• Meisell-Lehmer算法(统计较大数据里的素数)


    http://acm.hdu.edu.cn/showproblem.php?pid=5901

    1e11的数据量,这道题用这个算法花了202ms.

     1 #include<bits/stdc++.h>  
     2   
     3 using namespace std;  
     4   
     5 typedef long long LL;  
     6 const int N = 5e6 + 2;  
     7 bool np[N];  
     8 int prime[N], pi[N];  
     9   
    10 int getprime() {  
    11     int cnt = 0;  
    12     np[0] = np[1] = true;  
    13     pi[0] = pi[1] = 0;  
    14     for(int i = 2; i < N; ++i) {  
    15         if(!np[i]) prime[++cnt] = i;  
    16         pi[i] = cnt;  
    17         for(int j = 1; j <= cnt && i * prime[j] < N; ++j) {  
    18             np[i * prime[j]] = true;  
    19             if(i % prime[j] == 0)   break;  
    20         }  
    21     }  
    22     return cnt;  
    23 }  
    24 const int M = 7;  
    25 const int PM = 2 * 3 * 5 * 7 * 11 * 13 * 17;  
    26 int phi[PM + 1][M + 1], sz[M + 1];  
    27 void init() {  
    28     getprime();  
    29     sz[0] = 1;  
    30     for(int i = 0; i <= PM; ++i)  phi[i][0] = i;  
    31     for(int i = 1; i <= M; ++i) {  
    32         sz[i] = prime[i] * sz[i - 1];  
    33         for(int j = 1; j <= PM; ++j) {  
    34             phi[j][i] = phi[j][i - 1] - phi[j / prime[i]][i - 1];  
    35         }  
    36     }  
    37 }  
    38 int sqrt2(LL x) {  
    39     LL r = (LL)sqrt(x - 0.1);  
    40     while(r * r <= x)   ++r;  
    41     return int(r - 1);  
    42 }  
    43 int sqrt3(LL x) {  
    44     LL r = (LL)cbrt(x - 0.1);  
    45     while(r * r * r <= x)   ++r;  
    46     return int(r - 1);  
    47 }  
    48 LL getphi(LL x, int s) {  
    49     if(s == 0)  return x;  
    50     if(s <= M)  return phi[x % sz[s]][s] + (x / sz[s]) * phi[sz[s]][s];  
    51     if(x <= prime[s]*prime[s])   return pi[x] - s + 1;  
    52     if(x <= prime[s]*prime[s]*prime[s] && x < N) {  
    53         int s2x = pi[sqrt2(x)];  
    54         LL ans = pi[x] - (s2x + s - 2) * (s2x - s + 1) / 2;  
    55         for(int i = s + 1; i <= s2x; ++i) {  
    56             ans += pi[x / prime[i]];  
    57         }  
    58         return ans;  
    59     }  
    60     return getphi(x, s - 1) - getphi(x / prime[s], s - 1);  
    61 }  
    62 LL getpi(LL x) {  
    63     if(x < N)   return pi[x];  
    64     LL ans = getphi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - 1;  
    65     for(int i = pi[sqrt3(x)] + 1, ed = pi[sqrt2(x)]; i <= ed; ++i) {  
    66         ans -= getpi(x / prime[i]) - i + 1;  
    67     }  
    68     return ans;  
    69 }  
    70 LL lehmer_pi(LL x) {  
    71     if(x < N)   return pi[x];  
    72     int a = (int)lehmer_pi(sqrt2(sqrt2(x)));  
    73     int b = (int)lehmer_pi(sqrt2(x));  
    74     int c = (int)lehmer_pi(sqrt3(x));  
    75     LL sum = getphi(x, a) + LL(b + a - 2) * (b - a + 1) / 2;  
    76     for (int i = a + 1; i <= b; i++) {  
    77         LL w = x / prime[i];  
    78         sum -= lehmer_pi(w);  
    79         if (i > c) continue;  
    80         LL lim = lehmer_pi(sqrt2(w));  
    81         for (int j = i; j <= lim; j++) {  
    82             sum -= lehmer_pi(w / prime[j]) - (j - 1);  
    83         }  
    84     }  
    85     return sum;  
    86 }  
    87   
    88 int main() {  
    89     init();  
    90     LL n;  
    91     while(cin >> n) {  
    92         cout << lehmer_pi(n) << endl;  
    93     }  
    94     return 0;  
    95 }  
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  • 原文地址:https://www.cnblogs.com/fightfordream/p/5885876.html
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