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.
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
T lines each of them containing information about game winner. Print
“John” if John will win the game or “Brother” in other case.
John
Brother
题意:在n堆中取糖吃,每次可以在其中的一堆中取至少一个,谁取完所有的糖果就输了,问最后谁会赢?
此游戏可以简化为nim博弈模型:
今有若干堆火柴，两人依次从中拿取，规定每次只能从一堆中取若干根， 可将一堆全取走，但不可不取，最后取完者为负，求必胜的方法。
解法:定义：若所有火柴数异或为0，则该状态被称为利他态，用字母T表示；否则， 为利己态，用S表示。
定义：T态中，若充裕堆的堆数大于等于2，则称为完全利他态，用T2表示；若充裕堆的堆数等于0，则称为部分利他态，用T0表示。
[定理5]：S0态，即仅有奇数个孤单堆，必败。T0态必胜。
[定理6]：S1态，只要方法正确，必胜。
中间过程不再赘述:
所以这题求出John的必输态即可.
#include <iostream>
#include <cstring>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <map>
using namespace std;
int main()
{
int tcase;
scanf("%d",&tcase);
while(tcase--){
int n,sum = 0,v,_cnt=0,_cnt1=0; ///_cnt 代表孤单堆的数量,_cnt1 代表充裕堆的数量
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%d",&v);
if(v==1) _cnt++;
if(v>1) _cnt1++;
sum^=v;
}
if(_cnt%2&&_cnt1==0||_cnt1>=2&&sum==0){
printf("Brother\n");
}else{
printf("John\n");
}
}
return 0;
}
