题意：求[1,n]有多少个素数，1<=n<=10^11。时限为6000ms。
官方题解：一个模板题, 具体方法参考wiki或者Four Divisors。
题解：给出两种代码。
第一种方法Meisell-Lehmer算法只需265ms。
第二种方法不能运行但是能AC，只需35行。
第一种：
//Meisell-Lehmer
#include<cstdio>
#include<cmath>
using namespace std;
#define LL long long
const int N = 5e6 + ;
bool np[N];
int prime[N], pi[N];
int getprime()
{
int cnt = ;
np[] = np[] = true;
pi[] = pi[] = ;
for(int i = ; i < N; ++i)
{
if(!np[i]) prime[++cnt] = i;
pi[i] = cnt;
for(int j = ; j <= cnt && i * prime[j] < N; ++j)
{
np[i * prime[j]] = true;
if(i % prime[j] == ) break;
}
}
return cnt;
}
const int M = ;
const int PM = * * * * * * ;
int phi[PM + ][M + ], sz[M + ];
void init()
{
getprime();
sz[] = ;
for(int i = ; i <= PM; ++i) phi[i][] = i;
for(int i = ; i <= M; ++i)
{
sz[i] = prime[i] * sz[i - ];
for(int j = ; j <= PM; ++j) phi[j][i] = phi[j][i - ] - phi[j / prime[i]][i - ];
}
}
int sqrt2(LL x)
{
LL r = (LL)sqrt(x - 0.1);
while(r * r <= x) ++r;
return int(r - );
}
int sqrt3(LL x)
{
LL r = (LL)cbrt(x - 0.1);
while(r * r * r <= x) ++r;
return int(r - );
}
LL getphi(LL x, int s)
{
if(s == ) 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 + ;
if(x <= prime[s]*prime[s]*prime[s] && x < N)
{
int s2x = pi[sqrt2(x)];
LL ans = pi[x] - (s2x + s - ) * (s2x - s + ) / ;
for(int i = s + ; i <= s2x; ++i) ans += pi[x / prime[i]];
return ans;
}
return getphi(x, s - ) - getphi(x / prime[s], s - );
}
LL getpi(LL x)
{
if(x < N) return pi[x];
LL ans = getphi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - ;
for(int i = pi[sqrt3(x)] + , ed = pi[sqrt2(x)]; i <= ed; ++i) ans -= getpi(x / prime[i]) - i + ;
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 - ) * (b - a + ) / ;
for (int i = a + ; 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 - );
}
return sum;
}
int main()
{
init();
LL n;
while(~scanf("%lld",&n))
{
printf("%lld\n",lehmer_pi(n));
}
return ;
}
第二种：
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll maxn=1e11;
const ll maxp=sqrt(maxn)+;
ll f[maxp],g[maxp];
ll solve(ll n)
{
ll i,j,m;
for(m=;m*m<=n;m++)
f[m]=n/m-;
for(i=;i<=m;i++)
g[i]=i-;
for(i=;i<=m;i++)
{
if(g[i]==g[i-]) continue;
for(j=;j<=min(m-,n/i/i);j++)
{
if(i*j<m)
f[j]-=f[i*j]-g[i-];
else
f[j]-=g[n/i/j]-g[i-];
}
for(j=m;j>=i*i;j--)
g[j]-=g[j/i]-g[i-];
}
return f[];
}
int main()
{
ll n;
while(scanf("%lld",&n)!=EOF)
printf("%lld\n",solve(n));
return ;
}