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HDU5901大素数模板

来源:程序员人生   发布时间:2016-11-20 16:35:11 阅读次数:2377次

题意:

求区间[1,N]的质数的个数(1N1011)。 

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; }



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