(hdu 1907)John (Misère Nim,Nim博弈的变形)

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
2
3
3 5 1
1
1
Sample Output
John
Brother

Misère Nim
当最后取的人为败者时，有如下结论：
(1)当所有石子堆都为一时，奇数堆必败，偶数堆必胜。
(2)当仅有一堆石子数大于一时，为必胜态。为了保证还剩下奇数堆，可以选择拿光这一堆还是留一个。
(3)当有超过一堆石子数大于一时，将各堆石子数求Nim和，若结果为0则必败，否则必胜。
注意(2)和(3)可以合并，直接求Nim和判定胜负。在(3)局面的获胜办法和常规nim博弈一样，直至进入(2)。
如果你采用最优策略，至少有两堆大于一，否则无法保证nim和为0让对手必败。同样对手无法把两堆大于一的石子同时拿成为石子都为一的两堆，所以必然进入（2）。

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;#define mem(a,n) memset(a,n,sizeof(a))const double INF=0x3f3f3f3f+1.0;const double eps=1e-6;const double PI=2.0*acos(0.0);typedef long long LL;const int N=55;int a[N];int main(){    int t,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        int ans=0,num=0;        for(int i=0; i<n; i++)        {            scanf("%d",&a[i]);            ans^=a[i];            if(a[i]>1) num++;        }        if(num==0)            printf("%s\n",ans?"Brother":"John");        else printf("%s\n",ans?"John":"Brother");    }    return 0;}