143

Prime Query

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Time Limit: 1 Second      Memory Limit: 196608 KB

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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

题解：线段树的单点更新，区间修改，区间查询

写的很挫

///1085422276
#include<bits/stdc++.h>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define meminf(a) memset(a,127,sizeof(a));
#define TS printf("111111
");
#define FOR(i,a,b) for( int i=a;i<=b;i++)
#define FORJ(i,a,b) for(int i=a;i>=b;i--)
#define READ(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define inf 100000
inline ll read()
{
    ll x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
//****************************************
#define maxn 1000000+5

bool P[10000101];
ll a[1000000+5];
struct ss
{
    ll l,r,v,tag,sum;
}tr[maxn*5];

void pushdown(int k)
{
    if(!tr[k].tag)return;
   if(tr[k].l==tr[k].r){
     tr[k].v=tr[k].tag;
     if(!P[tr[k].v])
            tr[k].sum=1;
        else tr[k].sum=0;
   }
   else {
    if(!P[tr[k].tag])tr[k].sum=tr[k].r-tr[k].l+1;
    else {tr[k].sum=0;}
    tr[k<<1].tag=tr[k].tag;
    tr[k<<1|1].tag=tr[k].tag;
   }
   tr[k].tag=0;
}
void pushup(int k)
{
    pushdown(k<<1);
    pushdown(k<<1|1);
    tr[k].sum=tr[k<<1].sum+tr[k<<1|1].sum;
}
void build(int k,int s,int t)
{
    tr[k].l=s;
    tr[k].r=t;
    tr[k].tag=0;
    tr[k].sum=0;
    if(s==t)
        {tr[k].v=a[s];if(!P[a[s]])tr[k].sum=1;else tr[k].sum=0;return;}
    int mid=(s+t)>>1;
    build(k<<1,s,mid);
    build(k<<1|1,mid+1,t);
    pushup(k);
}
void add(int k,int x,int y)
{
    pushdown(k);
    if(x==tr[k].l&&tr[k].l==tr[k].r)
    {
        tr[k].v+=y;
        if(!P[tr[k].v])
            tr[k].sum=1;
        else tr[k].sum=0;
        return ;
    }
    int mid=(tr[k].l+tr[k].r)>>1;
    if(x<=mid)
        add(k<<1,x,y);
    else add(k<<1|1,x,y);
   pushup(k);
}
void update(int k,int x,int y,int c)
{
    pushdown(k);
    if(x==tr[k].l&&y==tr[k].r)
    {
       tr[k].tag=c;
       return ;
    }
    int mid=(tr[k].l+tr[k].r)>>1;
    if(y<=mid)update(k<<1,x,y,c);
    else if(x>mid)update(k<<1|1,x,y,c);
    else {
        update(k<<1,x,mid,c);
        update(k<<1|1,mid+1,y,c);
    }
    pushup(k);
}
ll ask(int k,int x,int y)
{
    pushdown(k);
    if(x==tr[k].l&&y==tr[k].r)
    {
       return tr[k].sum;
    }
    int mid=(tr[k].l+tr[k].r)>>1;
    if(y<=mid)return ask(k<<1,x,y);
    else if(x>mid)return ask(k<<1|1,x,y);
    else {
        return (ask(k<<1,x,mid)+ask(k<<1|1,mid+1,y));
    } pushup(k);
}
int main()
{
    P[1]=1;
    for(int i=2;i<=10000000;i++)
    {
        if(!P[i])
        for(int j=i+i;j<=10000000;j+=i)
            P[j]=1;
    }
    ll n,q,c;
    int T=read();
    while(T--)
    {
        scanf("%lld%lld",&n,&q);
        FOR(i,1,n)
        {
            scanf("%lld",&a[i]);
        }
        char ch[30];
        ll x,y;
        build(1,1,n);
        FOR(i,1,q)
        {
            scanf("%s%lld%lld",&ch,&x,&y);
            if(ch[0]=='A')
            {
                add(1,y,x);
            }
            else if(ch[0]=='R')
            {
                scanf("%lld",&c);
                update(1,y,c,x);
            }
            else if(ch[0]=='Q')
            {
                printf("%lld
",ask(1,x,y));
            }
        }
    }
    return 0;
}