Georgia and Bob（尼姆博弈）

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

231 2 381 5 6 7 9 12 14 17

Sample Output

Bob will winGeorgia will win

解题思路：

对相邻棋子进行两两配对。比如有6个棋子，就配成3对，若是奇数个棋子，就把第一个单独拿出来。配对后，相邻两对棋子间的距离对问题的解决是不构成影响的。比如第一对的后一个棋子走了a步，第二对棋子的前一个可以先走a步抵消影响后再选择自己的步数。这样就转化成了给几段步数，选择一段后再选择步数，看谁先走完。典型的尼姆博弈，不过多赘述。

AC代码：

#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int maxn = 1005;int main(){    int t, n, sum, a[maxn];    scanf("%d", &t);    while(t--)    {        sum = 0;        scanf("%d", &n);        for(int i = 0; i < n; i++)            scanf("%d", &a[i]);        sort(a, a + n);        int ans = 0;        if(n % 2 == 0)            for(int i = 1; i < n; i += 2)                ans ^= a[i] - a[i - 1] - 1;        else        {            for(int i = 2; i < n; i += 2)                ans ^= a[i] - a[i - 1] - 1;            ans ^= a[0] - 1;        }        if(ans)            printf("Georgia will win\n");        else            printf("Bob will win\n");    }    return 0;}