Has anyone solve this problem through dp??
if yes pls share approach
I did this with dp. The idea is to maintain two variables (number of odd numbers and number of even numbers). Have a look at my submission 
My solution : CodeChef: Practical coding for everyone
There are two states .dp[i][0] indicates count of number having summation as even until current i and dp[i][1] indicates count having summation as odd.
now u can build recuurence relation as:
dp[i][0]=dp[i-1][1]* (count of odd numbers in current index)+dp[i-1][0]* (count of even numbers in current index)
dp[i][1]=dp[i-1][0]* (count of odd numbers in current index)+dp[i-1][1]* (count of even numbers in current index)
finally the ans will be dp[n][1]