题意：
求区间[1,N]的质数的个数(1≤N≤1011)。
思路：
模板题，上两个模板
相干知识见WIKI
代码1：
复杂度大概O（n^(3/4)）
#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll f[340000],g[340000],n;
void init(){
ll i,j,m;
for(m=1;m*m<=n;++m)f[m]=n/m⑴;
for(i=1;i<=m;++i)g[i]=i⑴;
for(i=2;i<=m;++i){
if(g[i]==g[i⑴])continue;
for(j=1;j<=min(m⑴,n/i/i);++j){
if(i*j<m)f[j]-=f[i*j]-g[i⑴];
else f[j]-=g[n/i/j]-g[i⑴];
}
for(j=m;j>=i*i;--j)g[j]-=g[j/i]-g[i⑴];
}
}
int main(){
while(scanf("%I64d",&n)!=EOF){
init();
cout<<f[1]<<endl;
}
return 0;
}
代码2：
复杂度O（n^(2/3)）
//Meisell-Lehmer
//G++ 218ms 43252k
#include<cstdio>
#include<cmath>
using namespace std;
#define LL long long
const int N = 5e6 + 2;
bool np[N];
int prime[N], pi[N];
int getprime()
{
int cnt = 0;
np[0] = np[1] = true;
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]] = true;
if(i % prime[j] == 0) break;
}
}
return cnt;
}
const int M = 7;
const int PM = 2 * 3 * 5 * 7 * 11 * 13 * 17;
int phi[PM + 1][M + 1], sz[M + 1];
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);
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()
{
init();
LL n;
while(~scanf("%lld",&n))
{
printf("%lld\n",lehmer_pi(n));
}
return 0;
}