HDU 5901 Count primes

论文题

1e11内找素数个数。

Meissel-Lehmer 黑科技板子

#pragma warning(disable:4996)

#include<iostream>
#include<algorithm>
#include<bitset>
#include<tuple>
#include<unordered_map>
#include<fstream>
#include<iomanip>
#include<string>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<list>
#include<queue>
#include<stack>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<cstdlib>
#define INF 0x3f3f3f3f
#define inf 0x7FFFFFFF
#define MOD 1000000007
#define moD 1000000003
#define pii pair<int,int>
#define eps 1e-7
#define equals(a,b) (fabs(a-b)<eps)
#define bug puts("bug")
#define re  register
#define fi first
#define se second
#define pb push_back
const int maxn = 2e5 + 5;
const double Inf = 10000.0;
const double PI = acos(-1.0);
typedef  long long ll;
typedef unsigned long long ull;
using namespace std;

const int N = 5e6 + 5;
const int M = 7;
const int PM = 2 * 3 * 5 * 7 * 11 * 13 * 17;
int np[N];
int prime[N], pi[N];
int phi[PM + 1][M + 1], sz[M + 1];
int getprime()
{
    int cnt = 0;
    np[0] = np[1] = 1;
    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]] = 1;
            if (i % prime[j] == 0) break;
        }
    }
    return cnt;
}
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);    //cbrt(x): x的立方根
    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()
{
    ll n; init();
    while (~scanf("%lld", &n))
    {
        printf("%lld\n", lehmer_pi(n));
    }
    return 0;
}

标签：Count,pii,HDU,int,1000000003,5901,include,define
来源： https://www.cnblogs.com/hznumqf/p/13346439.html