So, Facebook question is:
The beauty of a number X is the number of 1s in the binary representation of X.I have solved it with this perl-code:
Two players are plaing a game. There is number n written on a black board. The game is played as following:
Each time a player chooses an integer number (0 <= k) so that 2^k is less than n and (n-2^k) has as beautiful as n. Next he removes n from blackboard and writes n-2^k instead. The player that can not continue the game (there is no such k that satisfies the constrains) looses the game.
The First player starts the game and they play in turns alternatively. Knowing that both two players play optimally you have to specify the winner.
Input:
First line of the Input contains an integer T, the number of testcase. 0 <= T <= 5.
Then follow T lines, each containing an integer n.
n more than 0 and less than 10^9 +1.
Output
For each testcase print "First Player" if first player can win the game and "Second Player" otherwise.
Sample Input
7 1 2 8 16 42 1000 123
Sample Output
Second Player First Player First Player Second Player Second Player First Player Second Player
Explanation
In the first example n is 1 and first player can't change it so the winner is the second player.
In the second example n is 2, so the first player subtracts 1 (2^0) from n and the second player looses the game.
even :: B odd :: A
perl -e 'my $s = 8; my $n = 0; my $sum = 0; while($s) { if ($s&1) {$sum+=$n} else {++$n; } $s>>=1; }; print $sum&1 ? "First\n" : "Second\n"; '
Beautiful, isn't it?
