POJ Georgia and Bob(Nim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
2
3
1 2 3
8
1 5 6 7 9 12 14 17
Sample Output
Bob will win

Georgia will win

假设零点处有一个石子，与已输入的n个石子共有n+1个石子，两两之间构成n个区间。

但n为奇数时，把第奇数个区间 的距离设为Nim博弈中的石子，当n为偶数时把第偶数个区间的石子设为Nim博弈中的石子。

具体为什么要有图才好解释，留给大家思考

AC代码：

# include <cstdio> # include <algorithm>using namespace std;int a[1010];int main(){int t, n, i, j, k, x;scanf("%d", &t);for(i=1; i<=t; i++){scanf("%d", &n);for(j=1; j<=n; j++){scanf("%d", &a[j]);}sort(a+1, a+1+n);a[0]=0;x=0;if(n%2==1){for(j=1; j<=n; j=j+2){x=(x^(a[j]-a[j-1]-1));} }else{for(j=2; j<=n; j=j+2){x=(x^(a[j]-a[j-1]-1));} }if(x){printf("Georgia will win\n");}else{printf("Bob will win\n");}}return 0;}