# HDU 3032 Nim or not Nim?(Multi-Nim)

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2508 Accepted Submission(s):
1297**

Problem Description

Nim is a two-player mathematic game of strategy in
which players take turns removing objects from distinct heaps. On each turn, a
player must remove at least one object, and may remove any number of objects
provided they all come from the same heap.

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.

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.

Input

Input contains multiple test cases. The first line is
an integer 1 ≤ T ≤ 100, the number of test cases. Each case begins with an
integer N, indicating the number of the heaps, the next line contains N integers
s[0], s[1], ...., s[N-1], representing heaps with s[0], s[1], ..., s[N-1]
objects respectively.(1 ≤ N ≤ 10^6, 1 ≤ S[i] ≤ 2^31 - 1)

Output

For each test case, output a line which contains either
"Alice" or "Bob", which is the winner of this game. Alice will play first. You
may asume they never make mistakes.

Sample Input

2
3
2 2 3
2
3 3

Sample Output

Alice
Bob

Source

Recommend

Multi-SG游戏的裸题

$$sg(x) = \begin{cases} x-1, & \text{$x\%4=0$}\\ x, & \text{$x\%4=1\lor x\%4=2$ }\\ x+1, & \text{$x\%4=3$} \end{cases}$$

#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; }

作者：自为风月马前卒

本文版权归作者和博客园共有，欢迎转载，但未经作者同意必须保留此段声明，且在文章页面明显位置给出原文连接，否则保留追究法律责任的权利。