\(\mathcal{Description}\)

  求出處 owo。
  給定一個長度為 \(n\)，僅包含小寫字母的字串 \(s\)，問是否存在長度為 \(n\)，僅包含小寫字母的字串 \(t\)，使得 \(t<s\) 且 \(s,t\) 的字尾陣列（\(\text{Suffix Array}\)，sa[]）相同。
  \(n\le50\)。（建議開到 \(n\le2\times10^5\)。

\(\mathcal{Solution}\)

奇怪的結論

  若存在 \(t\)，則存在一個 \(t\)，其與 \(s\) 僅相差一個字元。考試的時候我猜出來了 w！

演算法

  有了上面的結論就非常簡單了。首先一個顯而易見的特判：若 \(s\)

\(\mathcal{Code}\)

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>

#define YES() ( puts ( "Exists" ), exit ( 0 ) )
#define NO() ( puts ( "Does not exist" ), exit ( 0 ) )

const int MAXN = 50;
int n, sa[MAXN + 5];
char s[MAXN + 5];

inline void suffixSort ( int* sa ) {
	int m = 'z', x[200] {}, y[200] {}, c[200] {};
	for ( int i = 1; i <= n; ++ i ) c[i] = 0;
	for ( int i = 1; i <= n; ++ i ) ++ c[x[i] = s[i]];
	for ( int i = 1; i <= m; ++ i ) c[i] += c[i - 1];
	for ( int i = n; i; -- i ) sa[c[x[i]] --] = i;
	for ( int j = 1; j <= n; j <<= 1 ) {
		int cnt = 0;
		for ( int i = n - j + 1; i <= n; ++ i ) y[++ cnt] = i;
		for ( int i = 1; i <= n; ++ i ) if ( sa[i] > j ) y[++ cnt] = sa[i] - j;
		for ( int i = 1; i <= m; ++ i ) c[i] = 0;
		for ( int i = 1; i <= n; ++ i ) ++ c[x[i]];
		for ( int i = 1; i <= m; ++ i ) c[i] += c[i - 1];
		for ( int i = n; i; -- i ) sa[c[x[y[i]]] --] = y[i], y[i] = 0;
		std::swap ( x, y ), x[sa[1]] = 1, cnt = 1;
		for ( int i = 2; i <= n; ++ i ) {
			x[sa[i]] = cnt += y[sa[i]] ^ y[sa[i - 1]] || y[sa[i] + j] ^ y[sa[i - 1] + j];
		}
		if ( ( m = cnt ) == n ) break;
	}
}

inline void precheck () {
	bool used[26] {};
	for ( int i = 1; i <= n; ++ i ) used[s[i] - 'a'] = true;
	for ( int i = 0, ue = false; i < 26; ++ i ) {
		if ( ue && used[i] ) YES ();
		ue |= ! used[i];
	}
}

inline void check () {
	int nsa[MAXN + 5] {};
	suffixSort ( nsa );
	for ( int i = 1; i <= n; ++ i ) if ( nsa[i] ^ sa[i] ) return ;
	YES ();
}

int main () {
	freopen ( "sa.in", "r", stdin );
	freopen ( "sa.out", "w", stdout );
	scanf ( "%s", s + 1 ), n = strlen ( s + 1 );
	precheck (), suffixSort ( sa );
	for ( int i = 1, las = 'a', t; i <= n; ++ i ) {
		if ( s[sa[i]] == las ) continue;
		t = s[sa[i]], s[sa[i]] = las, check (), las = s[sa[i]] = t;
	}
	return NO (), 0;
}
```cpp