Count N | Java program for the above approach ; Function to find the value of x ^ y ; Stores the value of x ^ y ; Iterate until y is positive ; If y is odd ; Divide y by 2 ; Return the value of x ^ y ; Function to find the number of N - digit integers satisfying the given criteria ; Count of even positions ; Count of odd positions ; Return the resultant count ; Driver Code
import java . io . * ; class GFG { static int m = 1000000007 ; static int power ( int x , int y ) { int res = 1 ; while ( y > 0 ) { if ( ( y & 1 ) != 0 ) res = ( res * x ) % m ; y = y >> 1 ; x = ( x * x ) % m ; } return res ; } static int countNDigitNumber ( int N ) { int ne = N / 2 + N % 2 ; int no = ( int ) Math . floor ( N / 2 ) ; return power ( 4 , ne ) * power ( 5 , no ) ; } public static void main ( String [ ] args ) { int N = 5 ; System . out . println ( countNDigitNumber ( N ) % m ) ; } }