From 1de1a3e4f313f226747e73afc76d76954b19b061 Mon Sep 17 00:00:00 2001
From: Michael Lee <41327586+chl131@users.noreply.github.com>
Date: Sun, 8 Jun 2025 17:52:55 -0700
Subject: [PATCH 1/2] upper bound of TotalNumbersBeginWith() should be (prefix
 * exp + exp - 1) in problem 440.K-th-Smallest-in-Lexicographical-Order.cpp

---
 .../440.K-th-Smallest-in-Lexicographical-Order.cpp              | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/Others/440.K-th-Smallest-in-Lexicographical-Order/440.K-th-Smallest-in-Lexicographical-Order.cpp b/Others/440.K-th-Smallest-in-Lexicographical-Order/440.K-th-Smallest-in-Lexicographical-Order.cpp
index fe5afc34a..a6a2f39ac 100644
--- a/Others/440.K-th-Smallest-in-Lexicographical-Order/440.K-th-Smallest-in-Lexicographical-Order.cpp
+++ b/Others/440.K-th-Smallest-in-Lexicographical-Order/440.K-th-Smallest-in-Lexicographical-Order.cpp
@@ -33,7 +33,7 @@ class Solution {
         while (1)
         {
             long lower = prefix * exp;
-            long upper = prefix * exp + exp + 1;
+            long upper = prefix * exp + exp - 1;
             if (lower > n) break;
             if (lower <= n && upper >= n)
             {

From fd58a351e18285424d718de3280a1d314683fcfd Mon Sep 17 00:00:00 2001
From: wisdompeak <guan.huifeng@gmail.com>
Date: Fri, 13 Jun 2025 20:07:11 -0700
Subject: [PATCH 2/2] Update Readme.md

---
 .../Readme.md                                               | 6 ++++++
 1 file changed, 6 insertions(+)

diff --git a/Binary_Search/3553.Minimum-Weighted-Subgraph-With-the-Required-Paths-II/Readme.md b/Binary_Search/3553.Minimum-Weighted-Subgraph-With-the-Required-Paths-II/Readme.md
index 21b7d3d06..f210532f1 100644
--- a/Binary_Search/3553.Minimum-Weighted-Subgraph-With-the-Required-Paths-II/Readme.md
+++ b/Binary_Search/3553.Minimum-Weighted-Subgraph-With-the-Required-Paths-II/Readme.md
@@ -1,3 +1,9 @@
 ### 3553.Minimum-Weighted-Subgraph-With-the-Required-Paths-II
 
 本题的第一个知识点是：在一棵树里，联通u,v,w三个节点的最小子树的权重和，就是`[dist(u,v)+dist(u,w)+dist(v,w)]/2`.
+
+本体的第二个知识点是：在一棵树里，联通x,y两点的路径长度，等于`dist(r,x)+dist(r,y)-2*dist(r,c)`，其中r是整棵树的根节点，c是x和y的LCA（lowest common ancester）。
+
+任意一点到距离根节点的距离dist(r,x)可以通过DFS得到。于是本题的关键点就是求任意两点的LCA，于是就是一个binary list经典题。
+
+我们需要处理得到一个数组up[v][k]，表示节点v往上（朝根节点方向）走2^k步能够得到的位置。转移方程就是`up[v][k] = up[up[v][k-1]][k-1]`. 边界条件就是对于一对父子节点a->b，有`up[b][0]=a`.