1134. Armstrong Number The k-digit number N is an Armstrong number if and only if the kth power of each digit sums to N. Given a positive integer N, return true if and only if it is an Armstrong number.
| 8 | * |
| 9 | */ |
| 10 | class ArmstrongNum { |
| 11 | |
| 12 | /* driver function commented |
| 13 | public static void main(String[] args) { |
| 14 | |
| 15 | Scanner kb = new Scanner(System.in); |
| 16 | int N = kb.nextInt(); |
| 17 | |
| 18 | int k = length(N); |
| 19 | |
| 20 | System.out.println(isArmstrongNumber(N, k)); |
| 21 | |
| 22 | kb.close(); |
| 23 | } |
| 24 | */ |
| 25 | |
| 26 | public static int length(int N) { |
| 27 | int length = 0; |
| 28 | |
| 29 | while (N != 0) { |
| 30 | N = N / 10; |
| 31 | length++; |
| 32 | } |
| 33 | |
| 34 | return length; |
| 35 | } |
| 36 | |
| 37 | public static boolean isArmstrongNumber(int N, int k) { |
| 38 | long powerSum = 0; |
| 39 | int n = N; |
| 40 | |
| 41 | while (n != 0) { |
| 42 | int rem = n % 10; |
| 43 | |
| 44 | powerSum += (int) Math.pow(rem, k); |
| 45 | |
| 46 | n /= 10; |
| 47 | } |
| 48 | System.out.println(powerSum); |
| 49 | return (powerSum==N); |
| 50 | } |
| 51 | |
| 52 | } |
nothing calls this directly
no outgoing calls
no test coverage detected