知识点：最小表示法
这个没什么好说的，最小表示法就是用线性的时间求出一个字符串的最小表示，我感觉这个算法还是比较简单的，但是洛谷的模板题的难度都给标的蓝题，有点出人意料
#include <bits/stdc++.h>using namespace std;const int N = 1e6 + 5;string s1, s2;int solve(string s, int n) {int l = 1, r = 2;while (l <= n && r <= n) {int k;for (k = 0; k < n && s[l + k] == s[r + k]; k++);if (k == n) break;if (s[l + k] > s[r + k]) l += k + 1;else r += k + 1;if (l == r) l++;}return min(l, r);}int main() {cin >> s1 >> s2;int len = (int) s1.size();s1 = " " + s1;s2 = " " + s2;for (int i = 1; i <= len; i++) {s1 += s1[i];s2 += s2[i];}int pos1 = solve(s1, len);int pos2 = solve(s2, len);int ok = 1;for (int i = 0; i < len; i++) {if (s1[pos1 + i] != s2[pos2 + i]) ok = 0;}if (!ok) cout << "No";else cout << "Yes\n" << s1.substr(pos1, len);return 0;}
