BZOJ 1022 小约翰的游戏(anti-sg)
这是个anti-sg问题,套用sj定理即可解。
SJ定理
对于任意一个Anti-SG游戏,如果定义所有子游戏的SG值为0时游戏结束,先手必胜的条件:
1、游戏的SG值为0且所有子游戏SG值均不超过1。
2、游戏的SG值不为0且至少一个子游戏SG值超过1。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi acos(-1.0) # define eps 1e-3 # define MOD 1000000007 # define INF (LL)1<<60 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int res=0, flag=0; char ch; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0'); return flag?-res:res; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=105; //Code begin... int main () { int T, n, x; scanf("%d",&T); while (T--) { int ans=0, flag=0; scanf("%d",&n); FOR(i,1,n) { scanf("%d",&x); ans^=x; if (x>1) flag=1; } if ((flag==0&&ans==0)||(flag&&ans)) puts("John"); else puts("Brother"); } return 0; }