HDU 3032 Nim or not Nim?(Multi-Nim)
2018-06-17 21:10:30来源:未知 阅读 ()
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2508 Accepted Submission(s):
1297
Nim is usually played as a misere game, in which the player to take the last object loses. Nim can also be played as a normal play game, which means that the person who makes the last move (i.e., who takes the last object) wins. This is called normal play because most games follow this convention, even though Nim usually does not.
Alice and Bob is tired of playing Nim under the standard rule, so they make a difference by also allowing the player to separate one of the heaps into two smaller ones. That is, each turn the player may either remove any number of objects from a heap or separate a heap into two smaller ones, and the one who takes the last object wins.
#include<cstdio> #include<cstring> using namespace std; const int MAXN=1001; int read() { char c=getchar();int x=0,f=1; while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();} while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();} return x*f; } int a[MAXN],SG[MAXN]; int main() { #ifdef WIN32 freopen("a.in","r",stdin); #else #endif int QWQ=read(); while(QWQ--) { int N=read(); for(int i=1;i<=N;i++) a[i]=read(); for(int i=1;i<=N;i++) if(a[i] % 4 == 0) SG[i] = a[i]-1; else if(a[i]%4==1||a[i]%4==2) SG[i] = a[i]; else SG[i] = a[i]+1; int ans=0; for(int i=1;i<=N;i++) ans^=SG[i]; puts(ans?"Alice":"Bob"); } return 0; }
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