GG，，，g艹

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <vector>
#include <queue>
#include <math.h>
using namespace std;
vector<int>G[21][7];//G[i][j] 表示n=i k=j的情况下 二进制的状态
int n, k, l;
int a[21], d[6], val[6];
int vis[150], tim;
int work(int x){
    int i = 0, j = 0;
    while(x) {
        if(x&1)    d[i++] = j;//a[j]
        j++;
        x>>=1;
    }
    int ans = 0;
    do
    {
        memset(val, 0, sizeof val);
        tim ++;
        int st = 0;
        for(int num = 0; num < k; num++) {
            i = st; j = 0;
            while(1) {
                val[j] ^= a[d[i]];
                vis[val[j]] = tim;
                i++; j++;if(i>=k)i=0;
                if(i==st)break;
            }
            st++;
        }
        for(i = l; ; i++)
            if(vis[i]!=tim)
            {
                ans = max(ans, (i-1)>=l? (i-1):0);
                break;
            }
    } while (next_permutation(d + 1, d + k));
    return ans;
}
void dfs(int dep, int cnt, int num) {
    if (dep > 20 || cnt > 6) return ;
    G[dep][cnt].push_back(num);
    dfs(dep + 1, cnt + 1, num | (1 << dep));
    dfs(dep + 1, cnt, num);
}

struct Node {
    int cnt, idx;
    bool operator < (const Node &rhs) const {
        return cnt > rhs.cnt;
    }
};

Node qq[24];
int b[24];

int cmp(int a, int b) {
    return a > b;
}

int main(){
    int i, j;
    for(i = 1; i <= 20; i++)
        for(j = 1; j <= 6; j++) G[i][j].clear();
    dfs(0, 0, 0);
    tim = 100;
    while(~scanf("%d %d %d",&n, &k, &l)) {
        for(i = 0; i < n; i++)scanf("%d",&a[i]);
        
        int ans = 0, siz = (int)G[n][k].size();
        if (siz <= 12000) {
            for (int i = 0; i < siz; ++i) {
                int cur = G[n][k][i];
                ans = max(ans, work(cur));
              }
        } else if (siz <= 24001) {
            for (int i = 1; i < siz; i += 2) {
                int cur = G[n][k][i];
                ans = max(ans, work(cur));
            }
        } else {
            for (int i = 0; i < siz; i += 3) {
                int cur = G[n][k][i];
                ans = max(ans, work(cur));
            }
        }
        
        printf("%d\n",ans);
    }
    return 0;
}