ZOJ 3911Prime Query [素数处理 + 线段树]

Time Limit: 5 Seconds 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 <cstdlib>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#define INF 0x3f3f3f3f
#define EPS 0.00000001
#define lowbit(x) (x&(-x))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;

const int maxn = 1e5+7;
const int N = 1e7+7;
int t[maxn << 2],in[maxn << 2],lazy[maxn << 2];
int prime[N];

void getprime()
{
    for(int i=2;i<sqrt(1.0*N);i++)
        if(!prime[i])
            for(int j=i+i;j<N;j+=i)
                prime[j] = 1;
    prime[1] = 1;
    prime[0] = 1;
}

void pushdown(int i,int b,int e)
{
    if(lazy[i])
    {
        int mid = (b + e) / 2;
        in[i << 1] = !prime[lazy[i]] * (mid - b + 1);
        in[i << 1 | 1] = !prime[lazy[i]] * (e - mid);

        t[i << 1] = lazy[i];
        t[i << 1 | 1] = lazy[i];

        lazy[i << 1] = lazy[i];
        lazy[i << 1 | 1] = lazy[i];

        lazy[i] = 0;
    }
}

void add(int i,int b,int e,int pos,int val)
{
    if(b == e)
    {
        in[i] -= !prime[t[i]];
        t[i] += val;
        in[i] += !prime[t[i]];
        return ;
    }
    pushdown(i, b, e);

    int mid = (b + e) / 2;
    if(pos <= mid) add(i << 1, b, mid, pos, val);
    else add(i << 1 | 1, mid + 1, e, pos, val);

    in[i] = in[i << 1] + in[i << 1 | 1];
}

void replace(int i,int b,int e,int l,int r,int val)
{
    if(b >= l && e <= r)
    {
        in[i] = (!prime[val]) * (e - b + 1);
        t[i] = val;
        lazy[i] = val;
        return ;
    }
    pushdown(i, b, e);

    int mid = (b + e) / 2;
    if(r <= mid) replace(i << 1, b, mid, l, r, val);
    else if(l > mid) replace(i<< 1 | 1, mid + 1, e, l, r, val);
    else
    {
        replace(i << 1, b, mid, l, r, val);
        replace(i << 1 | 1, mid + 1, e, l, r, val);
    }

    in[i] = in[i << 1] + in[i << 1 | 1];
}

int query(int i,int b,int e,int l,int r)
{
    if(b >= l && e <= r)
        return in[i];
    pushdown(i, b, e);

    int mid = (b + e) / 2;
    if(r <= mid) return query(i << 1, b, mid, l, r);
    else if(l > mid) return query(i << 1 | 1, mid + 1, e, l, r);
    else return query(i << 1, b, mid, l, r) + query(i << 1 | 1, mid + 1, e, l, r);
}

int main()
{
    getprime();

    int T;
    scanf("%d",&T);
    while(T--)
    {
        char s[5];
        int n,m,x,a,b,c;

        scanf("%d%d",&n,&m);

        memset(t,0,sizeof(t));
        memset(in,0,sizeof(in));
        memset(lazy,0,sizeof(lazy));

        for(int i=1;i<=n;i++)
        {
            scanf("%d",&x);
            add(1,1,n,i,x);
        }

        for(int i=0;i<m;i++)
        {
            scanf("%s%d%d",s,&a,&b);
            if(s[0] == 'A') add(1,1,n,b,a);

            else if(s[0] == 'Q') printf("%d
", query(1,1,n,a,b));

            else
            {
                scanf("%d",&c);
                replace(1,1,n,b,c,a);
            }

        }
    }
}