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 }