//This code is solution to Armstrong numbers task. //An Armstrong number is a number that is the sum of its own digits each raised //to the power of the number of digits. For example, 153 is an Armstrong //number: 1^3+5^3+3^3=153. Program checks whether a given number is an Armstrong //number or not. #include #include #include #include using namespace std; void armstrong(int a, vector& arm_num) { vector digits; for (int b = 0; b <= a; b++) { digits.clear(); int n = b; while (n) { digits.push_back(n % 10); n = (n - (n % 10)) / 10; } n = digits.size(); int sum = 0; for (int i = 0; i < n; i++) sum += pow(digits[i], n); if (b == sum) arm_num.push_back(b); } } int main() { cout << "Write an integer.\n"; int a; cin >> a; a = (a < 0 ? -a : a); cout << "\nYour number is " << a << ".\n"; vector digits; int n = a; while (n) { digits.push_back(n % 10); n = (n - (n % 10)) / 10; } n = digits.size(); int sum = 0; for (int i = 0; i < n; i++) sum += pow(digits[i], n); if (a == sum) cout << "Number " << a << " is an Armstrong number:\n"; else cout << "Number " << a << " is not an Armstrong number:\n"; for (int i = n - 1; i >= 0; i--) { cout << digits[i] << "^" << n; if (i != 0) cout << " + "; else cout << " = " << sum; } cout << "\n\nAll Armstrong numbers from 0 to " << a << ":\n"; vector arm_num; armstrong(a, arm_num); n = 0; for (int i = 0; i < arm_num.size(); i++) { cout << arm_num[i] << " "; n++; if (n == 9) { n = 0; cout << endl; } } cout << endl; system("pause"); return 0; }

28. Armstrong numbers