diff --git a/TangDou/AcWing_TiGao/T1/LIS/1014.md b/TangDou/AcWing_TiGao/T1/LIS/1014.md
index 9b442b1..321012a 100644
--- a/TangDou/AcWing_TiGao/T1/LIS/1014.md
+++ b/TangDou/AcWing_TiGao/T1/LIS/1014.md
@@ -107,44 +107,49 @@ int res;
 
 int main() {
     cin >> n;
-    for (int i = 1; i <= n; i++) cin >> a[i];
+    for (int i = 0; i < n; i++) cin >> a[i];
 
     // 正向
-    f[++fl] = a[1];
-    p1[1] = 1;
+    f[0] = a[0];
+    p1[0] = 1;
 
-    for (int i = 2; i <= n; i++)
+    for (int i = 0; i < n; i++)
         if (a[i] > f[fl]) {
             f[++fl] = a[i];
             p1[i] = fl;
         } else {
-            int t = lower_bound(f + 1, f + fl + 1, a[i]) - f;
+            int t = lower_bound(f, f + fl, a[i]) - f;
             f[t] = a[i];
             p1[i] = t;
         }
 
     // 反向
-    g[++gl] = a[n];
-    p2[n] = 1;
+    g[0] = a[n - 1];
+    p2[n - 1] = 1;
 
-    for (int i = n - 1; i >= 1; i--)
+    for (int i = n - 1; i >= 0; i--)
         if (a[i] > g[gl]) {
             g[++gl] = a[i];
             p2[i] = gl;
         } else {
-            int t = lower_bound(g + 1, g + gl + 1, a[i]) - g;
+            int t = lower_bound(g, g + gl, a[i]) - g;
             g[t] = a[i];
             p2[i] = t;
         }
 
-    for (int i = 1; i <= n; i++) res = max(res, p2[i] + p1[i] - 1);
+    for (int i = 0; i < n; i++) res = max(res, p2[i] + p1[i] + 2 - 1);
 
     printf("%d\n", res);
     return 0;
 }
 ```
 
-### 六、状态机分析法
+>**总结**：
+① 朴素版本性能较差$O(N^2)$,但有一个先天的优势：可以知道每个数字作为最高点时，左边有多少个，右边有多少个，对于求左右最长这样的题目不用再拐弯了。
+② 贪心+二分版本的性能好$O(LogN*N)$,但有一个缺点，就是只能获取到最长上升序列的长度，不知道 **以每个数字为最高点时，左边有多少个，右边有多少个**，如果非得用这个算法不可的话，那么就需要对贪心+二分版本的代码进行改造：用数组记录第几个数字在上升序列中应该是排在第几位的，显得麻烦一些。
+
+
+### 六、状态机分析法【选读$O(N^2)$】
 这是我做过 $AcWing$第二场热身赛的$C$题——[$AcWing$ $3549$. 最长非递减子序列](https://www.acwing.com/problem/content/3552/) 后总结出来的一类模型
 
 那就是，**利用状态机模型$DP$解决最长$xxx$子序列模型** 的方法
diff --git a/TangDou/AcWing_TiGao/T1/LIS/1014_2.cpp b/TangDou/AcWing_TiGao/T1/LIS/1014_2.cpp
index ae3bead..4a9310c 100644
--- a/TangDou/AcWing_TiGao/T1/LIS/1014_2.cpp
+++ b/TangDou/AcWing_TiGao/T1/LIS/1014_2.cpp
@@ -16,37 +16,37 @@ int res;
 
 int main() {
     cin >> n;
-    for (int i = 1; i <= n; i++) cin >> a[i];
+    for (int i = 0; i < n; i++) cin >> a[i];
 
     // 正向
-    f[++fl] = a[1];
-    p1[1] = 1;
+    f[0] = a[0];
+    p1[0] = 1;
 
-    for (int i = 2; i <= n; i++)
+    for (int i = 0; i < n; i++)
         if (a[i] > f[fl]) {
             f[++fl] = a[i];
             p1[i] = fl;
         } else {
-            int t = lower_bound(f + 1, f + fl + 1, a[i]) - f;
+            int t = lower_bound(f, f + fl, a[i]) - f;
             f[t] = a[i];
             p1[i] = t;
         }
 
     // 反向
-    g[++gl] = a[n];
-    p2[n] = 1;
+    g[0] = a[n - 1];
+    p2[n - 1] = 1;
 
-    for (int i = n - 1; i >= 1; i--)
+    for (int i = n - 1; i >= 0; i--)
         if (a[i] > g[gl]) {
             g[++gl] = a[i];
             p2[i] = gl;
         } else {
-            int t = lower_bound(g + 1, g + gl + 1, a[i]) - g;
+            int t = lower_bound(g, g + gl, a[i]) - g;
             g[t] = a[i];
             p2[i] = t;
         }
 
-    for (int i = 1; i <= n; i++) res = max(res, p2[i] + p1[i] - 1);
+    for (int i = 0; i < n; i++) res = max(res, p2[i] + p1[i] + 2 - 1);
 
     printf("%d\n", res);
     return 0;