#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 1e6;

char s[N];

int main()
{
    scanf("%s", s + 1);
    //下标从1开始
    int n = strlen(s + 1);
    //复制自身拼接到后面
    for (int i = 1; i <= n; i ++ ) s[n + i] = s[i];
    //B[i],B[j]
    int i = 1, j = 2, k = 0;
    while (i <= n && j <= n){
        //相同的字符k++
        for (int k = 0; k < n && s[i + k] == s[j + k]; k ++ );
        //全部相同，则i，j较小的那个是答案
        if(k == n) break;
        if(s[i + k] > s[j + k]) {
            i = i + k + 1;

            if(i == j) i++;
        }else{
            j = j + k + 1;
            if(i == j) j ++;
        }
    }
    int ans = min(i, j);
    //B[ans]为答案
    for (int i = ans; i < ans + n; i ++ ){
        cout << s[i];
    }

}

在acwing上好像找不到对应的题，去洛谷搜了下，这里附上题目链接 problem
代码

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 1e6 + 10;

int s[N];

int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i ++ ){
        scanf("%d", &s[i]);
    }
    //复制自身拼接到后面
    for (int i = 1; i <= n; i ++ ) s[n + i] = s[i];
    //B[i],B[j]
    int i = 1, j = 2, k = 0;
    while (i <= n && j <= n){
        //相同的字符k++
        for (k = 0; k < n && s[i + k] == s[j + k]; k ++ );
        //全部相同，则i，j较小的那个是答案
        if(k == n) break;
        if(s[i + k] > s[j + k]) {
            i = i + k + 1;
            if(i == j) i++;
        }else{
            j = j + k + 1;
            if(i == j) j ++;
        }
    }
    int ans = min(i, j);
    //B[ans]为答案
    for (int i = ans; i < ans + n; i ++ ){
        printf("%d ", s[i]);
    }

}