Manthan, Codefest 16 G. Yash And Trees dfs序+线段树+bitset

G. Yash And Trees

题目连接：

http://www.codeforces.com/contest/633/problem/G

Description

Yash loves playing with trees and gets especially excited when they have something to do with prime numbers. On his 20th birthday he was granted with a rooted tree of n nodes to answer queries on. Hearing of prime numbers on trees, Yash gets too intoxicated with excitement and asks you to help out and answer queries on trees for him. Tree is rooted at node 1. Each node i has some value ai associated with it. Also, integer m is given.

There are queries of two types:

for given node v and integer value x, increase all ai in the subtree of node v by value x
for given node v, find the number of prime numbers p less than m, for which there exists a node u in the subtree of v and a non-negative integer value k, such that au = p + m·k.

Input

The first of the input contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ 1000) — the number of nodes in the tree and value m from the problem statement, respectively.

The second line consists of n integers ai (0 ≤ ai ≤ 109) — initial values of the nodes.

Then follow n - 1 lines that describe the tree. Each of them contains two integers ui and vi (1 ≤ ui, vi ≤ n) — indices of nodes connected by the i-th edge.

Next line contains a single integer q (1 ≤ q ≤ 100 000) — the number of queries to proceed.

Each of the last q lines is either 1 v x or 2 v (1 ≤ v ≤ n, 0 ≤ x ≤ 109), giving the query of the first or the second type, respectively. It's guaranteed that there will be at least one query of the second type.

Output

For each of the queries of the second type print the number of suitable prime numbers.

Sample Input

8 20
3 7 9 8 4 11 7 3
1 2
1 3
3 4
4 5
4 6
4 7
5 8
4
2 1
1 1 1
2 5
2 4

Sample Output

3
1
1

题意

给你一棵树，然后给你n,m

每个点有一个点权

有两个操作

1 x v 使得x子树里面的所有点的权值加v

2 x 查询x的子树里面有多少个质数p满足p+k*m，其中p是小于m的素数

题解：

这题感觉毫无头绪，看了大佬的博客才发一个神奇东西叫bitset。

先dfs将子树问题转化成区间问题。

然后线段树维护整个区间，每个节点挂一个bitset存这个子区间所含质数情况。

更新操作，就直接让这个bitset左右移动就好了，移动可以拆成两个步骤，一个向右边移动(x%m)，一个向左边移动(m-(x%m))，然后两个取并集。

查询操作，就最后得到那个区间的bitset和素数表&一下就好了

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define INF 0x7f7f7f7f
#define N 100050
#define M 200050
#define MAXN 1050
bitset<MAXN>Pri;
int n,m,k,tot,cnt;
int last[N],rk[N],kth[N],size[N],w[N];
struct Edge{int from,to,s;}edges[M];
struct Tree{int l,r,lazy;bitset<MAXN>sum;}tr[N<<2];
template<typename T>void read(T&x)
{
    ll k=0; char c=getchar();
    x=0;
    while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
    if (c==EOF)exit(0);
    while(isdigit(c))x=x*10+c-'0',c=getchar();
    x=k?-x:x;
}
void read_char(char &c)
{while(!isalpha(c=getchar())&&c!=EOF);}
void AddEdge(int x,int y)
{
    edges[++tot]=Edge{x,y,last[x]};
    last[x]=tot;
}
void get_pri()
{
    static int f[M];
    for(int i=2;i<m;i++)
    {
        if (f[i])continue;
        Pri[i]=1;
        int j=i;
        while(j+i<m)f[j+i]=1,j=j+i;
    }
}
void dfs(int x,int pre)
{
    kth[++cnt]=x;
    rk[x]=cnt;
    size[x]=1;
    for(int i=last[x];i;i=edges[i].s)
    {
        Edge & e=edges[i];
        if (e.to==pre)continue;
        dfs(e.to,x);
        size[x]+=size[e.to];
    }
}
void push_up(int x)
{
    tr[x].lazy%=m;
    if (tr[x].l!=tr[x].r)tr[x].sum=tr[x<<1].sum|tr[x<<1|1].sum;
    tr[x].sum=(tr[x].sum<<tr[x].lazy)|(tr[x].sum>>(m-tr[x].lazy));
    if (tr[x].l==tr[x].r)tr[x].lazy=0;   
}
void push_down(int x)
{
    tr[x<<1].lazy+=tr[x].lazy;
    tr[x<<1|1].lazy+=tr[x].lazy;
    tr[x].lazy=0;
    push_up(x<<1);
    push_up(x<<1|1);
}
void bt(int x,int l,int r)
{
    tr[x].l=l; tr[x].r=r; tr[x].lazy=0;
    if (l==r)
    {
        tr[x].sum[w[kth[l]]]=1;
        return ;
    }
    int mid=(l+r)>>1;
    bt(x<<1,l,mid);
    bt(x<<1|1,mid+1,r);
    push_up(x);
}
void update(int x,int l,int r,int tt)
{
    if (l<=tr[x].l&&tr[x].r<=r)
    {
        tr[x].lazy+=tt;
        push_up(x);
        return;
    }
    int mid=(tr[x].l+tr[x].r)>>1;
    if (l<=mid)update(x<<1,l,r,tt);
    if (mid<r)update(x<<1|1,l,r,tt);
    push_up(x);
}
bitset<MAXN> query(int x,int l,int r)
{
    if (l<=tr[x].l&&tr[x].r<=r)return tr[x].sum;
    int mid=(tr[x].l+tr[x].r)>>1;
    bitset<MAXN>ans(0);
    push_down(x);
    if (l<=mid)ans|=query(x<<1,l,r);
    if (mid<r)ans|=query(x<<1|1,l,r);
    push_up(x);
    return ans;
}
void work()
{
    read(n); read(m);
    get_pri();
    for(int i=1;i<=n;i++)read(w[i]),w[i]%=m;
    for(int i=1;i<n;i++)
    {
        int x,y;
        read(x); read(y);
        AddEdge(x,y);
        AddEdge(y,x);
    }
    dfs(1,0);
    bt(1,1,n);
    read(k);
    for(int i=1;i<=k;i++)
    {
        int id,x,tt;
        read(id);
        if (id==1){read(x);read(tt);update(1,rk[x],rk[x]+size[x]-1,tt);}
        if (id==2)
        {
            read(x);
            bitset<MAXN>ans=query(1,rk[x],rk[x]+size[x]-1)&Pri;
            printf("%d\n",(int)(ans.count()));
        }
    }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("aa.in","r",stdin);
#endif
    work();
}