hi_cf's blog

How to solve this problem(1) ..?
By hi_cf, 13 years ago, In English

The beauty of a number X is the number of 1s in the binary representation of X.

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.

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Comments Comments (13)
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13 years ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

Anyone Please ..

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13 years ago, hide # |
← Rev. 2  
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Usually solution to this type of problem goes like this: Do DP for numbers 1 — 200, print solutions and find pattern.

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13 years ago, hide # |
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You should only choose 0-s that are after 1-s, after that you sholdn't calculate beauty.

Ex:

100 - 010 = 010 OK. 1 moved right
100 - 001 = 011 bad, 1 moved to choosen 0 and all digits between them became 1s
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    13 years ago, hide # ^ |
     
    Vote: I like it +3 Vote: I do not like it

    After that it seems, it's not important how game are played and answer only depends of pairity (sum_of_number_of_position_of_ones — sum(0...n — 1)) where n is the number of ones.

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    13 years ago, hide # ^ |
     
    Vote: I like it 0 Vote: I do not like it

    "You should only choose 0-s that are after 1-s"

    Why should we leave other possible moves ..??

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      13 years ago, hide # ^ |
       
      Vote: I like it 0 Vote: I do not like it

      See, if we choose 1, it's disappered. It's bad.

      If we choose 0 consider nearest 1 on the left.

      (first)100000(second)
            ^ choose it

      After subtraction it will look like

      (first)011111(second)

      one 1 is disappeared, and [distance between chosen pos and pos of this 1] of 1's appered. So distance should be 1.

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13 years ago, hide # |
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actually choosing such k where (n-2^k) have the same beauty that means moving one of 1's bits one step to right for example let n=1000100 (in binary) you can either choose k=1 then n-2^k=1000010 or k=5 so n-2^k=0100100, there is no anothor valid value for k in the position.

I think the problem is too easy now :)

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