NIM游戏

（一）NIM simplfied

          现在有一堆棋子，共有n颗，两人轮流取子，每次能取1或2颗，谁取到最后一颗棋子谁就是loser，你先开始取

          输入：n

          输出：必胜第一次应该取出的棋子数量，如果必败，则输出为0

         分析：关键在于n%3——如果n%3=1则不可能获胜，必败；如果n%3=2则输出为1给最后对手剩一个；如果n%3=0则输出为2给最后对手剩1个

              错位组合~

#include <stdio.h>
#include <stdlib.h>

  int main()
{
int N,n;
int i;
scanf("%d",&N);
for(i=0;i<N;i++)
{
scanf("%d",&n);
int j;
j=n%3;
if(j==0)
printf("2\n");
else if(j==1)
printf("0\n");
else
printf("1\n");
}
return 0;
}

（二）NIM——POJ2975

    

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 removingone 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 arek₁ + k₂ + … + k_n possible moves. We write eachk_i in binary (base 2). Then, the Nim position is losing if and only if, among all thek_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 thexor 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, supposek₃ 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 withn = 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

分析：给定n堆石子，如果你必输，输出0，否则输出一个方案数，表明总共有多少总策略可以保证自己必胜

题目给出了必输的要求是n堆石子全部异或xor如果结果为0，则此状态必输。必胜的关键就是我们要在其中一堆石子中拿取一定量的石头使n堆石子异或结果为0。

题目要求输出在几堆石子中取子后能满足要求

array[i]^array[i]=0，则remain^array[i]是如果第i堆石头不加入异或时，其他石头总的异或值。

只要array[i]>remain^array[i]，则第i堆石头可行。

PS：注意是大于而不是大于等于，每一局都需要取一颗或以上的石头。

#include<stdio.h>
int main()
{
int maxn=1001;
int n,a[maxn];
    while(scanf("%d",&n)&&n)
    {
int t=0,i=0;
for(i=0;i<n;i++)
{
    scanf("%d",&a[i]);
    t=t^a[i];
}
int ans=0;
for(i=0;i<n;i++)
{
if((t^a[i])<a[i])
ans++;
}
  printf("%d\n",ans);
    }
    return 0;
}