HDU 5468 Puzzled Elena

Puzzled Elena

Time Limit: 2500ms

Memory Limit: 131072KB

This problem will be judged on HDU. Original ID: 5468
64-bit integer IO format: %I64d Java class name: Main

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 $(1\leq n\leq 10^5)$, after that has n−1 lines, each line has two numbers a and b$ (1\leq a,b\leq 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

Sample Output

Case #1: 1 1 0 0 0

Source

2015 ACM/ICPC Asia Regional Shanghai Online

解题：莫比乌斯反演

 #include <bits/stdc++.h>

 using namespace std;

 const int maxn = ;

 bool np[maxn] = {true,true};

 int mu[maxn],p[maxn],tot;

 vector<int>fac[maxn],g[maxn];

 void mobius(int n) {

     mu[] = ;

     for(int i = ; i <= n; ++i) {

         if(!np[i]) {

             p[tot++] = i;

             mu[i] = -;

         }

         for(int j = ; j < tot && p[j]*i <= n; ++j) {

             np[p[j]*i] = true;

             if(i%p[j] == ) {

                 mu[p[j]*i] = ;

                 break;

             }

             mu[p[j]*i] = -mu[i];

         }

     }

     for(int i = ; i <= n; ++i) if(mu[i])

             for(int j = i; j <= n; j += i)

                 fac[j].push_back(i);

 }

 int val[maxn],cnt[maxn],sz[maxn],ans[maxn];

 void dfs(int u,int fa) {

     sz[u] = ;

     vector<int>pre;

     for(int &c:fac[val[u]]) {

         pre.push_back(cnt[c]);

         ++cnt[c];

     }

     for(auto &v:g[u]) {

         if(v == fa) continue;

         dfs(v,u);

         sz[u] += sz[v];

     }

     ans[u] = sz[u];

     for(int i = ; i < fac[val[u]].size(); ++i) {

         int x = fac[val[u]][i];

         int y = cnt[x] - pre[i];

         ans[u] += mu[x]*y;

     }

 }

 int main() {

     int n,u,v,cs = ;

     mobius();

     while(~scanf("%d",&n)) {

         for(int i = ; i <= n; ++i) g[i].clear();

         for(int i = ; i < n; ++i) {

             scanf("%d%d",&u,&v);

             g[u].push_back(v);

             g[v].push_back(u);

         }

         for(int i = ; i <= n; ++i)

             scanf("%d",val + i);

         memset(cnt,,sizeof cnt);

         dfs(,);

         printf("Case #%d:",cs++);

         for(int i = ; i <= n; ++i)

             printf(" %d",ans[i]);

         putchar('\n');

     }

     return ;

 }