Portal

analysis

Minimum cyclic string problem
Given a string S S S. Each time, you can move its first character to the back to find the string with the smallest dictionary order.
T T Group T data (seemingly POJ data compared with water)

Here we first introduce the usage of suffix array idea
Size comparison of a string

If string A = a 1 + a 2 A = a_1+a_2 A=a1​+a2​， B = b 1 + b 2 B = b_1+b_2 B=b1​+b2​
hypothesis l e n ( a 1 ) = = l e n ( b 1 ) len(a_1) == len(b_1) len(a1​)==len(b1​)
If a 1 < b 1 a_1<b_1 a1 < B1 then string A < B A<B A < B and vice versa
If equal, compare b b The size of b, similarly

!!! Important part!!!
At this time, we can split a binary string into several small strings and compare the size recursively
Then we can use multiplication to solve this problem

set up d p [ i ] [ j ] dp[i][j] dp[i][j] refers to the current position i i i start, length 2 j 2^j 2j string size ranking. We specify that the ranking here is only for fixed length 2 j 2^j 2j, if the length is not enough, supplement it later 0 0 0, ensure the minimum number of empty characters
initialization d p [ i ] [ 0 ] dp[i][0] dp[i][0] is the value of each character a s c i i ascii ascii code value

But this transfer needs to be calculated by sorting
So we think of it as above A = a 1 + a 2 A = a_1 + a_2 A=a1 + a2 will have a length of 2 j 2^j 2j string split into two 2 ( j − 1 ) 2^(j-1) 2(j − 1) small string of length. Quickly sort to get the ranking of the current length. See the following code for details
Time complexity O ( n ∗ l o g n 2 ) O(n*logn^2) O(n∗logn2)
Details:

• Because it is circular, according to the usual routine, break the ring into a chain, and you can splice this string to the end
• Reduce the case of zero filling at the end of special judgment, and double the original size of the array
• Only if the binary split length > 2 N >2N >2n to overwrite all strings

Code [suffix array]

//P1509
/*
  @Author: YooQ
*/
//#include <bits/stdc++.h>
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
#define sc scanf
#define pr printf
#define ll long long
#define int long long
#define FILE_OUT freopen("out", "w", stdout);
#define FILE_IN freopen("in", "r", stdin);
#define debug(x) cout << #x << ": " << x << "\n";
#define max3(a, b, c) max(a, max(b, c))
#define min3(a, b, c) min(a, min(b, c))
#define MAX(a, b) (a >= b ? a : a = b)
#define MIN(a, b) (a <= b ? a : a = b)
#define AC 0
#define WA 1
#define INF 0x3f3f3f3f
const ll MAX_N = 1e6+5;
const ll MOD = 1e9+7;
const ll UP = 30;
int N, M, K;

char arr[MAX_N];
int dp[40005][31];

struct Node {
	int x, y, id;
	bool operator < (const Node& B) const {
		return x == B.x ? y < B.y : x < B.x;
	}
}brr[MAX_N];
int limit = 0;

bool cmp(int x, int y) {
	int len = N;
	for (int i = limit; i+1; --i) {
		if (len >= (1ll<<i)) {
			if (dp[x][i] != dp[y][i]) {
				return dp[x][i] > dp[y][i];
			}
			len -= (1ll<<i);
		}
	}
	return false;
}

void solve() {
	cin >> arr+1;
	N = strlen(arr+1);
	limit = 0;
	for (int i = 1; i <= N; ++i) {
		dp[N+i][0] = dp[i][0] = arr[N+i] = arr[i];
	}
	for (int i = 1; (1ll<<(i-1)) <= N; limit = i++) {
		for (int j = 1; j <= 2*N; ++j) {
			brr[j].x = dp[j][i-1];
			brr[j].y = dp[j+(1ll<<(i-1))][i-1];
			brr[j].id = j;
		}
		sort(brr+1, brr+1+2*N);
		for (int j = 1; j <= 2*N; ++j) {
			dp[brr[j].id][i] = (brr[j-1].x == brr[j].x && brr[j-1].y == brr[j].y) ? dp[brr[j-1].id][i] : j;
		}
	}
	
	int ans = 1;
	for (int i = 2; i <= N; ++i) {
		if (cmp(ans, i)) {
			ans = i;
		}
	}
	
	cout << ans << "\n";
}

signed main() {
	#ifndef ONLINE_JUDGE
	//FILE_IN
	FILE_OUT
	#endif
	int T = 1; cin >> T;
	while (T--) solve();

	return AC;
}

Code [AC automata] - to be supplemented

//todo SAM to solve this problem