diff --git a/GESP/NoBreak.cpp b/GESP/NoBreak.cpp
new file mode 100644
index 0000000..1fac6df
--- /dev/null
+++ b/GESP/NoBreak.cpp
@@ -0,0 +1,33 @@
+#include <iostream>
+using namespace std;
+// C++入门经典-例3.13-不加break的switch判断语句
+// https://blog.csdn.net/lovemi93/article/details/90702825
+
+int main() {
+    char c;
+    while (1) {
+        cin >> c;
+        if (c == 'q') break;
+
+        switch (c) {
+        case 'A':
+            cout << "1 ";
+            break;
+        case 'B':
+            cout << "3 ";
+        case 'C':
+            cout << "3 ";
+        case 'D':
+            cout << "Here!";
+            cout << "5 ";
+            break;
+        case 'E':
+            cout << "5 ";
+            break;
+        default:
+            cout << "9 " << endl;
+        }
+        cout << endl;
+    }
+    return 0;
+}
\ No newline at end of file
diff --git a/TangDou/AcWing/MinimalPath/920.cpp b/TangDou/AcWing/MinimalPath/920.cpp
index 5d23372..6af0e6c 100644
--- a/TangDou/AcWing/MinimalPath/920.cpp
+++ b/TangDou/AcWing/MinimalPath/920.cpp
@@ -5,12 +5,12 @@ const int INF = 0x3f3f3f3f;
 typedef pair<int, int> PII;
 
 const int N = 1e5 + 10;
-const int M = 2 * N;
+const int M = N * 2;
 
 int n; // 总共有N个车站
 int m; // 开通了M条单程巴士线路
 int h[N], e[M], w[M], ne[M], idx;
-int dist[N]; // 最小距离数组
+int dis[N];  // 最小距离数组
 int stop[N]; // 站点数组
 bool st[N];  // 是否在队列中
 
@@ -19,8 +19,8 @@ void add(int a, int b, int c) {
 }
 // 求1号点到n号点的最短路距离，如果从1号点无法走到n号点则返回-1
 void dijkstra() {
-    memset(dist, 0x3f, sizeof dist);                  // 求最小设最大
-    dist[1] = 0;                                      // 1到自己，乘车数0
+    memset(dis, 0x3f, sizeof dis);                    // 求最小设最大
+    dis[1] = 0;                                       // 1到自己，乘车数0
     priority_queue<PII, vector<PII>, greater<PII>> q; // 小顶堆
     q.push({0, 1});                                   // 1号入队列
 
@@ -28,14 +28,14 @@ void dijkstra() {
         auto t = q.top();
         q.pop();
         int u = t.second;
-        int d = t.first; // 此处 d=t.first没有用上，经测试，其实d=dist[u]，用哪个都是一样的
+        // int d = t.first; // 此处 d=t.first没有用上，经测试，其实d=dis[u]，用哪个都是一样的
         if (st[u]) continue;
         st[u] = true;
         for (int i = h[u]; ~i; i = ne[i]) {
             int v = e[i];
-            if (dist[v] > dist[u] + w[i]) {
-                dist[v] = dist[u] + w[i];
-                q.push({dist[v], v});
+            if (dis[v] > dis[u] + w[i]) {
+                dis[v] = dis[u] + w[i];
+                q.push({dis[v], v});
             }
         }
     }
@@ -43,16 +43,16 @@ void dijkstra() {
 
 int main() {
     memset(h, -1, sizeof h); // 初始化邻接表
-    scanf("%d%d", &m, &n);   // 总共有N个车站,开通了M条单程巴士线路
+    cin >> m >> n;           // 总共有N个车站,开通了M条单程巴士线路
     while (m--) {            // m条边
         // ① 先读入第一个数字
         int cnt = 0; // cnt一定要清零
-        scanf("%d", &stop[++cnt]);
+        cin >> stop[++cnt];
         char ch = getchar();
         while (ch == ' ') {
             // ② 读入其它数字
-            scanf("%d", &stop[++cnt]); // 还有就继续读
-            ch = getchar();            // 为下一次做准备
+            cin >> stop[++cnt]; // 还有就继续读
+            ch = getchar();     // 为下一次做准备
         }
         // 这个建图建的妙啊！
         // 通过多条边，成功映射了问题，将一趟车问题转化为多个点之间边是1问题
@@ -62,9 +62,9 @@ int main() {
     }
 
     dijkstra();
-    if (dist[n] == INF)
+    if (dis[n] == INF)
         puts("NO");
     else
-        printf("%d\n", dist[n] - 1);
+        printf("%d\n", dis[n] - 1);
     return 0;
 }
\ No newline at end of file
diff --git a/TangDou/AcWing/Test/796.cpp b/TangDou/AcWing/Test/796.cpp
new file mode 100644
index 0000000..4e5d998
--- /dev/null
+++ b/TangDou/AcWing/Test/796.cpp
@@ -0,0 +1,36 @@
+#include <bits/stdc++.h>
+using namespace std;
+const int N = 1010;
+int a[N][N];
+int s[N][N];
+
+int main() {
+    int n, m, q;
+    cin >> n >> m >> q;
+    for (int i = 1; i <= n; i++)
+        for (int j = 1; j <= m; j++) {
+            cin >> a[i][j];
+            s[i][j] = s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1] + a[i][j];
+        }
+
+    while (q--) {
+        int x1, y1, x2, y2;
+        cin >> x1 >> y1 >> x2 >> y2;
+        printf("%d\n", s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1]);
+    }
+    return 0;
+/*
+输入样例：
+3 4 3
+1 7 2 4
+3 6 2 8
+2 1 2 3
+1 1 2 2
+2 1 3 4
+1 3 3 4
+输出样例：
+17
+27
+21
+    */
+}
diff --git a/TangDou/LuoGuBook/P2615.cpp b/TangDou/LuoGuBook/P2615.cpp
index da53ea9..b085a9c 100644
--- a/TangDou/LuoGuBook/P2615.cpp
+++ b/TangDou/LuoGuBook/P2615.cpp
@@ -6,7 +6,7 @@ int n;
 int g[N][N];
 
 int main() {
-    scanf("%d", &n);
+    cin >> n;
     int r = 1;
     int c = n / 2 + 1;
     g[r][c] = 1;