提供三种解法
第一种为非递归型代码
第二种为递归代码
第三种同为非递归型代码，但相对暴力，不过也能AC

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

inline ll read() {
    ll f=1,x=0;
    char ch;
    do {
        ch=getchar();
        if(ch=='-')f=-1;
    } while(ch>'9'||ch<'0');
    do {
        x=x*10+ch-'0';
        ch=getchar();
    } while(ch>='0'&&ch<='9');
    return f*x;
}

ll n;

struct sth{
    ll pos,k,state;
};

int main() {
    n=read();
    stack<sth>S;
    sth a;a.pos=0,a.k=0,a.state=0;
    S.push(a);
    while(!S.empty())
    {
        sth e=S.top();
            S.pop();
        if(e.pos==0)
        {
            if(e.k==n){
                for(int i=0;i<n;i++)
                    if(e.state>>i&1)
                        printf("%d ",i+1);
                puts("");   
                continue;
            }
            e.pos=1;
            S.push(e);
            S.push({0,e.k+1,e.state});
        }
        else if(e.pos==1){
            e.pos=2;
            S.push(e);
            S.push({0,e.k+1,e.state|(1<<e.k)});
        }
        else continue;
    }
    return 0;
}

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

inline ll read() {
    ll f=1,x=0;
    char ch;
    do {
        ch=getchar();
        if(ch=='-')f=-1;
    } while(ch>'9'||ch<'0');
    do {
        x=x*10+ch-'0';
        ch=getchar();
    } while(ch>='0'&&ch<='9');
    return f*x;
}

ll n;

void dfs(int k,int S){
    if(k==n){
        for(int i=0;i<n;i++)
            if(S>>i&1)
             printf("%d ",i+1);
        puts("");   
        return ;
    }
    dfs(k+1,S); 
    dfs(k+1,S|(1<<k));
} 

int main() {
    n=read();
//  puts("");
    dfs(0,0);
    return 0;
}

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

inline ll read() {
    ll f=1,x=0;
    char ch;
    do {
        ch=getchar();
        if(ch=='-')f=-1;
    } while(ch>'9'||ch<'0');
    do {
        x=x*10+ch-'0';
        ch=getchar();
    } while(ch>='0'&&ch<='9');
    return f*x;
}

ll n,vis[20],AC[200];

inline void dfs(ll k,ll m)
{
    if(k==m+1){
        for(int i=1;i<=m;i++){
            printf("%lld ",AC[i]);
        }
        puts("");
    }
    for(int i=1;i<=n;i++){
        if(!vis[i]&&i>AC[k-1]){
            vis[i]=1;
            AC[k]=i;
            dfs(k+1,m);
            vis[i]=0;
        }
    }
}

int main() {
    n=read();
    puts("");
    for(int i=1;i<=n;i++){
        memset(vis,0,sizeof(vis));
        dfs(1,i);
    }
    return 0;
}