大素数区间快判模板

题意：求区间的质数的个数
按照网上来说是个模板题，按照论文积分来做的，复杂度O(n^(2/3))
看懂是不可能看懂的，就给记下来吧。。。
 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define LL long long
 4 const int N = 5e6 + 2;
 5 bool np[N];
 6 int prime[N], pi[N];
 7 int getprime()
 8 {
 9     int cnt = 0;
10     np[0] = np[1] = true;
11     pi[0] = pi[1] = 0;
12     for(int i = 2; i < N; ++i)
13     {
14         if(!np[i]) prime[++cnt] = i;
15         pi[i] = cnt;
16         for(int j = 1; j <= cnt && i * prime[j] < N; ++j)
17         {
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 {
29     getprime();
30     sz[0] = 1;
31     for(int i = 0; i <= PM; ++i)  phi[i][0] = i;
32     for(int i = 1; i <= M; ++i)
33     {
34         sz[i] = prime[i] * sz[i - 1];
35         for(int j = 1; j <= PM; ++j) phi[j][i] = phi[j][i - 1] - phi[j / prime[i]][i - 1];
36     }
37 }
38 int sqrt2(LL x)
39 {
40     LL r = (LL)sqrt(x - 0.1);
41     while(r * r <= x)   ++r;
42     return int(r - 1);
43 }
44 int sqrt3(LL x)
45 {
46     LL r = (LL)cbrt(x - 0.1);
47     while(r * r * r <= x)   ++r;
48     return int(r - 1);
49 }
50 LL getphi(LL x, int s)
51 {
52     if(s == 0)  return x;
53     if(s <= M)  return phi[x % sz[s]][s] + (x / sz[s]) * phi[sz[s]][s];
54     if(x <= prime[s]*prime[s])   return pi[x] - s + 1;
55     if(x <= prime[s]*prime[s]*prime[s] && x < N)
56     {
57         int s2x = pi[sqrt2(x)];
58         LL ans = pi[x] - (s2x + s - 2) * (s2x - s + 1) / 2;
59         for(int i = s + 1; i <= s2x; ++i) ans += pi[x / prime[i]];
60         return ans;
61     }
62     return getphi(x, s - 1) - getphi(x / prime[s], s - 1);
63 }
64 LL getpi(LL x)
65 {
66     if(x < N)   return pi[x];
67     LL ans = getphi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - 1;
68     for(int i = pi[sqrt3(x)] + 1, ed = pi[sqrt2(x)]; i <= ed; ++i) ans -= getpi(x / prime[i]) - i + 1;
69     return ans;
70 }
71 LL lehmer_pi(LL x)
72 {
73     if(x < N)   return pi[x];
74     int a = (int)lehmer_pi(sqrt2(sqrt2(x)));
75     int b = (int)lehmer_pi(sqrt2(x));
76     int c = (int)lehmer_pi(sqrt3(x));
77     LL sum = getphi(x, a) +(LL)(b + a - 2) * (b - a + 1) / 2;
78     for (int i = a + 1; i <= b; i++)
79     {
80         LL w = x / prime[i];
81         sum -= lehmer_pi(w);
82         if (i > c) continue;
83         LL lim = lehmer_pi(sqrt2(w));
84         for (int j = i; j <= lim; j++) sum -= lehmer_pi(w / prime[j]) - (j - 1);
85     }
86     return sum;
87 }
88 int main()
89 {
90     init();
91     LL n;
92     while(~scanf("%lld",&n))
93     {
94         printf("%lld
",lehmer_pi(n));
95     }
96     return 0;
97 }