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 notbool 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 Programint 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!
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