阶梯博弈

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


题意：给你一个 1 * n 的棋盘，里面在不同的位置放置着棋子，每次只能将一个位置棋子向前移动若干个位置，但不能超出前一个棋子的位置，当一个人在面对棋盘不能在移动时，就表示这个人输掉了比赛。
思路分析 ：这个问题也叫作阶梯博弈，即将两个两个位置捆绑在一起，相邻的两个位置之间的空格数量视为袋中石子的数量，从而转变为Nim 问题。当一对棋子的后一个向前移动，代表袋中石子的数量变少，当前一个棋子向前移动，代表袋中石子数量增加，这时下一个玩家只要将后一个棋子向前移动相同的位置，即回到了最初的状态。
　　　　如果棋子的个数是奇数个的话，就将第一个棋子和棋盘捆绑即可。　　
　　　　我们可能还会考虑一种情况，就是某个玩家故意破坏，使得问题无法转换为取石子，例如前一个人将某对中的前者左移，而当前玩家不将这对中的另一移动，则会导致本堆石子增多了，不符合nim。但是这种情况是不会出现的。因为赢家只要按照取石子进行即可获胜，而输家无法主动脱离取石子状态。如果输家想要让某堆石子增多，那么赢家只需要让该堆减少回原状，这样输家又要面临跟上一回合同样的局面。
代码示例：

int a[1005];

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int t, n;
    
    cin >> t;
    while(t--){
        scanf("%d", &n);
        for(int i = 1; i <= n; i++){
            scanf("%d", &a[i]);
        }
        sort(a+1, a+1+n);
        int s, p;
        if (n%2) s = a[1]-1, p = 2;
        else s = 0, p = 1;
        
        for(int i = p; i <= n; i += 2){
            s ^= (a[i+1]-a[i]-1);
        }
        if (s!= 0) printf("Georgia will win\n");
        else printf("Bob will win\n");
    }
    return 0;
}