text: ( n & ( n - 1 ) ) = > Checking whether we can write 2 n as 2 ^ k if yes ( can 't represent 2n as 2^k) then answer 1  if no (can represent 2n as 2^k) then answer 0
code: return ( ( n & ( n - 1 ) ) != 0 ) ; } int main ( ) { long long int n = 10 ; printf ( " % lld " , canBeSumofConsec ( n ) ) ; }