hdu 5901（模板，10^11以内的素数个数）

本文链接：https://blog.youkuaiyun.com/YHYYXT/article/details/52638180

本文介绍了一种高效的素数计数方法，通过使用多种优化技巧实现对区间内素数数量的快速计算。提供了两种模板代码，分别具有O(n^(3/4))和O(n^(2/3))的时间复杂度，适用于不同规模的数据输入。

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Count primes

Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1079 Accepted Submission(s): 606

Problem Description

Easy question! Calculate how many primes between [1...n]!

Input

Each line contain one integer n(1 <= n <= 1e11).Process to end of file.

Output

For each case, output the number of primes in interval [1...n]

Sample Input

Sample Output

Source

2016 ACM/ICPC Asia Regional Shenyang Online

我是参考这个博客做的： http://m.blog.youkuaiyun.com/article/details?id=51314679

#include <bits/stdc++.h>
using namespace std;
#define ll __int64
const int N=320005; 
ll phi[10005][105], p2[N], ans[N];
int len, vis[N];

void init(){
	len = 0;
	for(int i=2; i<N; i++){
		if(!vis[i]){
			for(int j=i+i; j<N; j+=i) vis[j]=1;
			p2[len++] = i;
			ans[i] = ans[i-1]+1;
			continue;
		}
		ans[i] = ans[i-1];
	}	
	for(int i=0; i<=10000; i++){
		phi[i][0] = (ll)i;
		for(int j=1; j<=100; j++){
			phi[i][j] = phi[i][j-1] - phi[i/p2[j-1]][j-1];
		}
	}
}

ll solve_phi(ll m, ll n){
	if(!n) return m;
	if(p2[n - 1] >= m) return 1;
	if(m<=10000 && n<=100) return phi[m][n];
	return solve_phi(m, n-1) - solve_phi(m/p2[n-1], n-1);
}

ll solve_p2(ll m){
	if(m < (ll)N) return ans[m];
	
	ll y = (int)cbrt(m*1.0);
	ll n = ans[y];
	ll sum = solve_phi(m, n) + n -1;
	
	for(ll i=n; p2[i]*p2[i]<=m; i++) //参考博客中的范围有误 
		sum = sum - solve_p2(m/p2[i])+solve_p2(p2[i])-1;
	return sum;
}

int main(){
	
	init();
	
	ll n;
	while(scanf("%I64d", &n)!=EOF){
		printf("%I64d\n", solve_p2(n));
	}
	
	return 0;
}

这个博客中还有两个模板，其中第二个模板跑的比较快：http://blog.youkuaiyun.com/chaiwenjun000/article/details/52589457

以下两个模板代码引自上面连接中的模板

模板代码一：

复杂度大概O（n^(3/4)）

#include <bits/stdc++.h>  
#define ll long long  
using namespace std;  
ll f[340000],g[340000],n;  
void init(){  
    ll i,j,m;  
    for(m=1;m*m<=n;++m)f[m]=n/m-1;  
    for(i=1;i<=m;++i)g[i]=i-1;  
    for(i=2;i<=m;++i){  
        if(g[i]==g[i-1])continue;  
        for(j=1;j<=min(m-1,n/i/i);++j){  
            if(i*j<m)f[j]-=f[i*j]-g[i-1];  
            else f[j]-=g[n/i/j]-g[i-1];  
        }  
        for(j=m;j>=i*i;--j)g[j]-=g[j/i]-g[i-1];  
    }  
}  
int main(){  
    while(scanf("%I64d",&n)!=EOF){  
        init();  
        cout<<f[1]<<endl;  
    }  
    return 0;  
}

模板代码二：

复杂度O（n^(2/3)）

#include<cstdio>  
#include<cmath>  
using namespace std;  
#define LL long long  
const int N = 5e6 + 2;  
bool np[N];  
int prime[N], pi[N];  
int getprime()  
{  
    int cnt = 0;  
    np[0] = np[1] = true;  
    pi[0] = pi[1] = 0;  
    for(int i = 2; i < N; ++i)  
    {  
        if(!np[i]) prime[++cnt] = i;  
        pi[i] = cnt;  
        for(int j = 1; j <= cnt && i * prime[j] < N; ++j)  
        {  
            np[i * prime[j]] = true;  
            if(i % prime[j] == 0)   break;  
        }  
    }  
    return cnt;  
}  
const int M = 7;  
const int PM = 2 * 3 * 5 * 7 * 11 * 13 * 17;  
int phi[PM + 1][M + 1], sz[M + 1];  
void init()  
{  
    getprime();  
    sz[0] = 1;  
    for(int i = 0; i <= PM; ++i)  phi[i][0] = i;  
    for(int i = 1; i <= M; ++i)  
    {  
        sz[i] = prime[i] * sz[i - 1];  
        for(int j = 1; j <= PM; ++j) phi[j][i] = phi[j][i - 1] - phi[j / prime[i]][i - 1];  
    }  
}  
int sqrt2(LL x)  
{  
    LL r = (LL)sqrt(x - 0.1);  
    while(r * r <= x)   ++r;  
    return int(r - 1);  
}  
int sqrt3(LL x)  
{  
    LL r = (LL)cbrt(x - 0.1);  
    while(r * r * r <= x)   ++r;  
    return int(r - 1);  
}  
LL getphi(LL x, int s)  
{  
    if(s == 0)  return x;  
    if(s <= M)  return phi[x % sz[s]][s] + (x / sz[s]) * phi[sz[s]][s];  
    if(x <= prime[s]*prime[s])   return pi[x] - s + 1;  
    if(x <= prime[s]*prime[s]*prime[s] && x < N)  
    {  
        int s2x = pi[sqrt2(x)];  
        LL ans = pi[x] - (s2x + s - 2) * (s2x - s + 1) / 2;  
        for(int i = s + 1; i <= s2x; ++i) ans += pi[x / prime[i]];  
        return ans;  
    }  
    return getphi(x, s - 1) - getphi(x / prime[s], s - 1);  
}  
LL getpi(LL x)  
{  
    if(x < N)   return pi[x];  
    LL ans = getphi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - 1;  
    for(int i = pi[sqrt3(x)] + 1, ed = pi[sqrt2(x)]; i <= ed; ++i) ans -= getpi(x / prime[i]) - i + 1;  
    return ans;  
}  
LL lehmer_pi(LL x)  
{  
    if(x < N)   return pi[x];  
    int a = (int)lehmer_pi(sqrt2(sqrt2(x)));  
    int b = (int)lehmer_pi(sqrt2(x));  
    int c = (int)lehmer_pi(sqrt3(x));  
    LL sum = getphi(x, a) +(LL)(b + a - 2) * (b - a + 1) / 2;  
    for (int i = a + 1; i <= b; i++)  
    {  
        LL w = x / prime[i];  
        sum -= lehmer_pi(w);  
        if (i > c) continue;  
        LL lim = lehmer_pi(sqrt2(w));  
        for (int j = i; j <= lim; j++) sum -= lehmer_pi(w / prime[j]) - (j - 1);  
    }  
    return sum;  
}  
int main()  
{  
    init();  
    LL n;  
    while(~scanf("%lld",&n))  
    {  
        printf("%lld\n",lehmer_pi(n));  
    }  
    return 0;  
}