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hdu 5901 Count primes 素数计数模板

转自:http://blog.csdn.net/chaiwenjun000/article/details/52589457

 

计从1到n的素数个数

两个模板

 

时间复杂度O(n^(3/4))

 1 #include <bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 ll f[340000],g[340000],n; 5 void init(){ 6     ll i,j,m; 7     for(m=1;m*m<=n;++m)f[m]=n/m-1; 8     for(i=1;i<=m;++i)g[i]=i-1; 9     for(i=2;i<=m;++i){10         if(g[i]==g[i-1])continue;11         for(j=1;j<=min(m-1,n/i/i);++j){12             if(i*j<m)f[j]-=f[i*j]-g[i-1];13             else f[j]-=g[n/i/j]-g[i-1];14         }15         for(j=m;j>=i*i;--j)g[j]-=g[j/i]-g[i-1];16     }17 }18 int main(){19     while(scanf("%I64d",&n)!=EOF){20         init();21         cout<<f[1]<<endl;22     }23     return 0;24 }

 

 

第二个 时间复杂度O(n^(2/3))

  1 //Meisell-Lehmer  2 //G++ 218ms 43252k  3 #include<cstdio>  4 #include<cmath>  5 using namespace std;  6 #define LL long long  7 const int N = 5e6 + 2;  8 bool np[N];  9 int prime[N], pi[N]; 10 int getprime() 11 { 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     { 17         if(!np[i]) prime[++cnt] = i; 18         pi[i] = cnt; 19         for(int j = 1; j <= cnt && i * prime[j] < N; ++j) 20         { 21             np[i * prime[j]] = true; 22             if(i % prime[j] == 0)   break; 23         } 24     } 25     return cnt; 26 } 27 const int M = 7; 28 const int PM = 2 * 3 * 5 * 7 * 11 * 13 * 17; 29 int phi[PM + 1][M + 1], sz[M + 1]; 30 void init() 31 { 32     getprime(); 33     sz[0] = 1; 34     for(int i = 0; i <= PM; ++i)  phi[i][0] = i; 35     for(int i = 1; i <= M; ++i) 36     { 37         sz[i] = prime[i] * sz[i - 1]; 38         for(int j = 1; j <= PM; ++j) phi[j][i] = phi[j][i - 1] - phi[j / prime[i]][i - 1]; 39     } 40 } 41 int sqrt2(LL x) 42 { 43     LL r = (LL)sqrt(x - 0.1); 44     while(r * r <= x)   ++r; 45     return int(r - 1); 46 } 47 int sqrt3(LL x) 48 { 49     LL r = (LL)cbrt(x - 0.1); 50     while(r * r * r <= x)   ++r; 51     return int(r - 1); 52 } 53 LL getphi(LL x, int s) 54 { 55     if(s == 0)  return x; 56     if(s <= M)  return phi[x % sz[s]][s] + (x / sz[s]) * phi[sz[s]][s]; 57     if(x <= prime[s]*prime[s])   return pi[x] - s + 1; 58     if(x <= prime[s]*prime[s]*prime[s] && x < N) 59     { 60         int s2x = pi[sqrt2(x)]; 61         LL ans = pi[x] - (s2x + s - 2) * (s2x - s + 1) / 2; 62         for(int i = s + 1; i <= s2x; ++i) ans += pi[x / prime[i]]; 63         return ans; 64     } 65     return getphi(x, s - 1) - getphi(x / prime[s], s - 1); 66 } 67 LL getpi(LL x) 68 { 69     if(x < N)   return pi[x]; 70     LL ans = getphi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - 1; 71     for(int i = pi[sqrt3(x)] + 1, ed = pi[sqrt2(x)]; i <= ed; ++i) ans -= getpi(x / prime[i]) - i + 1; 72     return ans; 73 } 74 LL lehmer_pi(LL x) 75 { 76     if(x < N)   return pi[x]; 77     int a = (int)lehmer_pi(sqrt2(sqrt2(x))); 78     int b = (int)lehmer_pi(sqrt2(x)); 79     int c = (int)lehmer_pi(sqrt3(x)); 80     LL sum = getphi(x, a) +(LL)(b + a - 2) * (b - a + 1) / 2; 81     for (int i = a + 1; i <= b; i++) 82     { 83         LL w = x / prime[i]; 84         sum -= lehmer_pi(w); 85         if (i > c) continue; 86         LL lim = lehmer_pi(sqrt2(w)); 87         for (int j = i; j <= lim; j++) sum -= lehmer_pi(w / prime[j]) - (j - 1); 88     } 89     return sum; 90 } 91 int main() 92 { 93     init(); 94     LL n; 95     while(~scanf("%lld",&n)) 96     { 97         printf("%lld\n",lehmer_pi(n)); 98     } 99     return 0;100 }

 

hdu 5901 Count primes 素数计数模板