HDU5468(dfs序+容斥原理)
Puzzled Elena
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1162 Accepted Submission(s): 339
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.
For each test, the first line has a number n (1≤n≤105), 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 105.
#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;
const int MAXN=;
typedef long long LL;
vector<LL> divisor[MAXN];
void prep()
{
for(LL e=;e<MAXN;e++)
{
LL x=e;
for(LL i=;i*i<=x;i++)
{
if(x%i==)
{
divisor[e].push_back(i);
while(x%i==) x/=i;
}
}
if(x>) divisor[e].push_back(x);
}
} vector<int> arc[MAXN];
int n,val[MAXN];
int cnt[MAXN],res[MAXN];
int cal(int n,int type)//求集合S中与n不互素的数的个数
{
int ans=;
for(LL mark=;mark<(<<divisor[n].size());mark++)
{
LL odd=;
LL mul=;
for(LL i=;i<divisor[n].size();i++)
{
if(mark&(<<i))
{
odd++;
mul*=divisor[n][i];
}
}
if(odd&) ans+=cnt[mul];
else ans-=cnt[mul];
cnt[mul]+=type;
}
return ans;
}
int dfs(int u,int fa)
{
int pre=cal(val[u],);
int s=;
for(int i=;i<arc[u].size();i++)
{
int to=arc[u][i];
if(to!=fa)
{
s+=dfs(to,u);
}
}
int post=cal(val[u],);
res[u]=s-(post-pre);//以u为根的子树结点数目-(遍历u之前与u不互素的结点数目-遍历u之后与u不互素的结点数目)
if(val[u]==) res[u]++;//若u的值为1,那么u与自身互素
return s+;
}
int main()
{
prep();
int cas=;
while(scanf("%d",&n)!=EOF)
{
memset(cnt,,sizeof(cnt));
for(int i=;i<=n;i++) arc[i].clear();
for(int i=;i<n-;i++)
{
int u,v;
scanf("%d%d",&u,&v);
arc[u].push_back(v);
arc[v].push_back(u);
}
for(int i=;i<=n;i++)
{
scanf("%d",&val[i]);
}
dfs(,-);
printf("Case #%d: ",++cas);
for(int i=;i<n;i++)
{
printf("%d ",res[i]);
}
printf("%d\n",res[n-]);
}
return ;
}
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