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
题目大意：
给一个1*M的棋盘，上面有N颗棋子，每人每次只能向左移动一颗棋子，并且至少移动一步，两人轮流操作，如果某人不能移动棋子了，那他就输了。
解题思路：
将棋子分成两两一组，如果棋子数是奇数，前面补0。如果先手移动的是一组当中左边的棋子，那么后手只要将该组右边的棋子移动相同的步数，就行了；如果先手移动的是右边的棋子，将两颗棋子之间的距离看作是石子的数目，只需按照取石子的操作即可。本题是尼姆博弈类型，有N堆石子，两人轮流取，每人每次至少取一个，最后取完石子的人赢。
将棋子分组后，两两棋子之间的距离为L1，L2，L3，……，Ln，则L1^L2^L3^……^Ln=0时，Bob赢，否则，Georgia赢。
代码如下：
#include<stdio.h>
#include<algorithm>
using namespace std;
int main()
{
int i,j,t,N,T,P[1005];
scanf("%d",&T);
while (T--)
{
scanf("%d",&N);
for(i=1;i<=N;i++)
scanf("%d",&P[i]);
sort(P+1,P+1+N);//从小到大排序
t=0;
j=1;
if(N%2==1)//N是奇数时
{
t=P[1]-1;
j++;
}
for(i=j;i<=N;i+=2)
{
t^=P[i+1]-P[i]-1;
}
if(t==0)
puts("Bob will win");
else
puts("Georgia will win");
}
return 0;
}
B - Georgia and Bob