CF 665F Four Divisors 1e11内素数个数(模板)

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题意:D(n)定义为[1..n]中有多少个数,其因子个数正好是4个.给出n,求出D(n),n<=1e11.

x的因子正好4个则除去1,x之外 n的因式分解形式中只能为p^3,p*q,(p,q为不同的素数).
可以直接枚举n以内p^3个数.

对于p*q形式枚举小的因子p(p<=sqrt(n) 则p<q的范围在<n/p之间求出这段区间内的素数个数即可.

n<=1e11 套用模板prime counting function求[1,n]内素数个数即可.

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#include<bits/stdc++.h>typedef long long ll;#define MAXN 100#define MAXM 100010#define MAXP 666666#define MAX 10000010#define chkbit(ar, i) (((ar[(i) >> 6]) & (1 << (((i) >> 1) & 31))))#define setbit(ar, i) (((ar[(i) >> 6]) |= (1 << (((i) >> 1) & 31))))#define isprime(x) (( (x) && ((x)&1) && (!chkbit(ar, (x)))) || ((x) == 2))using namespace std;namespace pcf{    long long dp[MAXN][MAXM];    unsigned int ar[(MAX >> 6) + 5] = {0};    int len = 0, primes[MAXP], counter[MAX];    void Sieve(){        setbit(ar, 0), setbit(ar, 1);        for (int i = 3; (i * i) < MAX; i++, i++){            if (!chkbit(ar, i)){                int k = i << 1;                for (int j = (i * i); j < MAX; j += k) setbit(ar, j);            }        }        for (int i = 1; i < MAX; i++){            counter[i] = counter[i - 1];            if (isprime(i)) primes[len++] = i, counter[i]++;        }    }    void init(){        Sieve();        for (int n = 0; n < MAXN; n++){            for (int m = 0; m < MAXM; m++){                if (!n) dp[n][m] = m;                else dp[n][m] = dp[n - 1][m] - dp[n - 1][m / primes[n - 1]];            }        }    }    long long phi(long long m, int n){        if (n == 0) return m;        if (primes[n - 1] >= m) return 1;        if (m < MAXM && n < MAXN) return dp[n][m];        return phi(m, n - 1) - phi(m / primes[n - 1], n - 1);    }    long long Lehmer(long long m){        if (m < MAX) return counter[m];        long long w, res = 0;        int i, a, s, c, x, y;        s = sqrt(0.9 + m), y = c = cbrt(0.9 + m);        a = counter[y], res = phi(m, a) + a - 1;        for (i = a; primes[i] <= s; i++) res = res - Lehmer(m / primes[i]) + Lehmer(primes[i]) - 1;        return res;    }}ll solve(ll n){ll res=0;for(int i=0;i<pcf::len;i++){ll x=pcf::primes[i],y=n/x;if(x*x>n)break;res+=(pcf::Lehmer(y)-pcf::Lehmer(x));}for(int i=0;i<pcf::len;i++){ll x=pcf::primes[i];if(x*x*x>n)break;res++;}return res;}int main(){ll n;cin>>n;pcf::init();cout<<solve(n)<<endl;return 0;}

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