HDU5901 1e11以内素数的个数
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#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <vector>#include <queue>#include <algorithm>#include <set>#include <time.h>using namespace std;typedef long long LL;typedef unsigned long long ULL;const LL INF = 1e9+5;const LL MAXN = 1e6+5;const LL MOD = 1e9+7;const double eps = 1e-7;const double PI = acos(-1);using namespace std;LL Scan_LL()///输入外挂{ LL res=0,ch,flag=0; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+ch-'0'; return flag?-res:res;}void Out(LL a)///输出外挂{ if(a>9) Out(a/10); putchar(a%10+'0');}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 get_phi(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 get_phi(x, s - 1) - get_phi(x / prime[s], s - 1);}LL getpi(LL x){ if(x < N) return pi[x]; LL ans = get_phi(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 = get_phi(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("%I64d",&n)) { Out(lehmer_pi(n)); puts(""); } return 0;}复杂度略高，但是代码长度短#include <bits/stdc++.h>#define ll long longusing namespace std;ll f[340000],g[340000],n;void init(){ ll i,j,m; for(m=1;m*m<=n;++m)f[m]=n/m-1; for(i=1;i<=m;++i)g[i]=i-1; for(i=2;i<=m;++i){ if(g[i]==g[i-1])continue; for(j=1;j<=min(m-1,n/i/i);++j){ if(i*j<m)f[j]-=f[i*j]-g[i-1]; else f[j]-=g[n/i/j]-g[i-1]; } for(j=m;j>=i*i;--j)g[j]-=g[j/i]-g[i-1]; }}int main(){ while(scanf("%I64d",&n)!=EOF){ init(); cout<<f[1]<<endl; } return 0;}
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