No source of problem (? What seems to POI) string - minimal representation

Description] [Title
give you two of length \ (n-\) string asked whether a prefix of a string by the string to the back so that the two strings are equal. If so, you may be required output lexicographically smallest character string obtained by the above-described operation can be.
[Input format
two integers of the first row \ (n-, T \) .
The next two lines of a length \ (n-\) string.

[Output Format]
If so, output \ (TAK \) , otherwise the output \ (NIE \) . If \ (. 1 = T \) , the output line of the string you need lexicographically smallest next.

Data range []
\ (n \ le 1000000 \)

First, what is \ (TAK \) ?
Own Baidu translation

Think about the practice of violence will receive back a prefix of the string so the original string is equivalent to one cyclic shift, so you can enumerate all the cyclic shift.
Time complexity \ (O (n ^ 2) \)

在这里介绍一种\(O(n)\)求出一个字符串最小的循环移位的方法
先将原串复制一遍加在原串后面 即 将\(ABCD\) 变为 \(ABCDABCD\) 这样在这个新字符串中任意取一段长为\(n\)的子串一定是原串的一种循环移位。
维护两个指针\(i, j\)和当前子串长度\(len\)，表示现在正在比较从\(i\)开始的长度为\(len+1\)的子串和从\(j\)开始的长为\(len+1\)的子串。
若某一时刻，\(s[i+len] \neq s[j+len]\)
若\(s[i+len] < s[j+len]\)，则说明从\(i\)开始的子串比从\(j\)开始的子串小，所以最小子串一定不会从\(j\)开始。此时可以直接让\(j = j + len + 1\)，然后继续匹配。
反之让\(i = i + len + 1\)。
为什么这样可以保证正确性呢？
反证：如果此时 \(s[i+len] < s[j+len]\) ，即\(j\)被“淘汰”了，假设最小子串其实是从\(s[j+k](1 \le k \le len)\)开始的，那么一定存在从\(s[i+k]\)开始的子串会比它小。
举例："\(ABAABBC\)": 此时\(i\)在\(s[1]\)，\(j\)在\(s[4]\)，\(len=2\)时发现\(s[1+2] < s[4+2]\)，所以\(j\)可以直接跳到7。因为从\(S[5], S[6]\)开始的子串一定不会成为最小的。对于从\(s[5]\)开始的"\(BBC\)..."有从\(s[2]\)开始的"\(BA\)..."比它小，从\(s[6]\)开始的\(s[3]\)一定比它小。

【代码实现】

#include <bits/stdc++.h>
#define ri register int
using namespace std;

inline int read() {
    int ret = 0, flag = 1;
    char ch = getchar();
    while (ch > '9' || ch < '0') {
        if (ch == '-') flag = -1;
        ch = getchar();
    }
    while (ch <= '9' && ch >= '0') {
        ret = (ret << 1) + (ret << 3) + (ch ^ '0');
        ch = getchar();
    }
    return ret * flag;
}
 
int n, t;
int i, j;
int min1, min2;
char s1[2000005], s2[2000005];
bool yes = 1;

int main() {
    n = read(); t = read();
    scanf("%s", s1+1);
    scanf("%s", s2+1);
    for (ri a = 1; a <= n; a++) {
        s1[a + n] = s1[a]; 
        s2[a + n] = s2[a];
    }
    i = 1; j = 2;
    while (i <= n && j <= n) {
        if (i == j) j++;
        for (int len = 0; ; len++) {
            if (s1[i+len] < s1[j+len]) {
                j = j + len + 1;
                break;
            }
            if (s1[j+len] < s1[i+len]) {
                i = i + len + 1;
                break;
            }
        }
    }
    if (i <= n) min1 = i;
    else min1 = j;
    i = 1; j = 2;
    while (i <= n && j <= n) {
        if (i == j) j++;
        for (int len = 0; ; len++) {
            if (s2[i+len] < s2[j+len]) {
                j = j + len + 1;
                break;
            }
            if (s2[j+len] < s2[i+len]) {
                i = i + len + 1;
                break;
            }
        }
    }
    if (i <= n) min2 = i;
    else min2 = j;
    for (ri len = 0; len < n; len++) {
        if (s1[min1 + len] != s2[min2 + len]) {
            yes = 0;
            break;
        }
    }
    if (yes) {
        puts("TAK");
        if (t) {
            for (ri a = min1; a <= min1 + n - 1; a++) {
                putchar(s1[a]);
            }
            puts("");
        }
        
    } else puts("NIE");
    return 0;
}