[POJ1704]Georgia and Bob

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8552		Accepted: 2704

Description

Georgia and Bob decide to play a self-invented game. They draw a row of grids on paper, number the grids from left to right by 1, 2, 3, ..., and place N chessmen on different grids, as shown in the following figure for example: 

Georgia and Bob move the chessmen in turn. Every time a player will choose a chessman, and move it to the left without going over any other chessmen or across the left edge. The player can freely choose number of steps the chessman moves, with the constraint that the chessman must be moved at least ONE step and one grid can at most contains ONE single chessman. The player who cannot make a move loses the game. 

Georgia always plays first since "Lady first". Suppose that Georgia and Bob both do their best in the game, i.e., if one of them knows a way to win the game, he or she will be able to carry it out. 

Given the initial positions of the n chessmen, can you predict who will finally win the game? 

Input

The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. Each test case contains two lines. The first line consists of one integer N (1 <= N <= 1000), indicating the number of chessmen. The second line contains N different integers P1, P2 ... Pn (1 <= Pi <= 10000), which are the initial positions of the n chessmen.

Output

For each test case, prints a single line, "Georgia will win", if Georgia will win the game; "Bob will win", if Bob will win the game; otherwise 'Not sure'.

Sample Input

2
3
1 2 3
8
1 5 6 7 9 12 14 17

Sample Output

Bob will win
Georgia will win

Source

POJ Monthly--2004.07.18

Thinking

　　可以转变成一个Nim问题。把棋子两两成对进行考虑，分成两组，每组的Nim值即为棋子之间的间隔数。

　　右移视为减小Nim值，左边棋子左移视为增加Nim值，然而不管怎么增加，对手总有办法将其恢复，所以必胜态和必败态和Nim游戏相同。

　　当棋子个数为奇数时，需要把第一个棋子和他前面的空格分为一组，剩下的分为一组即可。

var a:array[0..10000] of longint;
    n,i,t:longint;

procedure sort(l,r: longint);
      var
         i,j,x,y: longint;
      begin
         i:=l;
         j:=r;
         x:=a[(l+r) div 2];
         repeat
           while a[i]<x do
            inc(i);
           while x<a[j] do
            dec(j);
           if not(i>j) then
             begin
                y:=a[i];
                a[i]:=a[j];
                a[j]:=y;
                inc(i);
                j:=j-1;
             end;
         until i>j;
         if l<j then
           sort(l,j);
         if i<r then
           sort(i,r);
      end;

procedure main();
var x,i,j:longint;
begin
    fillchar(a,sizeof(a),0);
    readln(n);
    for i:=1 to n do
        read(a[i]);
    sort(1,n);
    x:=0;
    j:=1;
    if (n mod 2=1) then
        begin
            x:=a[1]-1;
            inc(j);
        end;
    while j<=n do
        begin
            x:=x xor(a[j+1]-a[j]-1);
            inc(j,2);
        end;
    if x=0 then writeln('Bob will win')
    else writeln('Georgia will win');
end;

begin
    readln(t);
    for i:=1 to t do main();
end.