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Hdu5901

描述:

$求n以内素数个数,n最大(10)^{11}$。


思路:

论文题,套模板就好。


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#include <cstdio>
#include <cmath>
#define LL long long
using namespace std;
const int N = 5e6 + 5;
const int M = 7;
const int PM = 2 * 3 * 5 * 7 * 11 * 13 * 17;
int np[N];
int prime[N], pi[N];
int phi[PM+1][M+1], sz[M+1];
int getprime()
{
int cnt = 0;
np[0] = np[1] = 1;
pi[0] = pi[1] = 0;
for(int i = 2; i < N; ++i)
{
if(!np[i]) prime[++cnt] = i;
pi[i] = cnt;
for(int j = 1; j <= cnt && i * prime[j] < N; ++j)
{
np[i * prime[j]] = 1;
if(i % prime[j] == 0) break;
}
}
return cnt;
}
void init()
{
getprime();
sz[0] = 1;
for(int i = 0; i <= PM; ++i) phi[i][0] = i;
for(int i = 1; i <= M; ++i)
{
sz[i] = prime[i] * sz[i-1];
for(int j = 1; j <= PM; ++j) phi[j][i] = phi[j][i-1] - phi[j/prime[i]][i-1];
}
}
int sqrt2(LL x)
{
LL r = (LL)sqrt(x - 0.1);
while(r*r <= x) ++r;
return (int)(r-1);
}
int sqrt3(LL x)
{
LL r = (LL)cbrt(x - 0.1); //cbrt(x): x的立方根
while(r*r*r <= x) ++r;
return (int)(r-1);
}
LL getphi(LL x, int s)
{
if(s == 0) return x;
if(s <= M) return phi[x%sz[s]][s] + (x/sz[s]) * phi[sz[s]][s];
if(x <= prime[s]*prime[s]) return pi[x] - s + 1;
if(x <= prime[s]*prime[s]*prime[s] && x < N)
{
int s2x = pi[sqrt2(x)];
LL ans = pi[x] - (s2x+s-2)*(s2x-s+1)/2;
for(int i = s+1; i <= s2x; ++i) ans += pi[x/prime[i]];
return ans;
}
return getphi(x, s-1) - getphi(x/prime[s], s-1);
}
LL getpi(LL x)
{
if(x < N) return pi[x];
LL ans = getphi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - 1;
for(int i = pi[sqrt3(x)]+1, ed = pi[sqrt2(x)]; i <= ed; ++i) ans -= getpi(x/prime[i]) - i + 1;
return ans;
}
LL lehmer_pi(LL x)
{
if(x < N) return pi[x];
int a = (int)lehmer_pi(sqrt2(sqrt2(x)));
int b = (int)lehmer_pi(sqrt2(x));
int c = (int)lehmer_pi(sqrt3(x));
LL sum = getphi(x, a) + (LL)(b+a-2) * (b-a+1)/2;
for (int i = a+1; i <= b; ++i)
{
LL w = x/prime[i];
sum -= lehmer_pi(w);
if(i > c) continue;
LL lim = lehmer_pi(sqrt2(w));
for(int j = i; j <= lim; ++j) sum -= lehmer_pi(w/prime[j]) - (j-1);
}
return sum;
}
int main()
{
LL n; init();
while(~scanf("%lld",&n))
{
printf("%lld\n", lehmer_pi(n));
}
return 0;
}
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