HDU 5293(ZYB's Tree-DP)

给一棵树，对于每一个点,求出离每个点距离不超过K的点的个数.
1≤N≤500000,1≤K≤10

树分治太慢了，直接统计多好

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
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 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 (x<<1)
#define Rson ((x<<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)
typedef long long ll;
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 (500000+10)
#define MAXK (11)
int Pre[MAXN],Next[MAXN*2],edge[MAXN*2],siz=1;
void addedge(int u,int v){
    edge[++siz]=v;
    Next[siz]=Pre[u];
    Pre[u]=siz;
}
void addedge2(int u,int v) {addedge(u,v),addedge(v,u);}
ll c[MAXN];



int f[MAXN];

int n,K,A,B;
int g[MAXN][MAXK];
void solve(int root)
{
    Rep(i,K+1) g[root][i]=1;
    Forp(root) {
        int v=edge[p];
        solve(v);
        For(i,K) g[root][i] += g[v][i-1];  
    }

} 
void calc(int root,int k2)
{
    c[root]=g[root][k2];
    int p=root;
    while ((p^1) && (k2--) ) {
        c[root]+=g[f[p]][k2];
        if (k2) c[root]-=g[p][k2-1];

        p=f[p];
    }  
} 

int main()
{
// freopen("D.in","r",stdin);
// freopen(".out","w",stdout);

    int T=read();
    while(T--) {
        n=read(),K=read(),A=read(),B=read();
        For(i,n) Pre[i]=c[i]=0; siz=1;



        f[1]=0;
        Fork(i,2,n) {
            f[i]=((ll)A*(ll)i+B)%(i-1)+1;
            addedge(f[i],i);
        }

        solve(1);
        For(i,n) calc(i,K);
        ll t=0;
        For(i,n) t^=c[i];
        cout<<t<<endl;



    }

    return 0;
}