Armstrong Number in C++

Armstrong number in C++: 153 = 1³+5³+3³. Count digits, then sum powers.

IntermediateLoop ProgramsExample 10 of 14

armstrong-number.cpp

1#include <iostream>
2#include <cmath>
3using namespace std;
4
5int main() {
6    int num, original, remainder, result = 0, n = 0;
7    
8    cout << "Enter a number: ";
9    cin >> num;
10    
11    original = num;
12    
13    // Count number of digits
14    int temp = num;
15    while (temp != 0) {
16        temp /= 10;
17        n++;
18    }
19    
20    temp = num;
21    // Calculate sum of digits raised to power n
22    while (temp != 0) {
23        remainder = temp % 10;
24        result += pow(remainder, n);
25        temp /= 10;
26    }
27    
28    if (result == original) {
29        cout << original << " is an Armstrong number" << endl;
30    } else {
31        cout << original << " is not an Armstrong number" << endl;
32    }
33    
34    return 0;
35}

Output

Enter a number: 153
153 is an Armstrong number

What's going on

An Armstrong number equals the sum of its digits each raised to the count of digits. 153 has 3 digits: 1³ + 5³ + 3³ = 1 + 125 + 27 = 153.

original = num;

You'll destroy num / temp. Keep the value to compare.

First loop — count digits:

temp = num;

while (temp != 0) { temp /= 10; n++; }

153 → 15 → 1 → 0. n is 3.

Second loop — sum powers:

temp = num;

remainder = temp % 10;

result += pow(remainder, n);

temp /= 10;

153: 3³=27, 5³=125, 1³=1. result 153.

if (result == original) match.

All one-digit numbers are Armstrong (5¹ = 5). 0 is. 370: 3³+7³+0³ = 27+343+0 = 370. 371 and 407 too. 123 is not.

pow returns double. For three digits it's exact. Huge n can round. Homework sizes are fine.

Two loops are required. You don't know the exponent until you've counted digits.

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