题目:
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.
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
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
分析:
anti-nim游戏:谁最后走最后一步谁输,先手赢的必胜策略是
1)所有堆石子数都为1而SG值为0;
2)存在某堆石子数不为1且SG值不为0.
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cstdio> int main() { int t,n,a[50],ans; int flag; scanf("%d",&t); while(t--) { flag=1; ans=0; scanf("%d",&n); for(int i=0;i<n;i++) { scanf("%d",&a[i]); if(a[i]!=1) flag=0; ans^=a[i]; } if(flag) { if(n&1) printf("Brother "); else printf("John "); } else { if(ans) { printf("John "); } else printf("Brother "); } } return 0; }