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  • [素数个数模板] HDU 5901 Count primes

    #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
    ",lehmer_pi(n));  
        }  
        return 0;  
    }
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  • 原文地址:https://www.cnblogs.com/Kurokey/p/5888386.html
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