hdu1907
2014-07-07 20:08
477 查看
John
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 2659 Accepted Submission(s): 1462
[align=left]Problem Description[/align]
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.
[align=left]Input[/align]
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
[align=left]Output[/align]
Output T lines each of them containing information about game winner. Print “John” if John will win the game or “Brother” in other case.
[align=left]Sample Input[/align]
2
3
3 5 1
1
1
规定吃完最后一颗的就输了,anti_sg,我也不是很懂,就是当所有堆都是1的时候特判,然后有一堆大于1的时候就按照nim游戏来做
博弈实在弱爆了!被博弈各种爆菊!
#include<cstdio> #include<vector> #include<utility> #include<cstring> #include<iostream> #include<algorithm> #define LL long long #define MM 100010 using namespace std; int n; int main() { int T; //freopen("D:\\o.txt", "r", stdin); scanf("%d", &T); while (T--) { int sum = 0, num = 0; scanf("%d", &n); for (int i = 1; i <= n; i++) { int x; scanf("%d", &x); if (x > 1)num++; sum ^= x; } if (num == 0) { if (n % 2 == 1)puts("Brother"); else puts("John"); } else { if (sum != 0)puts("John"); else puts("Brother"); } } return 0; }
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