poj1704（变形Nim博弈）

地址：http://poj.org/problem?id=1704

Georgia and Bob

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 6618 Accepted: 1928

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

题意：向左移动棋子，移动旗子不能超其左边的棋子并且不能与之在同一格。移动最后一步的人胜利。

Nim博弈变形，将相邻两个棋子看成一个石子堆，两个棋子间距为石子的数目，进行Nim博弈。

原理：间距为石子数目，那么靠右的石子左移可看为从石子堆中取出石子，靠左的石子左移可看为从石子堆中添加石子，添加可被取出抵消，所以石子数目未变。而这题目胜利条件就是让所有棋子间距为0，即让石子数目为0。

#include<cstdio>#include<cmath>#include<iostream>#include<algorithm>using namespace std;int main(){int a[1010];int t,m,ans,i;scanf("%d",&t);while(t--){scanf("%d",&m);for(i=0;i<m;i++){if(i==0&&m%2==1){a[i]=0;      //当棋子有奇数个时可将0点也看为一个棋子m++;continue;}scanf("%d",&a[i]);}sort(a,a+m);ans=0;for(i=0;i<m;i+=2)ans^=(a[i+1]-a[i]-1);if(ans==0) puts("Bob will win");else puts("Georgia will win");}return 0;}