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[素数个数模板] HDU 5901 Count primes

题目链接:传送门

题目大意:给你一个 n(1 <= n <= 1e11),问1~n中素数个数

题目思路:(Meisell-Lehmer算法)

 1 #include<cstdio>   2 #include<cmath>   3 using namespace std;   4 #define LL long long   5 const int N = 5e6 + 2;   6 bool np[N];   7 int prime[N], pi[N];   8 int getprime()   9 {  10     int cnt = 0;  11     np[0] = np[1] = true;  12     pi[0] = pi[1] = 0;  13     for(int i = 2; i < N; ++i)  14     {  15         if(!np[i]) prime[++cnt] = i;  16         pi[i] = cnt;  17         for(int j = 1; j <= cnt && i * prime[j] < N; ++j)  18         {  19             np[i * prime[j]] = true;  20             if(i % prime[j] == 0)   break;  21         }  22     }  23     return cnt;  24 }  25 const int M = 7;  26 const int PM = 2 * 3 * 5 * 7 * 11 * 13 * 17;  27 int phi[PM + 1][M + 1], sz[M + 1];  28 void init()  29 {  30     getprime();  31     sz[0] = 1;  32     for(int i = 0; i <= PM; ++i)  phi[i][0] = i;  33     for(int i = 1; i <= M; ++i)  34     {  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 int sqrt2(LL x)  40 {  41     LL r = (LL)sqrt(x - 0.1);  42     while(r * r <= x)   ++r;  43     return int(r - 1);  44 }  45 int sqrt3(LL x)  46 {  47     LL r = (LL)cbrt(x - 0.1);  48     while(r * r * r <= x)   ++r;  49     return int(r - 1);  50 }  51 LL getphi(LL x, int s)  52 {  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     {  58         int s2x = pi[sqrt2(x)];  59         LL ans = pi[x] - (s2x + s - 2) * (s2x - s + 1) / 2;  60         for(int i = s + 1; i <= s2x; ++i) ans += pi[x / prime[i]];  61         return ans;  62     }  63     return getphi(x, s - 1) - getphi(x / prime[s], s - 1);  64 }  65 LL getpi(LL x)  66 {  67     if(x < N)   return pi[x];  68     LL ans = getphi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - 1;  69     for(int i = pi[sqrt3(x)] + 1, ed = pi[sqrt2(x)]; i <= ed; ++i) ans -= getpi(x / prime[i]) - i + 1;  70     return ans;  71 }  72 LL lehmer_pi(LL x)  73 {  74     if(x < N)   return pi[x];  75     int a = (int)lehmer_pi(sqrt2(sqrt2(x)));  76     int b = (int)lehmer_pi(sqrt2(x));  77     int c = (int)lehmer_pi(sqrt3(x));  78     LL sum = getphi(x, a) +(LL)(b + a - 2) * (b - a + 1) / 2;  79     for (int i = a + 1; i <= b; i++)  80     {  81         LL w = x / prime[i];  82         sum -= lehmer_pi(w);  83         if (i > c) continue;  84         LL lim = lehmer_pi(sqrt2(w));  85         for (int j = i; j <= lim; j++) sum -= lehmer_pi(w / prime[j]) - (j - 1);  86     }  87     return sum;  88 }  89 int main()   90 {  91     init();  92     LL n;  93     while(~scanf("%lld",&n))  94     {  95         printf("%lld\n",lehmer_pi(n));  96     }  97     return 0;  98 }  

 

[素数个数模板] HDU 5901 Count primes