How to use Brute-force Approach In C++

The idea is to first count the number of digits (or find order). Let the number of digits be order_n. For every digit curr in input number num, compute curr^order_n. If the sum of all such values is equal to num, then return true, else false.

Below is the C++ program to find Armstrong numbers between 1 to 1000 using a brute force approach:

C++

// C++ program to find Armstrong numbers 
// between 1 to 1000 using a brute force 
// approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the order of
// a number.
int order(int num)
{
    int count = 0;
    while (num > 0)
    {
        num /= 10;
        count++;
    }
    return count;
}
 
// Function to check whether the
// given number is Armstrong number
// or not
bool isArmstrong(int num)
{
    int order_n = order(num);
    int num_temp = num, sum = 0;
 
    while (num_temp > 0)
    {
        int curr = num_temp % 10;
        sum += pow(curr, order_n);
        num_temp /= 10;
    }
    if (sum == num)
    {
        return true;
    }
    else
    {
        return false;
    }
}
 
// Driver code
int main()
{
 
    cout << "Armstrong numbers between 1 to 1000 : ";
    // Loop which will run from 1 to 1000
    for (int num = 1; num <= 1000; ++num)
    {
 
        if (isArmstrong(num))
        {
            cout << num << " ";
        }
    }
    return 0;
}

Output

Armstrong numbers between 1 to 1000 : 1 2 3 4 5 6 7 8 9 153 370 371 407

Time Complexity: O(n*d), where d is the order of a number and n is the range(1, 1000).
Auxiliary Space: O(1).

C++ Program to Print Armstrong Numbers Between 1 to 1000

Here, we will see how to print Armstrong numbers between 1 to 1000 using a C++ program.

Tags:

#C Basic Programs #C++ #C++ Programs #CPP

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C++ Program to Print Armstrong Numbers Between 1 to 1000

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