HDU 3032 Nim or not Nim?（博弈 sg表）

Nim or not Nim?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1326    Accepted Submission(s): 656

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

 

Recommend

gaojie   |   We have carefully selected several similar problems for you:   3031  3033  3035  3034  3036 

 

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")

#define eps 1e-8
typedef __int64 ll;

using namespace std;

#define INF 0x3f3f3f3f
#define N 10005

int main()
{
     int i,j,t;
     int n;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&n);
         int ans=0;
         int x;
         while(n--)
         {
             scanf("%d",&x);
             if(x%4==1)
                ans^=x;
             if(x%4==2)
                ans^=x;
             if(x%4==0)
                ans^=(x-1);
             if(x%4==3)
                ans^=(x+1);
         }
         if(ans)
            printf("Alice\n");
         else
            printf("Bob\n");
     }
     return 0;
}



/*
 打sg 表看规律

 sg(4k)=4k-1;sg(4k+1)=4k+1;sg(4k+2)=4k+2;sg(4k+3)=4k+4;


#include<iostream>
#include<string.h>
#include<cstdio>
#define N 1000001
using namespace std;
int sg[N];
int g(int x)
{
    int mex[1000];
    memset(mex,0,sizeof(mex));
    if(sg[x]!=-1)  return sg[x];
    for(int i=x-1;i>=0;i--)
    {
        mex[g(i)]=1;
    }
    for(int i=1;i<=x/2;i++)
    {
        int ans=0;
        ans^=g(i);
        ans^=g(x-i);
        mex[ans]=1;
    }
    for(int i=0;;i++)
    if(!mex[i])  return sg[x]=i;
}
int main()
{
    int t , n ,x ;
    memset(sg,-1,sizeof(sg));
    sg[0]=0;
      n=100;
     for(int i=1;i<n;i++)
         g(i);
       for(int i=0;i<=20;i++)
       {
          printf("i=%d %d\n",i,sg[i]);
       }
    return 0;
}

/*


i=0 0
i=1 1
i=2 2
i=3 4
i=4 3
i=5 5
i=6 6
i=7 8
i=8 7
i=9 9
i=10 10
i=11 12
i=12 11
i=13 13
i=14 14
i=15 16
i=16 15
i=17 17
i=18 18
i=19 20
i=20 19

Process returned 0 (0x0)   execution time : 0.194 s
Press any key to continue.

*/