[原创]大素数统计1e11以内 【数论】
2016-09-20 17:32:57 Tabris_ 阅读数：653
博客爬取于2020-06-14 22:43:14
以下为正文
版权声明：本文为Tabris原创文章，未经博主允许不得私自转载。
https://blog.csdn.net/qq_33184171/article/details/52598186
算法原理：https://en.wikipedia.org/wiki/Prime-counting_function
素数统计
这是一篇对素数问题描述非常详细的一篇博客了
代码一:
复杂度大概O（n^(3/4)）
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| # 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/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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代码二:
复杂度大概O（n^(2/3)）
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| # 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;
}
|
4因子个数统计
代码一:
复杂度大概O（n^(3/4)）
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| # include <bits/stdc++.h>
# define ll long long
using namespace std;
ll f[340000],g[340000],n,k;//f[i] means pi(n/i),g[i] means pi(i)
ll dp(ll n){
ll i,j,m,res=0;
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];
}
for(i=2;i<m;++i){
if(g[i]==g[i-1])continue;
res+=f[i]-g[i];
}return res;
}
int main(){
scanf("%I64d",&n);k=pow(n,1.0/3)+1e-9;
printf("%I64d\n",dp(n)+g[k]);
}
|
代码二:
复杂度大概O（n^(2/3)）
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| # include <cstring>
# include <cstdio>
# include <cstdlib>
# include <ctime>
# include <cmath>
# include <iostream>
# include <algorithm>
# include <vector>
# include <map>
# include <set>
# include <queue>
# include <stack>
# include <string>
# include <bitset>
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
typedef pair<LL,LL> PLL;
# define clr(a,b) memset(a,b,sizeof(a))
# define MP make_pair
# define PB push_back
# define lrt rt<<1
# define rrt rt<<1|1
# define lson l,m,lrt
using namespace std;
/*------------------------- template ------------------------------*/
const int N = 5e6+2;
bool np[N];
int p[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]) p[++cnt] = i;
for(int j = 1;j <= cnt && i * p[j] < N; ++j) {
np[i * p[j]] = true;
}
pi[i]=cnt;
}
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]=p[i]*sz[i-1];
for(int j=1;j<=PM;++j){
phi[j][i]=phi[j][i-1]-phi[j/p[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 <= p[s]*p[s]) return pi[x]-s+1;
if(x <= p[s]*p[s]*p[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/p[i]];
}
return ans;
}
return getphi(x,s-1)-getphi(x/p[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/p[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 / p[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 / p[j]) - (j - 1);
}
}
return sum;
}
//照素数统计只多了这么个部分..
LL getans(LL x){ // x < 1e11
LL ans = pi[sqrt3(x)];
for(int i=1,ed=pi[sqrt2(x-1)];i<=ed;++i){
ans += lehmer_pi(x/p[i])-i;
}
return ans;
}
int main(){
init();
LL n;
while(cin>>n){
cout<<getans(n)<<endl;
}
return 0;
}
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