ZOJ 3911 Prime Query

Prime Query Time Limit: 1 Second       Memory Limit: 196608 KB

You are given a simple task. Given a sequence A[i] with N numbers. You have to perform Q operations on the given sequence.

Here are the operations:

• A v l, add the value v to element with index l.(1<=V<=1000)
• R a l r, replace all the elements of sequence with index i(l<=i<= r) with a(1<=a<=10^6) .
• Q l r, print the number of elements with index i(l<=i<=r) and A[i] is a prime number

Note that no number in sequence ever will exceed 10^7.

Input

The first line is a signer integer T which is the number of test cases.

For each test case, The first line contains two numbers N and Q (1 <= N, Q <= 100000) - the number of elements in sequence and the number of queries.

The second line contains N numbers - the elements of the sequence.

In next Q lines, each line contains an operation to be performed on the sequence.

Output

For each test case and each query,print the answer in one line.

Sample Input

1
5 10
1 2 3 4 5
A 3 1      
Q 1 3
R 5 2 4
A 1 1
Q 1 1
Q 1 2
Q 1 4
A 3 5
Q 5 5
Q 1 5

Sample Output

2
1
2
4
0
4

预处理一下一千万的素数，然后线段树搞定

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=10000005;
int T,n,m,p[maxn],u[maxn/10],tot=0;
int l,r,x;
char ch[2];

void pre()
{
	p[0]=p[1]=1;
	for (int i=2;i<maxn;i++)
	{
		if (p[i]==0) u[tot++]=i;
		for (int j=0;j<tot&&i*u[j]<maxn;j++)
		{
			p[i*u[j]]=1;
			if (i%u[j]==0) break;
		}
	}
}

struct ST
{
	const static int maxn=500005;
	int f[maxn],u[maxn];
	void update(int x)
	{
		u[x]=u[x+x]+u[x+x+1];
		f[x]=0;
	}
	void build(int x,int l,int r)
	{
		if (l==r) 
		{
			scanf("%d",&f[x]);
			u[x]=p[f[x]]^1;
		}
		else 
		{
			int mid=l+r>>1;
			build(x+x,l,mid);
			build(x+x+1,mid+1,r);
			update(x);
		}
	}
	void down(int x,int l,int r)
	{
		int mid=l+r>>1;
		f[x+x]=f[x+x+1]=f[x];	f[x]=0;
		u[x+x]=(p[f[x+x]]^1)*(mid-l+1);
		u[x+x+1]=(p[f[x+x+1]]^1)*(r-mid);
	}
	void add(int x,int l,int r,int a,int b)
	{
		if (l==r) {f[x]+=b; u[x]=p[f[x]]^1;}
		else
		{
			if (f[x]) down(x,l,r);
			int mid=l+r>>1;
			if (a<=mid) add(x+x,l,mid,a,b);
			else add(x+x+1,mid+1,r,a,b);
			update(x);
		}
	}
	void change(int x,int l,int r,int ll,int rr,int b)
	{
		if (ll<=l&&r<=rr) {f[x]=b; u[x]=(p[b]^1)*(r-l+1);}
		else 
		{
			int mid=l+r>>1;
			if (f[x]) down(x,l,r);
			if (ll<=mid) change(x+x,l,mid,ll,rr,b);
			if (rr>mid) change(x+x+1,mid+1,r,ll,rr,b);
			update(x);
		}
	}
	int query(int x,int l,int r,int ll,int rr)
	{
		int ans=0;
		if (ll<=l&&r<=rr) ans=u[x];
		else 
		{
			int mid=l+r>>1;
			if (f[x]) down(x,l,r);
			if (ll<=mid) ans+=query(x+x,l,mid,ll,rr);
			if (rr>mid) ans+=query(x+x+1,mid+1,r,ll,rr);
		}
		return ans;
	}
}st;

int main()
{
	pre();
	scanf("%d",&T);
	while (T--)
	{
		scanf("%d%d",&n,&m);
		st.build(1,1,n);
		while (m--)
		{
			scanf("%s",ch);
			switch (ch[0])
			{
				case 'A':scanf("%d%d",&x,&l);
				st.add(1,1,n,l,x); break;
				case 'Q':scanf("%d%d",&l,&r);
					printf("%d\n",st.query(1,1,n,l,r));break;
				case 'R':scanf("%d%d%d",&x,&l,&r);
					st.change(1,1,n,l,r,x); break;
			}
		}
	}
	return 0;
}