Let \(S_{N,M}\) be the set of integer sequences of length \(N\) consisting of integers between \(1\) and \(N\).

First, if \(N=1\), then \(S_{N,M}=(\{1\},\{2\},\dots,\{M\})\).

Now let \(N \ge 2\). We enumerate the last term of the integer sequences in \(S_{N,M}\).

If the last term is \(K\), then the set of possible first \((N-1)\) terms equals \(S_{N-1,K-1}\). Here, note that \(N-1 \le K-1\).

Therefore, we can find \(S_{N,M}\) by iterating \(K\) and adding the elements of \(S_{N-1,K-1}\) appending \(K\) to each of its element.

Therefore, the problem has been solved.

Alternatively, in C++, we can prepare a vector consisting of \(M\) zeros and \((N-M)\) ones in this order, and use next_permutation to output the indices of zero-valued elements in each iteration. The implementation of this solution is simpler.

Sample code (C++)

#include <bits/stdc++.h>
using namespace std;
vector<vector<int>> dp[10][10];
void solve(int n,int m){
  if(dp[n][m].size()) return;
  if(n==1){
    for(int i=1;i<=m;i++){
      dp[n][m].push_back({i});
    }
  }
  else{
    for(int k=1;k<=m;k++){
      if(n-1<=k-1){
        solve(n-1,k-1);
        for(int i=0;i<dp[n-1][k-1].size();i++){
          vector<int> a=dp[n-1][k-1][i];
          a.push_back(k);
          dp[n][m].push_back(a);
        }
      }
    }
  }
}
int main() {
  int n,m;
  cin>>n>>m;
  solve(n,m);
  sort(dp[n][m].begin(),dp[n][m].end());
  for(int i=0;i<dp[n][m].size();i++){
    for(int j=0;j<n;j++) cout<<dp[n][m][i][j]<<" ";
    cout<<endl;
  }
}

Sample code (next_permutation,C++)

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

int main() {
  int n,m;
  cin>>n>>m;
  vector<int> a;
  for(int i=0;i<n;i++) a.push_back(0);
  for(int i=0;i<m-n;i++) a.push_back(1);
  do{
    for(int i=0;i<m;i++){
      if(a[i]==0) cout<<i+1<<" ";
    }
    cout<<endl;
  }while(next_permutation(a.begin(),a.end()));
}