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/*Program to find whether given no. is Armstrong or not. Example : Input - 153 Output - 1^3 + 5^3 + 3^3 = 153, so it is Armstrong no. */ class Armstrong{ public static void main(String args[]){ int num = Integer.parseInt(args[0]); int n = num; //use to check at last time int check=0,remainder; while(num > 0){ remainder = num % 10; check = check + (int)Math.pow(remainder,3); num = num / 10; } if(check == n) System.out.println(n+" is an Armstrong Number"); else System.out.println(n+" is not a Armstrong Number"); } }
class neonn { ; public static void check()throws IOException ; { ; BufferedReader br = new BufferedReader ( new InputStreamReader(System.in)); ; int num; ; System.out.println("Enter the number to be checked!"); ; num=Integer.parseInt(br.readLine()); ; int square; ; square=num*num; //squaring the number ; ; int sum=0; ; ; String sqs=Integer.toString(square); ; for(int i=0; i<sqs.length(); i++) ; sum+=Integer.parseInt(sqs.substring ( i,i+1 ) ) ; ; ; if (sum==num) //checking if the sum of the square is equal to the number entered ; System.out.println(num+" is a Neon Number!"); // if yes. It is neon ; else ; System.out.println(num+" is not a Neon number!"); //otherwise not ; ; } ; }
Dim x As String x = "HELLO" Dim tmpString As String tmpString = Nothing For j = 0 To x.Length - 1 Dim int int=x.Length - 1 int=int-j tmpString=x(j) Next
#include #include void main() { int rev num=0; while(num>0) { rev num=rev num*10+num%10; num=num%10; } return rev_num; } int main(); { int num=4562; printf("reverse of number is%d",reverse digit(num)); getch(); return o; }
Function is_prime(ByVal num As Integer) As Boolean If num < 0 Then num = num * -1 If num < 2 Then Return False If num Mod 2 = 0 Then Return num = 2 Dim max As Integer = Math.Sqrt(num) For div As Integer = 3 To max Step 2 If num Mod div = 0 Then Return False Next Return True End Function Sub Main() For num As Integer = 0 To 100 If is_prime(num) Then Console.Write(num) Console.WriteLine(" is prime") End If Next End Sub