From 01c093f641697ed68dbcc045afb0a538acc7c13d Mon Sep 17 00:00:00 2001
From: 24OI-bot <15963390+24OI-bot@users.noreply.github.com>
Date: Wed, 11 Sep 2019 23:34:02 -0400
Subject: [PATCH] style: format markdown files with remark-lint

---
 docs/string/hash.md | 62 ++++++++++++++++++++++-------------------------------
 1 file changed, 26 insertions(+), 36 deletions(-)

diff --git a/docs/string/hash.md b/docs/string/hash.md
index cd5ff271..51332f96 100644
--- a/docs/string/hash.md
+++ b/docs/string/hash.md
@@ -88,17 +88,16 @@ void cmp(const string& s, const string& t) {
 
 ???+note "[CF1200E Compress Words](http://codeforces.com/contest/1200/problem/E)"
     题目大意：给你若干个字符串，答案串初始为空。第 $i$ 步将第 $i$ 个字符串加到答案串的后面，但是尽量地去掉重复部分（即去掉一个最长的、是原答案串的后缀、也是第 $i$ 个串的前缀的字符串），求最后得到的字符串。
-    
+
     每次需要求最长的、是原答案串的后缀、也是第 $i$ 个串的前缀的字符串。枚举这个串的长度，哈希比较即可。
-    
+
     当然，这道题也可以使用 [KMP 算法](./kmp.md) 解决。
-    
+
     ??? note "参考代码"
-        
         ```cpp
+        #include <algorithm>
         #include <cstdio>
         #include <cstring>
-        #include <algorithm>
         
         const int N = 1000010;
         const int m1 = 998244353;
@@ -110,64 +109,55 @@ void cmp(const string& s, const string& t) {
         int m, h1[N], h2[N], len, i1[N], i2[N];
         char s[N];
         
-        void add(char x)
-        {
-          h1[len + 1] = ((ll) h1[len] * K + x) % m1;
-          h2[len + 1] = ((ll) h2[len] * K + x) % m2;
+        void add(char x) {
+          h1[len + 1] = ((ll)h1[len] * K + x) % m1;
+          h2[len + 1] = ((ll)h2[len] * K + x) % m2;
           ++len;
         }
         
-        int get1(int l, int r) { return (ll) (h1[r] - h1[l - 1] + m1) *  i1[l - 1] % m1; }
-        int get2(int l, int r) { return (ll) (h2[r] - h2[l - 1] + m2) *  i2[l - 1] % m2; }
+        int get1(int l, int r) { return (ll)(h1[r] - h1[l - 1] + m1) * i1[l - 1] % m1; }
+        int get2(int l, int r) { return (ll)(h2[r] - h2[l - 1] + m2) * i2[l - 1] % m2; }
         
-        bool cmp(int l1, int r1, int l2, int r2)
-        {
+        bool cmp(int l1, int r1, int l2, int r2) {
           return get1(l1, r1) == get1(l2, r2) && get2(l1, r1) == get2(l2, r2);
         }
         
-        int qpow(int x, int y, int mod)
-        {
+        int qpow(int x, int y, int mod) {
           int out = 1;
-          while (y)
-          {
-            if (y & 1) out = (ll) out * x % mod;
-            x = (ll) x * x % mod;
+          while (y) {
+            if (y & 1) out = (ll)out * x % mod;
+            x = (ll)x * x % mod;
             y >>= 1;
           }
           return out;
         }
         
-        int main()
-        {
+        int main() {
           i1[0] = i2[0] = 1;
-          int k1 = qpow(K, m1 - 2, m1); // 求逆元
+          int k1 = qpow(K, m1 - 2, m1);  // 求逆元
           int k2 = qpow(K, m2 - 2, m2);
-          for (int i = 1; i < N; ++i)
-          {
-            i1[i] = (ll) i1[i - 1] * k1 % m1;
-            i2[i] = (ll) i2[i - 1] * k2 % m2;
+          for (int i = 1; i < N; ++i) {
+            i1[i] = (ll)i1[i - 1] * k1 % m1;
+            i2[i] = (ll)i2[i - 1] * k2 % m2;
           }
-          
+        
           scanf("%d", &m);
-          
-          while (m--)
-          {
+        
+          while (m--) {
             scanf("%s", s + 1);
             int n = strlen(s + 1);
             for (int i = 1; i <= n; ++i) add(s[i]);
             // 先把当前串加到答案串的后面，可以方便地求哈希
-            for (int i = std::min(n, len - n); i >= 0; --i)
-            {
-              if (cmp(len - n - i + 1, len - n, len - n + 1, len - n + i))
-              {
-                len -= n; // 确定了要加多长再真正地加进去
+            for (int i = std::min(n, len - n); i >= 0; --i) {
+              if (cmp(len - n - i + 1, len - n, len - n + 1, len - n + i)) {
+                len -= n;  // 确定了要加多长再真正地加进去
                 for (int j = i + 1; j <= n; ++j) add(s[j]);
                 printf("%s", s + i + 1);
                 break;
               }
             }
           }
-          
+        
           return 0;
         }
         ```
-- 
2.11.0