John
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2095 Accepted Submission(s): 1133
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
Source
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lcy
简单的nim博弈.....(但是要注意全部为1的情况)....那是就不是nim博弈啦！！
代码：我想说的是，要把名字看清楚，开始将Johe---打错了...打成jone ，哎！！，尼玛，那叫一个错呀！！
//http://acm.hdu.edu.cn/showproblem.php?pid=1907
#include<iostream>
#include<vector>
#include<cstdio>
using namespace std; int main()
{
int t,n,i,x;
bool flag ;
cin>>t;
while(t--)
{
scanf("%d",&n);
vector<int>num(n+);
x=,flag=false;
for(i=;i<n;i++)
{
scanf("%d",&num[i]);
x^=num[i];
}
for(i=;i<n;i++)
{
if(num[i]>)
{
flag=true;
break;
}
}
if(flag)
{
if(x) puts("John");
else puts("Brother");
}
else
{
if(n&)
puts("Brother"); //奇数 win
else
puts("John"); //偶数 win
}
}
return ;
}