POJ 1704 —— Nim博弈
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Georgia and Bob
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
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题意是有无数个排成直线的格子，上面有n枚棋子，告诉你棋子的位置，Georgia和Bob轮流选择一枚棋子，使之左移一定长度，不能反超前面的棋子，谁不能走谁输，Georgia先走，输出赢家。
思路：我们按棋子奇偶去分类讨论。
偶数时，我们把每两个相邻的棋子作为一组，相当于Nim中的一堆，大小为两点间距离。则右边棋子移动就相当于从Nim的一堆石子中取若干个石子，左边棋子移动相当于把这一堆加上若干石子，但是对手又可以取走同样多的石子，故两种状态等效，不用考虑。
奇数时，我们在序列最左端放一枚石子，然后就和偶数时的情况相同了。
Nim博弈的证明及结论可以参考http://www.cnblogs.com/exponent/articles/2141477.html
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>
#include <cassert>
using namespace std;
///#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 1000 + 50;
const int MAXS = 10000 + 50;
const int sigma_size = 26;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
const int inf = 1 << 30;
#define eps 1e-8
const long long MOD = 1000000000 + 7;
const int mod = 100000;
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pii;
#define Bug(s) cout << "s = " << s << endl;
///#pragma comment(linker, "/STACK:102400000,102400000")
int n , p[MAXN];
void solve()
{
if(n & 1)p[n++] = 0;
sort(p , p + n);
int ans = 0;
for(int i = 0 ; i < n - 1; i += 2)
{
ans ^= (p[i + 1] - p[i] - 1);
}
if(ans)puts("Georgia will win");
else puts("Bob will win");
}
int main()
{
int t;
cin >> t;
while(t--)
{
scanf("%d" , &n);
for(int i = 0 ; i < n ; i++)
{
scanf("%d",&p[i]);
}
solve();
}
return 0;
}