poj1704 Georgia and Bob------P/N分析

Georgia and Bob

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 5821		Accepted: 1637

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

Source
POJ Monthly--2004.07.18


题意：给一个1*M的棋盘，上面有N颗棋子，每次只能向左移动棋子，并且至少移动一步，两人轮流操作，谁不能移动谁就输了。
分析这题的P/N点时，和往常不一样，不是从小数据推测，而是直接从性质推测，从右往左，把棋子分为两两一组，如果是奇数颗棋子，就补第一颗的位置为0，当对手移动两颗棋子的左边的棋子时，我们只需要把右边这颗棋子移动相同的步数就行，当对手移动右边的棋子时，把两颗棋子之间的距离看为石子的个数，我们只需按照取石子的规则走就行。

#include<iostream>
#include<cstdlib>
#include<stdio.h>
#include<algorithm>
using namespace std;
int a[1010];
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        scanf("%d",&a[i]);
        int ans=0;
        sort(a,a+n);
        if(n&1)
        {
            ans^=(a[0]-1);
            for(int i=2;i<n;i+=2)
            ans^=(a[i]-a[i-1]-1);
        }
        else
        for(int i=1;i<n;i+=2)
        ans^=(a[i]-a[i-1]-1);
        //cout<<ans<<endl;
        if(ans!=0)
        puts("Georgia will win");
        else
        puts("Bob will win");
    }
}
/*
3
3
1 2 3
8
1 5 6 7 9 12 14 17
Bob will win
Georgia will win*/