#include <iostream>
#include <algorithm>
#define ss(a) scanf("%d",&a)
#define ss64(a) scanf("%I64d",&a)
using namespace std;
int main()
{
int i,n,m,sum,cnt,j;
int temp;
while(~ss(n)) //加个n=0跳出结果超时了,醉了
{
while(n--)
{
temp=0;
ss(j);
for (i=0,sum=0;i<j;i++)
{
ss(m);
if (m>1)
temp=1;
sum ^=m;
}
if ((sum && temp) || (!sum && !temp))
cout<<"John"<<endl;
else
cout<<"Brother"<<endl;
}
}
return 0;
}
基本的Nim博弈
看了博弈总结之后才坐这道题,有关博弈总结可以看这里
http://blog.csdn.net/acm_cxlove/article/details/7854530 长篇大论
John
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 3395 Accepted Submission(s): 1910
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
Sample Output
Source
Southeastern Europe 2007