Bits（）

// Problem: Bits
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/21763/1010
// Memory Limit: 262144 MB
// Time Limit: 2000 ms
// 2022-04-08 17:58:44
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a%mod;
        a = a * a %mod;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}
int x[(1<<11) + 10];
int n, m;
char ch[20][(1<<11) + 10];
int num;
void print() {
    //n+2层
    num ++;
    for (int i = 1; i<= n+ 3; i ++) {
        for (int j = 1; j<= m; j ++) {
            if (i == n + 3) {
                ch[i][j] = '-';
            }
            else {
                ch[i][j] = '.';
                if (i == 1) continue;
                if (j == n + 2)
                    ch[i][j] = '|';
                else if (j == n + 2 + 2 * n + 2)
                    ch[i][j] = '|';
                else if (j == n + 2 + 4 * n + 4)
                    ch[i][j] = '|';
            }
        }
    }
    int dep[(1<<11) + 1];
    memset(dep, 0 , sizeof dep);
    for (int i= 1; i <= 1<<11; i ++)
        dep[i] = n + 2;

    for (int i = n; i >= 1;  i--) {
        
        int be = dep[x[i]];
        int cnt = 2 * i  +1;
        int mid;
        if (x[i] == 1)
            mid = n + 2;
        else if (x[i] == 2)
            mid = n + 2 + 2 * n + 2;
        else if (x[i] == 3)
            mid = n + 2 + 4 * n + 4;
        int tmid = mid - 1;
        for (int j = 1; j <= (cnt + 1) / 2; j ++)
            ch[be][mid] = '*', mid ++;
        for (int j = 1; j <= cnt / 2; j ++)
            ch[be][tmid] = '*', tmid --;
        dep[x[i]] --;
    }
    
    for (int i = 1; i <= n + 2; i ++) {
        for (int j = 1;j <= m; j ++) {
            cout << ch[i][j];
        }
        puts("");
    }
    if (num <= (1 <<n) - 1)
    {
        for (int i = 1; i<= m; i ++)
        cout << ch[n + 3][i];
        puts("");
    }
}
void move (int a, int b, int c, int n) {
    if (n == 1) {
        x[n] = c;
        print();
        return;
    }
    move(a, c, b, n - 1);
    x[n] = c;
    print();
    move(b, a, c, n - 1);
}
void solve() {
    cin >> n;
    m = 3 * (2 * n + 1) + 4;
    for (int i = 1; i<= n; i ++)
        x[i] = 1;
    print();
      if (n % 2)
          move(1, 3, 2, n);
      else move(1, 2, 3, n);
    
}
int main () {
    // ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

首先递归的去移动 然后对于每一次移动输出对应的图案，找到规律 确定杠的位置就很好写了