对于任意一个 Anti-SG 游戏，如果我们规定当局面中所有的单一游戏的 SG 值为 0 时，游戏结束，则先手必胜当且仅当：
（1）游戏的 SG 函数不为 0 且游戏中某个单一游戏的 SG 函数大于 1；
（2）游戏的 SG 函数为 0 且游戏中没有单一游戏的 SG 函数大于 1。
John
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 2233 Accepted Submission(s): 1200
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
Recommend
lcy
#include <iostream> #include <cstdio> #include <cstring>
using namespace std;
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,XOR=0,a,k=0;
scanf("%d",&n);
while(n--)
{
scanf("%d",&a);
XOR^=a;
if(a>1) k++;
}
if((XOR==0&&k==0)||(XOR!=0&&k!=0))
puts("John");
else
puts("Brother");
}
return 0;
}
|
* This source code was highlighted by YcdoiT. ( style: Codeblocks )
