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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 2940 | Accepted: 1432 |
Description
Little John is playing very funny game with his younger brother. There is one big box filled with M&Ms of different colors. At first John has to eat several M&Ms of the same color. Then his opponent has to make a turn. And so on. Please note that each player has to eat at least one M&M during his turn. If John (or his brother) will eat the last M&M from the box he will be considered as a looser and he will have to buy a new candy box.
Both of players are using optimal game strategy. John starts first always. You will be given information about M&Ms and your task is to determine a winner of such a beautiful game.
Input
The first line of input will contain a single integer T – the number of test cases. Next T pairs of lines will describe tests in a following format. The first line of each test will contain an integer N – the amount of different M&M colors in a box. Next line will contain N integers Ai, separated by spaces – amount of M&Ms of i-th color.
Constraints:
1 <= T <= 474,
1 <= N <= 47,
1 <= Ai <= 4747
Output
Output T lines each of them containing information about game winner. Print “John” if John will win the game or “Brother” in other case.
Sample Input
2 3 3 5 1 1 1
Sample Output
John Brother
Source
题意：在一般的nim博弈上稍作修改，就是最后取完的输，其他一样；
分析：这就牵扯到一个牛逼的定理了....SJ定理，然后就和基础nim一样了。
我们提出定理里的两个限制：1、SG函数为不为0。2、有没有某单一游戏的SG函数大于1。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 1e9;
const int MOD = 1e9+7;
#define ll long long
#define CL(a,b) memset(a,b,sizeof(a))
#define lson (i<<1)
#define rson ((i<<1)|1)
#define N 50010
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int main()
{
int T,n,x;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
int ans=0,cnt=0;
for(int i=0; i<n; i++)
{
scanf("%d",&x);
ans ^= x;
if(x > 1) cnt++;
}
if(ans==0)
{
if(cnt) cout<<"Brother"<<endl;
else cout<<"John"<<endl;
}
else
{
if(cnt) cout<<"John"<<endl;
else cout<<"Brother"<<endl;
}
}
return 0;
}
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原文地址：http://blog.csdn.net/d_x_d/article/details/52130103
