hdu 1907 John

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

233 5 111

Sample Output

JohnBrother

【分析】

简单的NIM博弈，虽然简单，但是原理对于我这样蒟蒻来说还是不能理解，等什么时候真正的懂了在来更新.

【代码】

#include <cstdio>#include <cstdlib>#include <cstring>#include <map>#include <list>#include <algorithm>#include <iostream>using namespace std;int ans(void){    int len;    cin>>len;    int a=0,b=1,c;    for(int i=0;i<len;i++)    {        cin>>c;        if(c!=1)            b=0;        a^=c;    }    if(b) cout << ((len&1)?"Brother":"John")<<endl;    else cout<<(a?"John":"Brother")<<endl;}int main() {//    freopen("in.txt","r",stdin);    int T;    cin>>T;    while(T--)    {        ans();    }    return 0;}