BZOJ 4401(块的计数-树的划分)

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已知一棵树，请把它划分成k份，使得每个块中的点数都相同。求方案数。

首先有2个结论

一棵树划分成k份的方案是唯一的（但可以不存在）
一棵树可以划分成k份，当且仅当有k个子节点的子树大小是n/k的倍数

#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[n]<<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;}

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