John
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Problem 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
题意：Nim取石子，取到最后一个的输
若局面异或和为不为0，定义其为S态，否则，定义其为T态
若一堆石子只有1个，定义其为孤独堆，否则，定义其为充裕堆
S0：无充裕堆，异或和不为0
S1：有1个充裕堆，异或和不为0
S2：有>=2个充裕堆，异或和不为0
T0：无充裕堆，异或和为0
T1不存在
T2：有>=2个充裕堆，异或和为0
S0：一定是有奇数个孤独堆，所以必败
T0：一定是有偶数个孤独堆，必胜
S1：若孤独堆个数为奇数，则拿空充裕堆，那么留给对方的是S0态，所以S1必胜
S2：可以转到S1、T2
T2：可以转到S1、S2
若T2转到了S2，则S2有转回了T2
若T2转到了S1，则T2必败
所以S2必胜
#includeusing namespace std;int main(){ int T,n,x,sum,yh; scanf("%d",&T); while(T--) { scanf("%d",&n); sum=yh=0; for(int i=1;i<=n;i++) { scanf("%d",&x); yh^=x; if(x>1) sum++; } if(yh&&!sum) printf("Brother\n"); else if(!yh&&sum>=2) printf("Brother\n"); else printf("John\n"); }}