POJ 1704 Georgia and Bob (博弈)

题意：给定一个格子，有一些位置有球，每次可以向前移动一些位置，但不能超过前一个小球，问谁会胜利。

析：由于两个人都是足够聪明，所以我们可以把每两个相邻的球分成一个组，因为先手移动前一个移动多少位置，后手也可以移动第后一个，

如果是奇数，可以在0号位置放一个，没有影响，然后就成一个Nim组合游戏了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2000 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

int main(){
  int T;  cin >> T;
  while(T--){
    scanf("%d", &n);
    vector<int> v;
    v.push_back(0);
    for(int i = 0; i < n; ++i){
      scanf("%d", &m);
      v.push_back(m);
    }
    sort(v.begin(), v.end());
    int ans = 0;
    for(int i = n; i > 0; i -= 2)  ans ^= (v[i] - v[i-1] - 1);
    if(ans)  puts("Georgia will win");
    else puts("Bob will win");
  }
  return 0;
}