1316B - String Modification

// Problem: B. String Modification
// Contest: Codeforces - CodeCraft-20 (Div. 2)
// URL: https://codeforces.com/problemset/problem/1316/B
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 2022-03-02 17:02:14
// 
// 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 pii pair<int, int>
#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

const ll mod = 1e9 + 7;

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);
}
bool cmp (pair<string, int> a, pair<string, int> b) {
    return a.fi < b.fi;
}

string get (string s, int k) {
    int n = s.size();
    string suf = s.substr(0, k - 1);
    string pre = s.substr(k - 1, n - k + 1);
    if (n % 2 == k % 2)
        reverse(all(suf));
    return pre +suf;
}
void solve() {
    int n;
    cin >> n;
    string s;
    cin >> s;
    vector<pair<string, int>> temp;
    string t = s;
    string ans = t;
    int l = 1;
    for (int i = 1; i <= n; i ++) {
        if (ans > get (s, i)) {
            ans = get (s, i);
            l = i;
        }
    }
    cout << ans << endl;
    cout << l << endl;
}
int main () {
    int t;
    t =1;
    cin >> t;
    while (t --) solve();
    return 0;
}

找规律。有一个性质没找到。 就是如果同余就反转