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• 2

  nogugu LV 6 @ 2026-5-26 12:39:47

  #include <bits/stdc++.h>
  using namespace std;
  
  #define int long long
  #define endl '\n'
  
  const double EPS = 1e-8;
  const int N = 2e5 + 5;
  
  int a[N];
  
  signed main() {
      ios::sync_with_stdio(0);
      cin.tie(0), cout.tie(0);
  
      int n;
      cin >> n;
      vector<int> vec;
      for (int i = 1; i <= n; i++) {
          vec.push_back(i);
      }
  
      do {
          for (int i = 0; i < n; i++)
              cout << vec[i] << " ";
          cout << endl;
      } while (next_permutation(vec.begin(), vec.end()));
      return 0;
  }
  
• 1

  wangyiyang LV 10 @ 2026-9-26 15:01:48

  这题一共两个写法：

  1.$for$循环

      #include<bits/stdc++.h>
  using namespace std;
  int main(){
  	//freopen(".in","r",stdin);
  	//freopen(".out","w",stdout);
  	int n;
  	cin>>n;
  	for(int i=1;i<=n;i++){
  		for(int j=i;j<=n;j++){
  			for(int k=j;k<=n;k++){
  				cout<<i<<" "<<j<<" "<<k<<"\n";
  			}
  		}
  	}
  	return 0;
  }
  
  

  $$复制的人就走运了$$ $$运行一下就知道了$$

  正片开始：

  1.STL大法

  用c++库里的一个very牛X的函数：next__permutation 给代码：

  #include<bits/stdc++.h>
  using namespace std;
  int main(){
  	//freopen(".in","r",stdin);
  	//freopen(".out","w",stdout);
  	int n;
  	cin>>n;
  	vector<int> a(n);
  	for(int i=0;i<n;i++){
  		a[i]=i+1;
  	}
  	do{
  		【自己填吧，看懂了再写】
  	}while(next_permutation(a.begin(),a.end()));
  	return 0;
  }
  
  

  重点登场：

  $第二种：dfs+回溯$ c++代码：

  #include<bits/stdc++.h>
  using namespace std;
  int a[13],n;
  int vis[15];
  void dfs(int u){
  	if(u==n){
  		for(int i=0;i<n;i++){
  			cout<<a[i]<<" ";
  		}
  		cout<<"\n";
  		return;
  	}//边界
  	for(int i=1;i<=n;i++){
  		if(vis[i]==false){
  			a[u]=i;
  			vis[i]=true;
  			dfs(u+1);
  			vis[i]=false;
  			a[u]=0;
  		}
  	}
  }
  void solve(){
  	cin>>n;
  	dfs(0);
  }
  int main(){
  	//freopen(".in","r",stdin);
  	//freopen(".out","w",stdout);
  	solve();
  	return 0;
  }
  
  
• 1

  Asuka LV 7 @ 2026-5-27 19:44:32

  
  #include<bits/stdc++.h>
  using namespace std;
  typedef long long ll;
  int n;
  vector<int> v;
  bool st[10];
  
  void dfs(int num){
  	if(num==n){
  		for(int i=0;i<n;i++){
  			if(i==0) cout<<v[i];
  			else cout<<" "<<v[i];
  		}
  		cout<<"\n";
  		return;
  	}
  	
  	for(int i=1;i<=n;i++){
  		if(!st[i]){
  			st[i]=true;
  			v.push_back(i);
  			dfs(num+1);
  			st[i]=false;
  			v.pop_back();
  		}
  	}
  }
  
  int main(){
  	ios::sync_with_stdio(false);
  	cin.tie(0);
  	
  	cin>>n;
  	dfs(0);
  
  	
  	return 0;
  }
  
• 0

  li1005345206 LV 6 @ 2026-5-21 16:09:59

  n = int(input())
  def dfs(arr):
      if len(arr) >= n:
          print(*arr)
          return
      for i in range(n):
          i += 1
          if i in arr:
              continue
          dfs(arr[:] + [i])
  dfs([])
  
• 0

  fengdu7 LV 7 @ 2025-5-9 18:34:52

  JAVA ac代码:

  import java.util.*;
  import java.util.StringTokenizer;
  import java.io.*;
  
  public class Main {
      static int n;
      static boolean flag[];//这个数是否被用过
      public static void main(String[] args) {
          n=in.nextInt();
          flag=new boolean[n+1];
          int[] arr=new int[n];//当前数组
          dfs(arr,0);
          out.close();
      }
  
      
      private static void dfs(int[] arr, int i) {
          //终止条件:i==n,即arr数组中1到n-1位置都被填完了
          if(i==n) {
              for(int j=0;j<n;j++) out.print(arr[j]+" ");
              out.println();
              return;
          }
          
          for(int j=1;j<=n;j++) {
              //j还没被用过
              if(!flag[j]) {
                  flag[j]=true;
                  arr[i]=j;//赋值
                  dfs(arr,i+1);
                  //回溯
                  arr[i]=0;
                  flag[j]=false;
              }
          }
      }
  
  
      static FastReader in=new FastReader();
      static PrintWriter out=new PrintWriter(System.out);
      static class FastReader{
          static BufferedReader br;
          static StringTokenizer st;
          FastReader(){
              br=new BufferedReader(new InputStreamReader(System.in));
          }
          String next() {
              String str="";
              while(st==null||!st.hasMoreElements()) {
                  try {
                      str=br.readLine();
                  }catch(IOException e) {
                      throw new RuntimeException(e);
                  }
                  st=new StringTokenizer(str);
              }
              return st.nextToken();
          }
          int nextInt() {
              return Integer.parseInt(next());
          }
          double nextDouble() {
              return Double.parseDouble(next());
          }
          long nextLong() {
              return Long.parseLong(next());
          }
      }
      
  }
  

• 1