John

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 3762 Accepted Submission(s): 2127

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

3 5 1

Sample Output

John

Brother

Source

Southeastern Europe 2007

n堆物品，每次从一堆中取至少一个，取到最后一个的败，John先取，Brother后取，输出胜利者的名字

这种类型的尼姆博弈，必胜态有S2,S1,T0，必败态有T2,S0，具体参考前面关于三种博弈的总结。

int main() {

    int T, n;

    scanf("%d", &T);

    while (T --) {

        scanf("%d", &n);

        int res = , cnt = , tmp;

        for (int i = ; i <= n; i ++) {

            scanf("%d", &tmp);

            res ^= tmp;

            if (tmp > ) cnt ++;

        }

        if ((res ==  && cnt == ) || (res && cnt)) printf("John\n");

        else printf("Brother\n");

    }

    return ;

}

巴特西

HDU1907（尼姆博弈）

John

最新文章

热门文章