【HDU - 5468 】Puzzled Elena（容斥）

Since both Stefan and Damon fell in love with Elena, and it was really difficult for her to choose. Bonnie, her best friend, suggested her to throw a question to them, and she would choose the one who can solve it.

Suppose there is a tree with n vertices and n - 1 edges, and there is a value at each vertex. The root is vertex 1. Then for each vertex, could you tell me how many vertices of its subtree can be said to be co-prime with itself?
NOTES: Two vertices are said to be co-prime if their values’ GCD (greatest common divisor) equals 1.
Input
There are multiply tests (no more than 8).
For each test, the first line has a number n ($1≤n≤10^5$), after that has n−1 lines, each line has two numbers a and b $(1≤a,b≤n)$, representing that vertex a is connect with vertex b. Then the next line has n numbers, the ith number indicates the value of the ith vertex. Values of vertices are not less than $1$ and not more than $10^5$.
Output
For each test, at first, please output “Case #k: “, k is the number of test. Then, please output one line with n numbers (separated by spaces), representing the answer of each vertex.
Sample Input
5
1 2
1 3
2 4
2 5
6 2 3 4 5
Sample Output
Case #1: 1 1 0 0 0
思路：当时做这题的idea其实是联想到今年山东acm省赛热身赛的B题（当时做掉这题，我们的rank就直接到第5了，在正赛造了报应，，，），我们先考虑一个问题，先解决如何在一堆整数集合当中找到互质的个数，可以考虑枚举，但是在树上做这个必然会T,我换了一个角度而是去观察每个数具有什么属性，这个就是容斥，属性的个数是有多少个sqare-free number是他的因子。极端情况是$2^6-1$个，即这个数是$2*3*5*7*11*13$的连乘，这种情况是最坏的情况，复杂度虽然$O(n)$，但是常数就很大，但是目测还是可以过掉，优化的地方是因为case稍多，所以可以预处理一下每个数的sqare-free number，可以筛一下莫比乌斯函数来做，同时也需要莫比乌斯函数来容斥答案。
代码:

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#define ll long long
#define maxx 100005
using namespace std;
vector<int> fac[maxx];
int mu[maxx];
bool p[maxx];
int prime[maxx],cnt;
void init()
{
    p[1]=mu[1]=1;
    for(int i=2;i<maxx;i++)
    {
        if(!p[i])
        {
            prime[cnt++]=i;
            mu[i]=-1;
        }
        for(int j=0;j<cnt&&i*prime[j]<=maxx;j++)
        {
            p[i*prime[j]]=true;
            if(i%prime[j])
                mu[i*prime[j]]=-mu[i];
            else
            {
                mu[i*prime[j]]=0;
                break;
            }
        }
    }
    for(int i=2;i<maxx;i++)
    {
        if(mu[i]==0)continue;
        for(int j=i;j<maxx;j+=i)
            fac[j].push_back(i);
    }
}
int head[maxx],to[maxx<<1],_next[maxx<<1],edge;
int c[maxx];
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 n;
int a[maxx];
int total;
int ans[maxx];
void dfs(int u,int fa)
{
    int num=a[u];
    int temp=total;
    for(int i=0;i<fac[num].size();i++)
    {
        if(mu[fac[num][i]]==-1)temp-=c[fac[num][i]];
        else temp+=c[fac[num][i]];
    }
    for(int i=head[u];i;i=_next[i])
    {
        int v=to[i];
        if(v==fa)continue;
        dfs(v,u);
    }
    ++total;
    for(int i=0;i<fac[num].size();i++) ++c[fac[num][i]];
    ans[u]=total;
    for(int i=0;i<fac[num].size();i++)
    {
        int _fac=fac[num][i];
        if(mu[_fac]==-1)ans[u]-=c[_fac];
        else ans[u]+=c[_fac];
    }

    ans[u]-=temp;
}
void _init()
{
    total=0;
    edge=0;
    memset(head,0,sizeof(head));
    memset(c,0,sizeof(head));

}
int main()
{
    init();
    int cal=1;
    int x,y;
    while(scanf("%d",&n)==1)
    {
        _init();
        for(int i=1;i<n;i++)
        {
            scanf("%d%d",&x,&y);
            addEdge(x,y);
        }
        for(int i=1;i<=n;i++)
            scanf("%d",a+i);
        dfs(1,0);
        printf("Case #%d:",cal++);
        for(int i=1;i<n;i++)printf(" %d",ans[i]);
        printf(" %d\n",ans[n]);
    }
    return 0;
}