loj#6235. The number of prime numbers in the interval (Zhou Ge sieve)

topic here

Wrote it a while ago...and re-written it this time as a review

And then found that it was 2000+ms faster than the last time? ?Although still very slow. I don't seem to have added any optimizations.

(Xu is that the performance of the loj evaluation machine has become better...?

practice

Chau Court Sieve Template. There are detailed comments in the code.

code

=> Note that initialization // is not for this question, this question is fine if it is not initialized because there is only one set of data, but if there are multiple sets or multiple cals are used, you should pay attention

#include<bits/stdc++.h>
#define rep(i,x,y) for (int i=(x); i<=(y); i++)
#define per(i,x,y) for (int i=(x); i>=(y); i--)
#define ll long long
#define ld long double
#define inf 1000000000
using namespace std;
#define N 350005
int p[N/10],tot,res[N]; bitset<N> vis; ll n;
void pre(int n){
    rep (i,2,n){
        res[i]=res[i-1]; if (!vis[i]) p[++tot]=i,res[i]++;
        for (int j=1; j<=tot && (ll)i*p[j]<=n; j++){
            vis[i*p[j]]=1;
            if (i%p[j]==0) break;
        }
    }
}
#define M 350005
int sn,pos,cnt,last[M<<1]; ll g[M<<1],value[M<<1];
ll cal(ll n){//洲阁筛
    cnt=0; sn=(ll)sqrt((ld)n);//考虑<=n的数由<=sqrtn的质数筛出的情况
    pos=upper_bound(p+1,p+1+tot,sn)-p-1;//pos第一个小于等于sn的质数位置
    for (ll i=n; i>=1; i=n/(n/i+1)) value[++cnt]=n/i;//记录所有[n/i]的值，只有这样的数才会出现在转移中 //离散
    //g[i][j]表示1~j中与前i个质数互质的数的个数 //筛不掉的
    //g[i][j]=g[i-1][j]-g[i-1][j/p[i]]
    //当p[i+1]>j时，g[i][j]=1 //只有1
    //p[i]>j/p[i]时，g[i][j]=g[i-1][j]-1，可以O(1)计算
    ll k;
    rep (i,1,cnt) g[i]=value[i],last[i]=0;//注意初始化last[i]=0
    rep (i,1,pos) per (j,cnt,1){
        k=value[j]/p[i]; if (k<p[i]) break;//忽略那些-1的转移
        k=k<sn?k:cnt-n/k+1;//找到在value中的对应下标
        g[j]-=g[k]-(i-last[k]-1);//将g[k]的-1的转移补回去
        last[j]=i;
    }
    return res[sn]+g[cnt]-1;//-1是减去1的贡献
}
//#define local
int main(){
#ifdef local
    freopen("test.in","r",stdin); freopen("test.out","w",stdout);
#endif
    pre(350000); cin>>n; cout<<cal(n)<<endl;
    return 0;
}