C++-POJ2975-Nim

题目把Nim游戏为什么可以取异或和讲解得十分清楚，建议多读几次，理解一下

再一个，可以把每次异或视为一次取数，因此(k[i]^sg)<k[i]即为一种可行操作

/*
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, …, kn stones respectively;
in such a position, there are k1 + k2 + … + kn possible
moves. We write each ki in binary (base 2). Then, the Nim
position is losing if and only if, among all the ki’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 ki’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.
*/

#include <set>
#include <map>
#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int n,k[];
int main(){
    while(scanf("%d",&n)){
        if(n==)break;
        int sg=,ans=;
        for(int i=;i<=n;i++)scanf("%d",&k[i]),sg^=k[i];
        for(int i=;i<=n;i++)if((k[i]^sg)<k[i])ans++;
        printf("%d\n",ans);
    }
    return ;
}