題目描述
Niuniu likes playing games. He has n piles of stones. The i-th pile has ai stones. He wants to play with his good friend, UinUin. Niuniu can choose some piles out of the n piles. They will play with the chosen piles of stones. UinUin takes the first move. They take turns removing at least one stone from one chosen pile. The player who removes the last stone from the chosen piles wins the game. Niuniu wants to choose the maximum number of piles so that he can make sure he wins the game. Can you help Niuniu choose the piles?

輸入描述:
The first line contains one integer n (1 ≤ n ≤ 500000), which means the number of piles.
The second line describes the piles, containing n non-negative integers, a1 a2 … an, separated by a space. The integers are less than or equal to 500000.

輸出描述:
Print a single line with one number, which is the maximum number of piles Niuniu can choose to make sure he wins. If Niuniu cannot always win whatever piles he chooses, print 0.
示例1
輸入
複製
8
1 9 2 6 0 8 1 7
輸出
複製
7
說明
Niuniu can choose the piles {1,9,6,0,8,1,7} to make sure he wins the game.

賽後看了別人的題解才會，剛開始不知道被誰榜帶偏了，所有人都在做這道題，呵呵，沒多少做出來的。
我們可以知道，若一堆數的異或和為0，先手一定會打破這種平衡，後手則可以修補這種平衡，所以在異或為0的情況下，後手勝，之後就是通過dp來找出最少需要去掉幾個數才能使異或和為0；

#include<bits/stdc++.h>
using namespace std;
#define maxn 600005
#define inf 0x3f3f3f3f
int dp[2][maxn],vis[maxn];
vector<int>vec;
int main()
{
    int n;
    scanf("%d
 
",&n);
    int a;
    int sum=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a);
        if(!vis[a])
            vec.push_back(a);
        vis[a]=1;
        dp[0][a]=1;
        sum^=a;
    }
    if(sum==0)
        printf("%d\n",n);
    else if(vis[sum])
        printf("%d\n",n-1);
    else
    {
        int flag=0;
        for(int i=0;i<vec.size();i++)
            if(vis[sum^vec[i]])
            {
                flag=1;
                break;
            }
        if(flag)
            printf("%d\n",n-2);
        else
        {
            int pos=0;
            memset(dp,inf,sizeof(dp));
            dp[0][0]=0;
            for(int i=0;i<vec.size();i++)
            {
                pos^=1;
                for(int j=0;j<=maxn;j++)
                {
                    if(dp[pos^1][j]>=inf)
                        continue;
                    dp[pos][j]=min(dp[pos][j],dp[pos^1][j]);
                    dp[pos][j^vec[i]]=min(dp[pos^1][j]+1,dp[pos][j^vec[i]]);
                }
            }
            printf("%d\n",max(n-dp[pos][sum],0));
        }
    }
    return 0;
}