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 }