Prime Query (ZOJ 3911 线段树)

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

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <string>

#include <algorithm>

#include <set>

#include <map>

#include <vector>

#include <cmath>

#define VI vector<int>

#define VP vector<Point>

#define pr pair<int,int>

#define LL long long

#define fread() freopen("../in.in","r",stdin)

#define fwrite() freopen("out.out","w",stdout)

using namespace std;

const int Max = 1e7;

const int Maxn = 100000;

typedef struct Tree
{
int num;//记录所在区间的素数的个数

int lazy;//标记所在的区间有没有被整体更新
} Tree;

int vis[Max+1000];

Tree Tr[Maxn*5];

int a[Maxn+100];

void init()//素数表
{
memset(vis,0,sizeof(vis));

int  m= (int)sqrt(Max);

vis[0]=1;

vis[1]=1;

for(LL i=2; i<=m; i++)
{
if(!vis[i])
{
for(LL j=i*i; j<=Max; j+=i)
{
vis[j]=1;
}
}
}
}

void Pushup(int st,int L,int R)//线段树区间向上更新
{
if(Tr[st<<1|1].lazy&&Tr[st<<1].lazy&&Tr[st<<1|1].lazy==Tr[st<<1].lazy)//
{
Tr[st].lazy=Tr[st<<1].lazy;
}
else
{
Tr[st].lazy=0;
}

Tr[st].num=Tr[st<<1].num+Tr[st<<1|1].num;
}

void Pushdown(int st,int L,int R)//线段树区间向下更新
{
if(Tr[st].lazy&&L!=R)
{
int mid =(L+R)>>1;

Tr[st<<1].lazy=Tr[st<<1|1].lazy=Tr[st].lazy;

if(Tr[st].num)
{
Tr[st<<1|1].num=R-mid;

Tr[st<<1].num=mid+1-L;
}
else
{
Tr[st<<1|1].num=Tr[st<<1].num=0;
}

Tr[st].lazy=0;
}
}

void Build(int L,int R,int st)//初始化线段树
{
Tr[st].lazy=0;

Tr[st].num=0;

if(L==R)
{
Tr[st].lazy=a[L];

Tr[st].num=(!vis[a[L]]);

return ;
}
int mid=(L+R)>>1;

Build(L,mid,st<<1);

Build(mid+1,R,st<<1|1);

Pushup(st,L,R);
}

void Add(int L,int R,int st,int s,int d)//单点更新
{
Pushdown(st,L,R);

if(L==s&&R==s)
{
Tr[st].lazy+=d;

Tr[st].num=(!vis[Tr[st].lazy]);

return ;
}
int mid =(L+R)>>1;

if(s<=mid)
{
Add(L,mid,st<<1,s,d);
}
else
{
Add(mid+1,R,st<<1|1,s,d);
}

Pushup(st,L,R);
}

void Update(int L,int R,int st,int l,int r,int d)//区间更新
{
if(L>r||R<l)
{
return ;
}

if(L>=l&&R<=r)
{
Tr[st].lazy=d;

Tr[st].num=(!vis[d])*(R-L+1);

return ;
}

Pushdown(st,L,R);

int mid = (L+R)>>1;

if(l<=mid)
{
Update(L,mid,st<<1,l,r,d);
}

if(r>mid)
{
Update(mid+1,R,st<<1|1,l,r,d);
}

Pushup(st,L,R);
}

int Query(int L,int R,int st,int l,int r)//区间查询
{
if(L>r||R<l)
{
return 0;
}

Pushdown(st,L,R);

if(L>=l&&R<=r)
{
return Tr[st].num;

}
int mid=(L+R)>>1;

int sum=0;

if(l<=mid)
{
sum+=Query(L,mid,st<<1,l,r);
}

if(r>mid)
{
sum+=Query(mid+1,R,st<<1|1,l,r);
}

Pushup(st,L,R);

return sum;
}
int main()
{
int T,n,q;

int l,r,s,d;

char op[3];

init();

scanf("%d",&T);

while(T--)
{
scanf("%d %d",&n,&q);

for(int i=1; i<=n; i++)
{
scanf("%d",&a[i]);
}

Build(1,n,1);

while(q--)
{
scanf("%s",op);

if(op[0]=='A')
{
scanf("%d %d",&d,&s);

Add(1,n,1,s,d);
}
else if(op[0]=='Q')
{
scanf("%d %d",&l,&r);

printf("%d\n",Query(1,n,1,l,r));
}
else if(op[0]=='R')
{
scanf("%d %d %d",&d,&l,&r);

Update(1,n,1,l,r,d);
}
}
}
return 0;
}