算法笔记--区间素数筛

筛[l,r]之间的合数

l<=r<=1e12

r-l<=1e6

筛小于等于r的合数所需的质因子大小最多不会超过根号r（<=1e6）

模板:

const int N=1e6+5;
bool not_prime_small[N]={false};
bool not_prime_big[N]={false};
void segment_seive(ll a,ll b)
{
    mem(not_prime_big,false);
    if(a==1)not_prime_big[0]=true;
    for(ll i=2;i*i<=b;i++)
    {
        if(!not_prime_small[i])
        {
            for(ll j=i+i;j*j<=b;j+=i)
            {
                not_prime_small[j]=true;
            }
            for(ll j=max(2ll,(a+i-1)/i)*i;j<=b;j+=i)
            not_prime_big[j-a]=true;
        }
    }    
}

例题1：POJ 2689 Prime Distance

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath> 
using namespace std;
#define pb push_back
#define ll long  long
#define mem(a,b) memset(a,b,sizeof(a)) 

const int N=1e6+5;
bool not_prime_small[N]={false};
bool not_prime_big[N]={false};
vector<ll>s;
void segment_seive(ll a,ll b)
{
    s.clear();
    mem(not_prime_big,false);
    if(a==1)not_prime_big[0]=true;
    for(ll i=2;i*i<=b;i++)
    {
        if(!not_prime_small[i])
        {
            for(ll j=i+i;j*j<=b;j+=i)
            {
                not_prime_small[j]=true;
            }
            for(ll j=max(2ll,(a+i-1)/i)*i;j<=b;j+=i)
            not_prime_big[j-a]=true;
        }
    }    
    for(ll i=a;i<=b;i++)
    if(!not_prime_big[i-a])s.pb(i);
    if(s.size()<=1)cout<<"There are no adjacent primes."<<endl;
    else
    {
        pair<ll,ll>ans,_ans; 
        int mx=0,mn=0x3f3f3f3f;
        for(int i=0;i<s.size()-1;i++)
        {
            //cout<<s[i]<<endl;
            if(s[i+1]-s[i]>mx)
            {
                mx=s[i+1]-s[i];
                ans.first=s[i];
                ans.second=s[i+1];    
            } 
            if(s[i+1]-s[i]<mn)
            {
                mn=s[i+1]-s[i];
                _ans.first=s[i];
                _ans.second=s[i+1];
            }
        }
        cout<<_ans.first<<","<<_ans.second<<" are closest, "<<ans.first<<","<<ans.second<<" are most distant."<<endl;
    }
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll l,u;
    while(cin>>l>>u)
    {
        segment_seive(l,u); 
    }
    return 0;
}

 例题2：HDU 6069 Counting Divisors

要运用到约数个数定理

代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))

const int N=1e6+5;
const int MOD=998244353;
bool not_prime[N]={false};
ll ans[N];
ll val[N];
vector<int>prime;
void prime_seive()
{
    for(int i=2;i<N;i++)
    {
        if(!not_prime[i])
        {
            prime.pb(i);
        }
        for(int j=0;i*prime[j]<N;j++)
        {
            not_prime[i*prime[j]]=true;
            if(i%prime[j]==0)break;
        }
    }
}
void solve(ll l,ll r,ll k)
{
    for(ll i=l;i<=r;i++)
    {
        ans[i-l]=1;
        val[i-l]=i;
    }
    for(int i=0;i<prime.size();i++)
    {
        for(ll j=max(2ll,(l+prime[i]-1)/prime[i])*prime[i];j<=r;j+=prime[i])
        {
            int tot=0;
            if(val[j-l]%prime[i])continue;
            while(val[j-l]%prime[i]==0)tot++,val[j-l]/=prime[i];
            ans[j-l]=(1+tot*k)%MOD*ans[j-l]%MOD; 
        }
    }
    
    for(int i=0;i<=r-l;i++)
    {
        if(val[i]>1)ans[i]=(1+k)%MOD*ans[i]%MOD;
    }
    ll _ans=0;
    for(int i=0;i<=r-l;i++)
    _ans=(_ans+ans[i])%MOD;
    cout<<_ans<<endl;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll l,r,k;
    int T;
    prime_seive(); 
    cin>>T;
    while(T--)
    {
        cin>>l>>r>>k;
        solve(l,r,k);
    }
    return 0;
}