
while(n!=0){
n =n /10; //0
count ++;
}
while(n1!=0){
rem = n1 % 10;
sum= sum + pow(rem,count);
n1 = n1/10;
}
if (n2==sum )
cout<<"Armstrong number ";
else
cout<<"not armstrong number ";
return 0;
}
This is dummy text. It is not meant to be read. Accordingly, it is difficult to figure out when to end it. But then, this is dummy text. It is not meant to be read. Period.
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