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

Hint

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点的权值满足p+k*m，其中p是小于m的素数

题解：

就直接dfs序跑一发就好了

我们用bitset维护每一个点他下面出现过哪些数

更新操作，就直接让这个bitset循环移动就好了，注意循环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)，然后两个并起来就好了

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;
const int maxn = 200005;
const int MAXN = 1005;
typedef bitset<MAXN> SgTreeDataType;
vector<int> E[maxn];
bitset<MAXN>pri;
int st[maxn],ed[maxn],tot=0,n,m,v[1020],a[maxn],c[maxn];
struct treenode
{
    int L , R  ;
    SgTreeDataType sum;
    int lazy;
    void update(int x)
    {
        lazy=(lazy+x)%m;
        sum=(sum<<x)|(sum>>(m-x));
    }
};

treenode tree[maxn*4];

inline void push_down(int o)
{
    int lazyval = tree[o].lazy;
	if(lazyval)
    {
        tree[2*o].update(lazyval) ; tree[2*o+1].update(lazyval);
        tree[o].lazy = 0;
    }
}

inline void push_up(int o)
{
	tree[o].sum = tree[2*o].sum | tree[2*o+1].sum;
}

inline void build_tree(int L , int R , int o)
{
	tree[o].L = L , tree[o].R = R,tree[o].sum.reset(), tree[o].lazy = 0;
	if (L == R)
    {
        tree[o].sum[c[L]]=1;
        return;
    }
	if (R > L)
	{
		int mid = (L+R) >> 1;
		build_tree(L,mid,o*2);
		build_tree(mid+1,R,o*2+1);
		push_up(o);
	}
}

inline void update(int QL,int QR,int v,int o)
{
	int L = tree[o].L , R = tree[o].R;
	if (QL <= L && R <= QR) tree[o].update(v);
	else
	{
		push_down(o);
		int mid = (L+R)>>1;
		if (QL <= mid) update(QL,QR,v,o*2);
		if (QR >  mid) update(QL,QR,v,o*2+1);
		push_up(o);
	}
}

inline SgTreeDataType query(int QL,int QR,int o)
{
	int L = tree[o].L , R = tree[o].R;
	if (QL <= L && R <= QR) return tree[o].sum;
	else
	{
		push_down(o);
		int mid = (L+R)>>1;
		SgTreeDataType res;
		res.reset();
		if (QL <= mid) res |= query(QL,QR,2*o);
		if (QR > mid) res |= query(QL,QR,2*o+1);
		push_up(o);
		return res;
	}
}

void dfs(int x,int fa)
{
    st[x]=++tot;
    c[tot]=a[x];
    for(int i=0;i<E[x].size();i++)
    {
        if(E[x][i]==fa)continue;
        dfs(E[x][i],x);
    }
    ed[x]=tot;
}
void get_pri()
{
    for(int i=2;i<m;i++)
    {
        if(v[i])continue;
        pri[i]=1;
        for(int j=i;j<m;j+=i)
            v[j]=1;
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    get_pri();
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]),a[i]%=m;
    for(int i=1;i<n;i++)
    {
        int x,y;scanf("%d%d",&x,&y);
        E[x].push_back(y);
        E[y].push_back(x);
    }
    dfs(1,-1);
    build_tree(1,n,1);
    int q;scanf("%d",&q);
    while(q--)
    {
        int op;scanf("%d",&op);
        if(op==1)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            update(st[x],ed[x],y%m,1);
        }
        else
        {
            int x;scanf("%d",&x);
            bitset<MAXN> ans=query(st[x],ed[x],1);
            printf("%d\n",(ans&pri).count());
        }
    }
    return 0;
}