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
5 1 2 1 3 2 4 2 5 6 2 3 4 5
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
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int maxn = 100010;
4 bool np[maxn] = {true,true};
5 int mu[maxn],p[maxn],tot;
6 vector<int>fac[maxn],g[maxn];
7 void mobius(int n) {
8 mu[1] = 1;
9 for(int i = 2; i <= n; ++i) {
10 if(!np[i]) {
11 p[tot++] = i;
12 mu[i] = -1;
13 }
14 for(int j = 0; j < tot && p[j]*i <= n; ++j) {
15 np[p[j]*i] = true;
16 if(i%p[j] == 0) {
17 mu[p[j]*i] = 0;
18 break;
19 }
20 mu[p[j]*i] = -mu[i];
21 }
22 }
23 for(int i = 2; i <= n; ++i) if(mu[i])
24 for(int j = i; j <= n; j += i)
25 fac[j].push_back(i);
26 }
27 int val[maxn],cnt[maxn],sz[maxn],ans[maxn];
28 void dfs(int u,int fa) {
29 sz[u] = 1;
30 vector<int>pre;
31 for(int &c:fac[val[u]]) {
32 pre.push_back(cnt[c]);
33 ++cnt[c];
34 }
35 for(auto &v:g[u]) {
36 if(v == fa) continue;
37 dfs(v,u);
38 sz[u] += sz[v];
39 }
40 ans[u] = sz[u];
41 for(int i = 0; i < fac[val[u]].size(); ++i) {
42 int x = fac[val[u]][i];
43 int y = cnt[x] - pre[i];
44 ans[u] += mu[x]*y;
45 }
46 }
47 int main() {
48 int n,u,v,cs = 1;
49 mobius(100000);
50 while(~scanf("%d",&n)) {
51 for(int i = 1; i <= n; ++i) g[i].clear();
52 for(int i = 1; i < n; ++i) {
53 scanf("%d%d",&u,&v);
54 g[u].push_back(v);
55 g[v].push_back(u);
56 }
57 for(int i = 1; i <= n; ++i)
58 scanf("%d",val + i);
59 memset(cnt,0,sizeof cnt);
60 dfs(1,0);
61 printf("Case #%d:",cs++);
62 for(int i = 1; i <= n; ++i)
63 printf(" %d",ans[i]);
64 putchar(‘\n‘);
65 }
66 return 0;
67 }