POJ2975 Nim

                                                                                               Nim

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3591		Accepted: 1623

Description

Nim is a 2-player game featuring several piles of stones. Players alternate turns, and on his/her turn, a player’s move consists of removing one or more stones from any single pile. Play ends when all the stones have been removed, at which point the last player to have moved is declared the winner. Given a position in Nim, your task is to determine how many winning moves there are in that position.

A position in Nim is called “losing” if the first player to move from that position would lose if both sides played perfectly. A “winning move,” then, is a move that leaves the game in a losing position. There is a famous theorem that classifies all losing positions. Suppose a Nim position contains n piles having k₁, k₂, …, k_n stones respectively; in such a position, there are k₁ + k₂ + … + k_n possible moves. We write each k_i in binary (base 2). Then, the Nim position is losing if and only if, among all the k_i’s, there are an even number of 1’s in each digit position. In other words, the Nim position is losing if and only if the xor of the k_i’s is 0.

Consider the position with three piles given by k₁ = 7, k₂ = 11, and k₃ = 13. In binary, these values are as follows:

 111
1011
1101
 

There are an odd number of 1’s among the rightmost digits, so this position is not losing. However, suppose k₃ were changed to be 12. Then, there would be exactly two 1’s in each digit position, and thus, the Nim position would become losing. Since a winning move is any move that leaves the game in a losing position, it follows that removing one stone from the third pile is a winning move when k₁ = 7, k₂ = 11, and k₃ = 13. In fact, there are exactly three winning moves from this position: namely removing one stone from any of the three piles.

Input

The input test file will contain multiple test cases, each of which begins with a line indicating the number of piles, 1 ≤ n ≤ 1000. On the next line, there are n positive integers, 1 ≤ k_i ≤ 1, 000, 000, 000, indicating the number of stones in each pile. The end-of-file is marked by a test case with n = 0 and should not be processed.

Output

For each test case, write a single line with an integer indicating the number of winning moves from the given Nim position.

Sample Input

3
7 11 13
2
1000000000 1000000000
0

Sample Output

3
0

Source

Stanford Local 2005

 

 

思路：本题的大概意思是，给出N堆石子（K1，K2，K3，K4，......，KN)，两个人轮流从任一堆中可以拿去1或更多的石子，最后拿完的人为胜利者。面对N堆石子，这堆石子都会对应一个相应的状态，必胜态或必败态。问当面对给出的N堆石子的数量，有多少种方法可以将这N堆石子的状态变为必胜态。

        这是典型的尼姆博奕（Nimm Game）题，而如果有K1^K2^K3....^KN==0则该状态为必胜态。故对于第i堆石子，我们可以从中拿掉Ki-(k1^k2...^k(i-1)^k(i+1)......^kn),只要这个式子的结果大于0就可以。所以让i从1~n逐个试探即可。

 

 

#include <cstdlib>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <map>
#include <string>
#include <algorithm>




using namespace std;


int data[1000+4];

int main(int argc, char *argv[])
{
    int n;
    int i,j,k;
    
    while(scanf("%d",&n)!=EOF)
    {
       if(n==0)
       break;
       
       for(i=0;i<n;i++)
       {scanf("%d",&data[i]);}
       sort(data,data+n);
       
       int sum=0;
       
       for(i=0;i<n;i++)
       {sum^=data[i];}
       
      int num=0;
      
      for(i=0;i<n;i++)
      {
                      k=data[i]^sum;
                      if(data[i]>k)
                      num++;
      }
      
      
      printf("%d\n",num);
    }
                      
       
       

                                    
                    
    
   // system("PAUSE");
    return EXIT_SUCCESS;
}