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.

 

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

2009
Multi-University Training Contest 13 - Host by HIT

 

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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=;
int read()
{
    char c=getchar();int x=,f=;
    while(c<''||c>''){if(c=='-')f=-;c=getchar();}
    while(c>=''&&c<=''){x=x*+c-'';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=;i<=N;i++) a[i]=read();
        for(int i=;i<=N;i++)
            if(a[i] %  == ) SG[i] = a[i]-;
            else if(a[i]%==||a[i]%==) SG[i] = a[i];
            else SG[i] = a[i]+;
        int ans=;
        for(int i=;i<=N;i++)
            ans^=SG[i];
        puts(ans?"Alice":"Bob");
    }
    return ;
}