BZOJ 4401(块的计数-树的划分)

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=Pre[x];p;p=Next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=Next[p])
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case %d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a
<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
typedef long long ll;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
#define MAXN (1000000+10)
vector<int> G[MAXN];
int siz[MAXN]={0};
void dfs(int x,int fa){
siz[x]=1;
int m=SI(G[x]);
Rep(i,m) {
int v=G[x][i];
if (v==fa) continue;
dfs(v,x);
siz[x]+=siz[v];
}
}
int cnt[MAXN]={0};
int main()
{
//  freopen("bzoj4401.in","r",stdin);
//  freopen(".out","w",stdout);
int n=read();
For(i,n-1) {
int a=read(),b=read();
G[a].pb(b);
G[b].pb(a);
}
dfs(1,0);
For(i,n) cnt[siz[i]]++;
int ans=0;
For(i,n) {
for(int j=2*i;j<=n;j+=i)
cnt[i]+=cnt[j];
if (cnt[i]*i==n) ans++;
}
cout<<ans<<endl;
return 0;
}