HDU 5901 Count primes (1e11内的素数个数) -2016 ICPC沈阳赛区网络赛

Posted 2020-08-09 Ritchie丶

题目链接

题意：求[1,n]有多少个素数，1<=n<=10^11。时限为6000ms。

官方题解：一个模板题, 具体方法参考wiki或者Four Divisors。

题解：给出两种代码。

　　　第一种方法Meisell-Lehmer算法只需265ms。

　　　第二种方法不能运行但是能AC，只需35行。

第一种：

//Meisell-Lehmer
#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;
}

 

第二种：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll maxn=1e11;
const ll maxp=sqrt(maxn)+10;
ll f[maxp],g[maxp];
ll solve(ll n)
{
    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];
    }
    return f[1];
}
int main()
{
    ll n;
    while(scanf("%lld",&n)!=EOF)
    printf("%lld\n",solve(n));
    return 0;
}