求1-1e11 以内的素数。
原理以后再搞，先把板子弄上：
1 #include<cstdio>
2 #include<cmath>
3
4 using namespace std;
5
6 #define LL long long
7 const int N = 5e6 + 2;
8 bool np[N];
9 int prime[N], pi[N];
10
11 int getprime(){
12 int cnt = 0;
13 np[0] = np[1] = true;
14 pi[0] = pi[1] = 0;
15 for(int i = 2; i < N; ++i){
16 if(!np[i]) prime[++cnt] = i;
17 pi[i] = cnt;
18 for(int j = 1; j <= cnt && i * prime[j] < N; ++j){
19 np[i * prime[j]] = true;
20 if(i % prime[j] == 0) break;
21 }
22 }
23 return cnt;
24 }
25
26 const int M = 7;
27 const int PM = 2 * 3 * 5 * 7 * 11 * 13 * 17;
28 int phi[PM + 1][M + 1], sz[M + 1];
29
30 void init(){
31 getprime();
32 sz[0] = 1;
33 for(int i = 0; i <= PM; ++i) phi[i][0] = i;
34 for(int i = 1; i <= M; ++i){
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
40 int sqrt2(LL x){
41 LL r = (LL)sqrt(x - 0.1);
42 while(r * r <= x) ++r;
43 return int(r - 1);
44 }
45
46 int sqrt3(LL x){
47 LL r = (LL)cbrt(x - 0.1);
48 while(r * r * r <= x) ++r;
49 return int(r - 1);
50 }
51
52 LL getphi(LL x, int s){
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 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
65 LL getpi(LL x){
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
72 LL lehmer_pi(LL x){
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 LL w = x / prime[i];
80 sum -= lehmer_pi(w);
81 if (i > c) continue;
82 LL lim = lehmer_pi(sqrt2(w));
83 for (int j = i; j <= lim; j++) sum -= lehmer_pi(w / prime[j]) - (j - 1);
84 }
85 return sum;
86 }
87
88 int main(){
89 init();
90 LL n;
91 while(~scanf("%lld",&n)){
92 printf("%lld
",lehmer_pi(n));
93 }
94 return 0;
95 }