hdu3032 Nim or not Nim?-----sg時計を打って法則を探して1山の石を2つの小さい山に分けることを許可します

Nim or not Nim?
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 406    Accepted Submission(s): 185
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
 
 
标题:n堆石、1堆石を少なくとも1つ取るか、ある堆石を2つに分けて、最後に石を取る人が勝つ.
まず表を打って法則を探さなければならないとsg[4 k]=4 k-1が発見される   sg[4k+1]=4k+1    sg[4k+2]=4k+2    sg[4k+3]=4k+1
 
打表コード:

  ：
void take_part(int n)
{
     for(int i = 1; i <= n / 2; i++)
     {
         int yihuo = 0;
         yihuo ^= sg[i] ^ sg[n - i];
         vst[yihuo] = true;
     }
}

 void get_sg()
 {
     memset(sg, 0, sizeof(sg));
     for(int i = 0; i < MAX; i++)
     {
         memset(vst, false, sizeof(vst));
         int j = 0;
         while(j++ < i)
         {
             vst[sg[j]] = true;
         }
         take_part(i);
         for(int j = 0; j < MAX; j++)
         {
             if(!vst[j])
             {
                 sg[i] = j;
                 break;
             }
         }
     }
     return ;
 }

A題コード

#include<iostream>
#include<cstdlib>
#include<stdio.h>
#include<memory.h>
using namespace std;
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        int ans=0;
        int num;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&num);
            if(num%4==0) ans^=(num-1);
            else if(num%4==2||num%4==1) ans^=num;
            else  ans^=(num+1);
        }
        if(ans!=0) puts("Alice");
        else puts("Bob");
    }
}

/*
Sample Input
2
3
2 2 3
2
3 3
 

Sample Output
Alice
Bob
*/