HDU_1907_基础博弈nim游戏
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John
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 4440 Accepted Submission(s):
2541
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.
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
哎呀卧槽。。。
之前看nim游戏的论文,性子太急,看了一半(拿最后一个胜的情况),然后以为自己弄懂了,然后遇到这个题就有点懵逼,再把论文看完,发现之前自己理解得并不是很透彻。
无论拿最后一个胜还是拿最后一个败,都是在争夺S1状态(只有一个充裕堆)。仔细一想,确实是这样,当到达S1状态,玩家可以决定下一个状态,即把下一个状态变为T0或S0
也就是说谁争夺到S1,谁就必胜。
在拿最后一个败的情况中,当所有堆为孤单堆时,需要特殊判断一下,原因容易想到。
博弈这方面,还需多多练习,多多揣摩!!!
#include<iostream> #include<cstdio> #include<cstring> #include<stdlib.h> #include<algorithm> #include<cmath> using namespace std; int main() { int t,n; scanf("%d",&t); while(t--) { scanf("%d",&n); int res=0,num; int flag=0; for(int i=0; i<n; i++) { scanf("%d",&num); if(num>1) flag=1; res^=num; } if(flag) { if(res>0) printf("John\n"); else printf("Brother\n"); } else { if(n%2==0) printf("John\n"); else printf("Brother\n"); } } return 0; }
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