Question

from 1 to n how many armstrongs are present; n is an integer.

Answer 1

#include <bits/stdc++.h>

using namespace std;

  

// Prints Armstrong Numbers in given range

void findArmstrong(int low, int high)

{

    for (int i = low+1; i < high; ++i) {

  

        // number of digits calculation

        int x = i;

        int n = 0;

        while (x != 0) {

            x /= 10;

            ++n;

        }

  

        // compute sum of nth power of

        // its digits

        int pow_sum = 0;

        x = i;

        while (x != 0) {

            int digit = x % 10;

            pow_sum += pow(digit, n);

            x /= 10;

        }

  

        // checks if number i is equal to the

        // sum of nth power of its digits

        if (pow_sum == i)

            cout << i << " ";     

    }

}

  

// Driver code

int main()

{

    int num1 = 100;

    int num2 = 400;

    findArmstrong(num1, num2);

    cout << '\n';

    return 0;

}