题干：
Momiji has got a rooted tree, consisting of n nodes. The tree nodes are numbered by integers from 1 to n. The root has number 1. Momiji decided to play a game on this tree.
The game consists of several steps. On each step, Momiji chooses one of the remaining tree nodes (let's denote it by v) and removes all the subtree nodes with the root in node v from the tree. Node v gets deleted as well. The game finishes when the tree has no nodes left. In other words, the game finishes after the step that chooses the node number 1.
Each time Momiji chooses a new node uniformly among all the remaining nodes. Your task is to find the expectation of the number of steps in the described game.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of nodes in the tree. The next n - 1 lines contain the tree edges. The i-th line contains integers ai, bi(1 ≤ ai, bi ≤ n; ai ≠ bi) — the numbers of the nodes that are connected by the i-th edge.
It is guaranteed that the given graph is a tree.
Output
Print a single real number — the expectation of the number of steps in the described game.
The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.
Examples
Input
2
1 2
Output
1.50000000000000000000
Input
3
1 2
1 3
Output
2.00000000000000000000
Note
In the first sample, there are two cases. One is directly remove the root and another is remove the root after one step. Thus the expected steps are:
1 × (1 / 2) + 2 × (1 / 2) = 1.5
In the second sample, things get more complex. There are two cases that reduce to the first sample, and one case cleaned at once. Thus the expected steps are:
1 × (1 / 3) + (1 + 1.5) × (2 / 3) = (1 / 3) + (5 / 3) = 2
题目大意：
给一颗n个点有根的树，每次任意删一个当前还存在的点，并删掉其子树，问删完整颗树的删点次数的数学期望。
解题报告：
琢磨了半天，似懂非懂，先记下来以后再加深理解吧。
考虑等价问题：对于每一个点，假设考虑第i个点，选择删第i个点的平均次数 的和。
既然最终第i个点肯定要被删除的，那么肯定是他和他到根节点这一条链上的节点 都可以做到，那么其中假设深度为d，那么选择这个点的概率就是，那么对所有的点求个和就是答案。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define F first
#define S second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
int dep[MAX],n;
vector<int> vv[MAX];
void dfs(int cur,int fa,int d) {dep[cur] = d;for(int i = 0; i<vv[cur].size(); i++) {if(vv[cur][i] == fa) continue;dfs(vv[cur][i],cur,d+1);}
}
int main()
{cin>>n;for(int a,b,i = 1; i<=n-1; i++) cin>>a>>b,vv[a].pb(b),vv[b].pb(a);dfs(1,-1,1);double ans = 0;for(int i = 1; i<=n; i++) {ans += 1.0/dep[i];}printf("%.10f\n",ans);return 0 ;
}
