HDU 5468 Puzzled Elena (dfs + 莫比乌斯反演)

题意：给定一棵带权树，求每个点与其子树结点的权值互质的个数。

析：首先先要进行 dfs 遍历，len[i] 表示能够整除 i 的个数，在遍历的前和遍历后的差值就是子树的len值，有了这个值，就可以使用莫比斯反演了。注意如果子树的权值是1，还要加上它本身。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++")
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
//#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int maxm = 2e4 + 10;
const LL mod = 100000007;
const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};
const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r >= 0 && r < n && c >= 0 && c < m;
}

bool vis[maxn];
int prime[maxn], mu[maxn];
vector<int> factor[maxn];
void Moblus(){
  mu[1] = 1;  int tot = 0;
  for(int i = 2; i < maxn; ++i){
    if(!vis[i])  prime[tot++] = i, mu[i] = -1;
    for(int j = 0; j < tot; ++j){
      int t = i * prime[j];
      if(t >= maxn)  break;
      vis[t] = 1;
      if(i % prime[j] == 0)  break;
      mu[t] = -mu[i];
    }
  }
  for(int i = 1; i < maxn; ++i)  if(mu[i])
    for(int j = i; j < maxn; j += i)  factor[j].pb(i);
}

struct Edge{
  int to, next;
};
Edge edges[maxn<<1];
int head[maxn], cnt;

void addEdge(int u, int v){
  edges[cnt].to = v;
  edges[cnt].next = head[u];
  head[u] = cnt++;
}

int val[maxn], len[maxn];
int ans[maxn];

void dfs(int u, int fa){
  vector<int> tmp;
  for(int &i : factor[val[u]])  tmp.pb(len[i]);
  for(int i = head[u]; ~i; i = edges[i].next){
    int v = edges[i].to;
    if(v == fa)  continue;
    dfs(v, u);
  }
  ans[u] = 0;
  for(int i = 0; i < tmp.sz; ++i){
    int t = factor[val[u]][i];
    ans[u] += mu[t] * (len[t]-tmp[i]);
    ++len[t];
  }
  if(val[u] == 1)  ++ans[u];
  tmp.cl;
}

int main(){
  Moblus();
  int kase = 0;
  while(scanf("%d", &n) == 1){
    ms(head, -1);  cnt = 0;  ms(len, 0);
    for(int i = 1; i < n; ++i){
      int u, v;  scanf("%d %d", &u, &v);
      addEdge(u, v);
      addEdge(v, u);
    }
    for(int i = 1; i <= n; ++i)  scanf("%d", val + i);
    dfs(1, -1);
    printf("Case #%d:", ++kase);
    for(int i = 1; i <= n; ++i)  printf(" %d", ans[i]);
    printf("\n");
  }
  return 0;
}