For string $$$S$$$ of length $$$n$$$

For each new word $$$x$$$ from $$$S$$$ that $$$\forall i$$$ we have $$$x[i] = s[i mod n]$$$ (mean $$$x$$$ is a sub-string of some cyclic strings that shifted from initial $$$S$$$)

While the last symbol of $$$x$$$ is in sorted order, remove it to make a shorter word

Replace the last remained symbol of $$$x$$$ by its successor in sorted order.

#include <iostream>

using namespace std;

void duval(const string &s)
{
    int n = s.size();
    for (int l = 0; l < n; )
    {
        int r = l, p = l + 1;
        for (; r < n && s[r] <= s[p]; ++r, ++p)
            if (s[r] < s[p]) r = l - 1;

        while (l <= r)
        {
            cout << s.substr(l, p - r) << '\n';
            l += p - r;
        }
    }
}

int main()
{
    string s;
    cin >> s;
    duval(s);
    return 0;
}

#include <iostream>

using namespace std;

/// Factorize all lyndon word in s
void duval(const string &s)
{
    int n = s.size();

    ///
    /// s = s1 + s2 + s3
    /// s1 = s[1..l-1] is handled
    /// s2 = s[l..r]   is handling
    /// s3 = s[p..n]   is going to be handled
    /// 

    for (int l = 0; l < n; )
    {
        /// Extend as much as possible lyndon word s2 = s[l..r]
        int r = l, p = l + 1;
        while (r < n)
        {
            /// (s2 + s[p]) is not a lyndon word
            if (s[r] > s[p]) 
            {
                break;
            }

            /// (s2 + s[p]) is stil a lyndon word, hence extend s2
            if (s[r] == s[p]) 
            {
                ++r;
                ++p;
                continue;
            }
            
            /// (s2 + s[p]) is a lyndon word, but it may be a repeated string
            if (s[r] < s[p]) 
            {
                r = l;
                ++p;
                continue;
            }
        }

        /// The lyndon word may have the form of s2 = sx + sx + .. + sx like "12312123"
        while (l <= r) 
        {
            /// s[l..l + p - r] is sx
            cout << s.substr(l, p - r) << '\n';
            l += p - r;
        }
    }
}

int main()
{
    string s;
    cin >> s;
    duval(s);
    return 0;
}

#include <algorithm>
#include <iostream>
#include <map>

using namespace std;
    
int min_cyc(string s)
{
    map<string, int> M;
    for (int i = 0; i < s.size(); ++i)
    {
        if (!M.count(s)) M[s] = i;
        rotate(s.begin(), s.begin() + 1, s.end());
    }
    return (M.begin() -> second);
}

int main()
{
    string s;
    cin >> s;
    cout << min_cyc(s);
    return 0;
}

• The idea is that when we factorize duplicated string $$$t = s + s$$$

• Then the answer will be a substring of maximum starting position $$$p$$$ not exceed $$$|s|$$$

• The proves is already inside the code

#include <algorithm>
#include <iostream>

using namespace std;

/// Find starting position of minimum acyclic string in (s)
int min_cyc(string s)
{
    int n = s.size(); /// the real size of the string
    s += s; /// for convention since we are deadling with acyclic

    ///
    /// s = s1 + s2 + s3
    /// s1 = s[1..l-1] is handled
    /// s2 = s[l..r]   is handling
    /// s3 = s[p..n]   is going to be handled
    /// 

    int res = 0; /// minimum acyclic string
    /// while (s2) is a lyndon word, try to add s2 with s[p]
    for (int l = 0; l < n; )
    {
        ///
        /// - Case 1: 
        ///     If (s) is fully ordered, then return 0
        ///     Surely will this loop make [l..r] = [0..n-1]
        ///     Ans it is currently that (l = 0) 
        ///     => res = l is a correct answer
        ///
        /// - Case 2:
        ///     Minimum acyclic string s' = s[l..r] that 0 <= l < n <= r < 2n
        ///     Also if s2 is s', then the loop will extend its (r >= n)
        ///     Since l < n, the latest (l) will create s'    
        ///     => res = l is a correct answer
        /// 
        /// Hence in both cases, res = last(l) will return a correct answer
        ///
        res = l;

        /// Extend as much as possible lyndon word s2 = s[l..r]
        int r = l, p = l + 1;
        while (p < s.size())
        {
            /// (s2 + s[p]) is not a lyndon word
            if (s[r] > s[p]) 
            {
                break;
            }

            /// (s2 + s[p]) is stil a lyndon word, hence extend s2
            if (s[r] == s[p]) 
            {
                ++r;
                ++p;
                continue;
            }
            
            /// (s2 + s[p]) is a lyndon word, but it may be a repeated string
            if (s[r] < s[p]) 
            {
                r = l;
                ++p;
                continue;
            }
        }

        /// The lyndon word may have the form of s2 = sx + sx + .. + sx like "12312123"
        while (l <= r) 
        {
            /// s[l..l + p - r] is sx
            l += p - r;
        }
    }

    /// Dont forget to return the value ;)
    return res;
}

/// If you wanna know about that minimum acyclic string
string cyc(string s, int k)
{
    rotate(s.begin(), s.begin() + k, s.end());
    return s;
}

int main()
{
    /// Input 
    string s;
    cin >> s;

    /// Output
    cout << min_cyc(s) << '\n';
//  cout << cyc(s, min_cyc(s));
    return 0;
}

#include <iostream>

using namespace std;

int min_cyc(string s)
{
    int n = s.size();
    s += s;

    int res = 0;
    for (int l = 0; l < n; )
    {
        res = l;
        int r = l, p = l + 1;
        for (; p < s.size() && s[r] <= s[p]; ++r, ++p)
            if (s[r] < s[p]) r = l - 1;

        while (l <= r) l += p - r;
    }

    return res;
}

int main()
{
    string s;
    cin >> s;
    cout << min_cyc(s) << '\n';
    return 0;
}

#include <iostream>
    
using namespace std;
    
int min_cyc(const string &s) 
{
    int n = s.size();
    int res = 0;
    for (int l = 0; l < n; )
    {
        res = l;
        int r = l, p = l + 1;
        for (; r < n; ++r, ++p) /// If there is such string found, then its length wont exceed |s|
        {
            char c = (p < n) ? s[p] : s[p - n]; /// to avoid modulo
            if (s[r] > c) break;
            if (s[r] < c) r = l - 1;
        }        
        l = max(r, l + p - r); /// just skip those (s2 = sx + sx + ... + sx) cases
    }
    return res;
}

int main()
{
    string s;
    cin >> s;
    cout << min_cyc(s);
    return 0;
}