From 8a6d90359679af8daf2292a97a0d985527733727 Mon Sep 17 00:00:00 2001
From: 24OI-bot <15963390+24OI-bot@users.noreply.github.com>
Date: Fri, 13 Sep 2019 22:27:26 -0400
Subject: [PATCH] style: format markdown files with remark-lint

---
 docs/string/seq-automaton.md | 40 ++++++++++++++++++++--------------------
 1 file changed, 20 insertions(+), 20 deletions(-)

diff --git a/docs/string/seq-automaton.md b/docs/string/seq-automaton.md
index 08d37715..69d57093 100644
--- a/docs/string/seq-automaton.md
+++ b/docs/string/seq-automaton.md
@@ -45,7 +45,7 @@ $$
 
 ## 例题
 
-### "[「HEOI2015」最短不公共子串](https://www.luogu.org/problem/P4112)"
+### " [「HEOI2015」最短不公共子串](https://www.luogu.org/problem/P4112) "
 
 这题的 (1) 和 (3) 两问需要后缀自动机，而且做法类似，在这里只讲解 (2) 和 (4) 两问。
 
@@ -53,9 +53,9 @@ $$
 
 (4) 需要 DP。令 $f(i, j)$ 表示在 A 的序列自动机中处于状态 $i$ ，在 B 的序列自动机中处于状态 $j$ ，需要再添加多少个字符能够不是公共子序列。
 
-$f(i, null)=0$
+ $f(i, null)=0$ 
 
-$f(i, j)=\min\limits_{\delta_A(i,c)\ne null}f(\delta_A(i, c), \delta_B(j, c))$ 
+ $f(i, j)=\min\limits_{\delta_A(i,c)\ne null}f(\delta_A(i, c), \delta_B(j, c))$ 
 
 ??? note "整道题的参考代码"
     ```cpp
@@ -105,29 +105,29 @@ $f(i, j)=\min\limits_{\delta_A(i,c)\ne null}f(\delta_A(i, c), \delta_B(j, c))$
     
     int main() {
       scanf("%s%s", s + 1, t + 1);
-      
+    
       n = strlen(s + 1);
       m = strlen(t + 1);
-      
+    
       for (int i = 1; i <= n; ++i) a[i] = s[i] - 'a';
       for (int i = 1; i <= m; ++i) b[i] = t[i] - 'a';
-      
+    
       for (int i = 1; i <= m; ++i) insert(b[i]);
-      
+    
       for (int i = 0; i < 26; ++i) nxt[i] = n + 1;
       for (int i = n; i >= 0; --i) {
         memcpy(na[i], nxt, sizeof(nxt));
         nxt[a[i]] = i;
       }
-      
+    
       for (int i = 0; i < 26; ++i) nxt[i] = m + 1;
       for (int i = m; i >= 0; --i) {
         memcpy(nb[i], nxt, sizeof(nxt));
         nxt[b[i]] = i;
       }
-      
+    
       int ans = N;
-      
+    
       for (int l = 1; l <= n; ++l) {
         for (int r = l, u = 1; r <= n; ++r) {
           u = sam[u].ch[a[r]];
@@ -137,11 +137,11 @@ $f(i, j)=\min\limits_{\delta_A(i,c)\ne null}f(\delta_A(i, c), \delta_B(j, c))$
           }
         }
       }
-      
+    
       printf("%d\n", ans == N ? -1 : ans);
-      
+    
       ans = N;
-      
+    
       for (int l = 1; l <= n; ++l) {
         for (int r = l, u = 0; r <= n; ++r) {
           u = nb[u][a[r]];
@@ -151,9 +151,9 @@ $f(i, j)=\min\limits_{\delta_A(i,c)\ne null}f(\delta_A(i, c), \delta_B(j, c))$
           }
         }
       }
-      
+    
       printf("%d\n", ans == N ? -1 : ans);
-      
+    
       for (int i = n; i >= 0; --i) {
         for (int j = 1; j <= tot; ++j) {
           f[i][j] = N;
@@ -164,11 +164,11 @@ $f(i, j)=\min\limits_{\delta_A(i,c)\ne null}f(\delta_A(i, c), \delta_B(j, c))$
           }
         }
       }
-      
+    
       printf("%d\n", f[0][1] == N ? -1 : f[0][1]);
-      
+    
       memset(f, 0, sizeof(f));
-      
+    
       for (int i = n; i >= 0; --i) {
         for (int j = 0; j <= m; ++j) {
           f[i][j] = N;
@@ -179,9 +179,9 @@ $f(i, j)=\min\limits_{\delta_A(i,c)\ne null}f(\delta_A(i, c), \delta_B(j, c))$
           }
         }
       }
-      
+    
       printf("%d\n", f[0][0] == N ? -1 : f[0][0]);
-      
+    
       return 0;
     }
     ```
-- 
2.11.0