【CodeForces-280C】Game on Tree(期望可加性)

outputstandard output
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 ≤ 100000) — 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
inputCopy
2
1 2
outputCopy
1.50000000000000000000
inputCopy
3
1 2
1 3
outputCopy
2.00000000000000000000
代码：

#include
#include
#include
#include
#include
#include
#define INF 0x7f7f7f7f
#define maxx 100005
#define mod 998244353
using namespace std;
typedef long long ll;
int head[maxx],to[maxx<<1],_next[maxx<<1];
int edge;
void addEdge(int x,int y)
{
    to[++edge]=y,_next[edge]=head[x],head[x]=edge;
    to[++edge]=x,_next[edge]=head[y],head[y]=edge;
}
int num[maxx];
void dfs(int u,int fa)
{
    num[u]++;
    for(int i=head[u];i;i=_next[i])
    {
        int v=to[i];
        if(v==fa)continue;
        num[v]+=num[u];
        dfs(v,u);
    }
}
int main()
{
    int n;cin>>n;
    int x,y;
    for(int i=1;i<n;i++)
    {
        scanf("%d%d",&x,&y);
        addEdge(x,y);
    }
    dfs(1,0);
    double ans=0;
    for(int i=1;i<=n;i++)
        ans+=1.0/num[i];
    printf("%.7lf\n",ans);
    return 0;
}