[素数个数模板] HDU 5901 Count primes

题目链接：传送门

题目大意：给你一个 n(1 <= n <= 1e11)，问1~n中素数个数

题目思路：（Meisell-Lehmer算法）

 1 #include<cstdio>
 2 #include<cmath>
 3 using namespace std;
 4 #define LL long long
 5 const int N = 5e6 + 2;
 6 bool np[N];
 7 int prime[N], pi[N];
 8 int getprime()
 9 {
10     int cnt = 0;
11     np[0] = np[1] = true;
12     pi[0] = pi[1] = 0;
13     for(int i = 2; i < N; ++i)
14     {
15         if(!np[i]) prime[++cnt] = i;
16         pi[i] = cnt;
17         for(int j = 1; j <= cnt && i * prime[j] < N; ++j)
18         {
19             np[i * prime[j]] = true;
20             if(i % prime[j] == 0)   break;
21         }
22     }
23     return cnt;
24 }
25 const int M = 7;
26 const int PM = 2 * 3 * 5 * 7 * 11 * 13 * 17;
27 int phi[PM + 1][M + 1], sz[M + 1];
28 void init()
29 {
30     getprime();
31     sz[0] = 1;
32     for(int i = 0; i <= PM; ++i)  phi[i][0] = i;
33     for(int i = 1; i <= M; ++i)
34     {
35         sz[i] = prime[i] * sz[i - 1];
36         for(int j = 1; j <= PM; ++j) phi[j][i] = phi[j][i - 1] - phi[j / prime[i]][i - 1];
37     }
38 }
39 int sqrt2(LL x)
40 {
41     LL r = (LL)sqrt(x - 0.1);
42     while(r * r <= x)   ++r;
43     return int(r - 1);
44 }
45 int sqrt3(LL x)
46 {
47     LL r = (LL)cbrt(x - 0.1);
48     while(r * r * r <= x)   ++r;
49     return int(r - 1);
50 }
51 LL getphi(LL x, int s)
52 {
53     if(s == 0)  return x;
54     if(s <= M)  return phi[x % sz[s]][s] + (x / sz[s]) * phi[sz[s]][s];
55     if(x <= prime[s]*prime[s])   return pi[x] - s + 1;
56     if(x <= prime[s]*prime[s]*prime[s] && x < N)
57     {
58         int s2x = pi[sqrt2(x)];
59         LL ans = pi[x] - (s2x + s - 2) * (s2x - s + 1) / 2;
60         for(int i = s + 1; i <= s2x; ++i) ans += pi[x / prime[i]];
61         return ans;
62     }
63     return getphi(x, s - 1) - getphi(x / prime[s], s - 1);
64 }
65 LL getpi(LL x)
66 {
67     if(x < N)   return pi[x];
68     LL ans = getphi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - 1;
69     for(int i = pi[sqrt3(x)] + 1, ed = pi[sqrt2(x)]; i <= ed; ++i) ans -= getpi(x / prime[i]) - i + 1;
70     return ans;
71 }
72 LL lehmer_pi(LL x)
73 {
74     if(x < N)   return pi[x];
75     int a = (int)lehmer_pi(sqrt2(sqrt2(x)));
76     int b = (int)lehmer_pi(sqrt2(x));
77     int c = (int)lehmer_pi(sqrt3(x));
78     LL sum = getphi(x, a) +(LL)(b + a - 2) * (b - a + 1) / 2;
79     for (int i = a + 1; i <= b; i++)
80     {
81         LL w = x / prime[i];
82         sum -= lehmer_pi(w);
83         if (i > c) continue;
84         LL lim = lehmer_pi(sqrt2(w));
85         for (int j = i; j <= lim; j++) sum -= lehmer_pi(w / prime[j]) - (j - 1);
86     }
87     return sum;
88 }
89 int main()
90 {
91     init();
92     LL n;
93     while(~scanf("%lld",&n))
94     {
95         printf("%lld\n",lehmer_pi(n));
96     }
97     return 0;
98 }