style: format markdown files with remark-lint

diff --git a/docs/basic/prefix-sum.md b/docs/basic/prefix-sum.md
index 85d58fc..3e188ea 100644 (file)
--- a/docs/basic/prefix-sum.md
+++ b/docs/basic/prefix-sum.md
@@ -56,7 +56,7 @@ int main() {
 -   [洛谷 U53525 前缀和（例题）](https://www.luogu.org/problemnew/show/U53525)
 -   [洛谷 U69096 前缀和的逆](https://www.luogu.org/problemnew/show/U69096)
 -   [AT2412 最大の和](https://www.luogu.org/problemnew/show/AT2412)
--   [洛谷 P3131 [USACO16JAN]子共七Subsequences Summing to Sevens](https://www.luogu.org/problemnew/show/P3131)
+-   [洛谷 P3131 \[USACO16JAN\] 子共七 Subsequences Summing to Sevens](https://www.luogu.org/problemnew/show/P3131)
 
 ### 参考
 
@@ -90,48 +90,42 @@ int main() {
 第二个问题就是如何应用，譬如求 $(x1,y1) - (x2,y2)$ 子矩阵的和。  
 那么，根据类似的思考过程，易得答案为 $sum_{x2,y2} - sum_{x1 - 1,y2} - sum_{x2,y1 - 1} + sum_{x1 - 1,y1 - 1}$ 。
 
-下面给出 [洛谷 P1387 最大正方形](https://www.luogu.org/problemnew/show/P1387) 这道题目的参考程序来帮助大家理解二维前缀和。
+下面给出[洛谷 P1387 最大正方形](https://www.luogu.org/problemnew/show/P1387)这道题目的参考程序来帮助大家理解二维前缀和。
 
 ```cpp
-#include <iostream>
 #include <algorithm>
+#include <iostream>
 using namespace std;
 int a[103][103];
-int b[103][103];     // 前缀和数组，相当于上文的 sum[]
-int main()
-{
-    int n, m;
-    cin >> n >> m;
-
-    for (int i = 1; i <= n; i++)
-    {
-        for (int j = 1; j <= m; j++)
-        {
-            cin >> a[i][j];
-            b[i][j] = b[i][j-1] + b[i-1][j] - b[i-1][j-1] + a[i][j];   // 求前缀和
-        }
+int b[103][103];  // 前缀和数组，相当于上文的 sum[]
+int main() {
+  int n, m;
+  cin >> n >> m;
+
+  for (int i = 1; i <= n; i++) {
+    for (int j = 1; j <= m; j++) {
+      cin >> a[i][j];
+      b[i][j] =
+          b[i][j - 1] + b[i - 1][j] - b[i - 1][j - 1] + a[i][j];  // 求前缀和
     }
+  }
 
-    int ans = 1;
-
-    int l = 2;
-    while (l <= min(n, m))
-    {
-        for (int i = l; i <= n; i++)
-        {
-            for (int j = l; j <= m; j++)
-            {
-                if (b[i][j] - b[i-l][j] - b[i][j-l] + b[i-l][j-l] == l*l)
-                {
-                    ans = max(ans, l);
-                }
-            }
+  int ans = 1;
+
+  int l = 2;
+  while (l <= min(n, m)) {
+    for (int i = l; i <= n; i++) {
+      for (int j = l; j <= m; j++) {
+        if (b[i][j] - b[i - l][j] - b[i][j - l] + b[i - l][j - l] == l * l) {
+          ans = max(ans, l);
         }
-        l++;
+      }
     }
-    
-    cout << ans << endl;
-    return 0;
+    l++;
+  }
+
+  cout << ans << endl;
+  return 0;
 }
 ```