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Merged

pull merged 3 commits into AlgorithmAndLeetCode:master from wisdompeak:master

Jun 14, 2025

Merged

[pull] master from wisdompeak:master #323

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upper bound of TotalNumbersBeginWith() should be (prefix * exp + exp ...

chl131 Jun 9, 2025

Update Readme.md

wisdompeak Jun 14, 2025

Merge pull request #113 from chl131/patch-1

wisdompeak Jun 14, 2025

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6 changes: 6 additions & 0 deletions Binary_Search/3553.Minimum-Weighted-Subgraph-With-the-Required-Paths-II/Readme.md

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Original file line number	Diff line number	Diff line change
		@@ -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`.

2 changes: 1 addition & 1 deletion ...440.K-th-Smallest-in-Lexicographical-Order/440.K-th-Smallest-in-Lexicographical-Order.cpp

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -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)
		{
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