HDU 5901 Count primes (区间素数个数)

题目链接

题意：计算区间[1,1e11]中的素数的个数

题解上给出了一些相关的知识

1,icpc

2, cf ,F. Four Divisors

3,cf,题解

是一道论文题目了，拿来当做模板好了

本题目：

模板1

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/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();
printf("%lld\n",f[1]);
}
return 0;
}

模板2

//Meisell-Lehmer
/o(n^(2/3))
#include<cstdio>
#include<cmath>
using namespace std;
#define LL long long
const int N = 5e6 + 2;
bool np
;
int prime
, pi
;
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;
}

模板3

#include<bits/stdc++.h>
using namespace std;

#define MAXN 100
#define MAXM 50010
#define MAXP 166666
#define MAX 1000010
#define clr(ar) memset(ar, 0, sizeof(ar))
#define read() freopen("lol.txt", "r", stdin)
#define dbg(x) cout << #x << " = " << x << endl
#define chkbit(ar, i) (((ar[(i) >> 6]) & (1 << (((i) >> 1) & 31))))
#define setbit(ar, i) (((ar[(i) >> 6]) |= (1 << (((i) >> 1) & 31))))
#define isprime(x) (( (x) && ((x)&1) && (!chkbit(ar, (x)))) || ((x) == 2))

using namespace std;

namespace pcf{
long long dp[MAXN][MAXM];
unsigned int ar[(MAX >> 6) + 5] = {0};
int len = 0, primes[MAXP], counter[MAX];

void Sieve(){
setbit(ar, 0), setbit(ar, 1);
for (int i = 3; (i * i) < MAX; i++, i++){
if (!chkbit(ar, i)){
int k = i << 1;
for (int j = (i * i); j < MAX; j += k) setbit(ar, j);
}
}

for (int i = 1; i < MAX; i++){
counter[i] = counter[i - 1];
if (isprime(i)) primes[len++] = i, counter[i]++;
}
}

void init(){
Sieve();
for (int n = 0; n < MAXN; n++){
for (int m = 0; m < MAXM; m++){
if (!n) dp
[m] = m;
else dp
[m] = dp[n - 1][m] - dp[n - 1][m / primes[n - 1]];
}
}
}

long long phi(long long m, int n){
if (n == 0) return m;
if (primes[n - 1] >= m) return 1;
if (m < MAXM && n < MAXN) return dp
[m];
return phi(m, n - 1) - phi(m / primes[n - 1], n - 1);
}

long long Lehmer(long long m){
if (m < MAX) return counter[m];

long long w, res = 0;
int i, a, s, c, x, y;
s = sqrt(0.9 + m), y = c = cbrt(0.9 + m);
a = counter[y], res = phi(m, a) + a - 1;
for (i = a; primes[i] <= s; i++) res = res - Lehmer(m / primes[i]) + Lehmer(primes[i]) - 1;
return res;
}
}

int main()
{
pcf::init();
long long n;
while(cin>>n)
cout<<pcf::Lehmer(n)<<endl;
}

cf

题目的数是形如pq,或者ppp这种类型的数。

ppp类型的可以直接计算出来

pq暴力枚举较小的p,然后计算出n/p之间的素数

#include<bits/stdc++.h>
using namespace std;

#define MAXN 100
#define MAXM 100010
#define MAXP 666666
#define MAX 10000010
#define clr(ar) memset(ar, 0, sizeof(ar))
#define read() freopen("lol.txt", "r", stdin)
#define dbg(x) cout << #x << " = " << x << endl
#define chkbit(ar, i) (((ar[(i) >> 6]) & (1 << (((i) >> 1) & 31))))
#define setbit(ar, i) (((ar[(i) >> 6]) |= (1 << (((i) >> 1) & 31))))
#define isprime(x) (( (x) && ((x)&1) && (!chkbit(ar, (x)))) || ((x) == 2))

using namespace std;

namespace pcf{
long long dp[MAXN][MAXM];
unsigned int ar[(MAX >> 6) + 5] = {0};
int len = 0, primes[MAXP], counter[MAX];

void Sieve(){
setbit(ar, 0), setbit(ar, 1);
for (int i = 3; (i * i) < MAX; i++, i++){
if (!chkbit(ar, i)){
int k = i << 1;
for (int j = (i * i); j < MAX; j += k) setbit(ar, j);
}
}

for (int i = 1; i < MAX; i++){
counter[i] = counter[i - 1];
if (isprime(i)) primes[len++] = i, counter[i]++;
}
}

void init(){
Sieve();
for (int n = 0; n < MAXN; n++){
for (int m = 0; m < MAXM; m++){
if (!n) dp
[m] = m;
else dp
[m] = dp[n - 1][m] - dp[n - 1][m / primes[n - 1]];
}
}
}

long long phi(long long m, int n){
if (n == 0) return m;
if (primes[n - 1] >= m) return 1;
if (m < MAXM && n < MAXN) return dp
[m];
return phi(m, n - 1) - phi(m / primes[n - 1], n - 1);
}

long long Lehmer(long long m){
if (m < MAX) return counter[m];

long long w, res = 0;
int i, a, s, c, x, y;
s = sqrt(0.9 + m), y = c = cbrt(0.9 + m);
a = counter[y], res = phi(m, a) + a - 1;
for (i = a; primes[i] <= s; i++) res = res - Lehmer(m / primes[i]) + Lehmer(primes[i]) - 1;
return res;
}
}

long long solve(long long n){
int i, j, k, l;
long long x, y, res = 0;

for (i = 0; i < pcf::len; i++){
x = pcf::primes[i], y = n / x;
if ((x * x) > n) break;
res += (pcf::Lehmer(y) - pcf::Lehmer(x));
}

for (i = 0; i < pcf::len; i++){
x = pcf::primes[i];
if ((x * x * x) > n) break;
res++;
}

return res;
}

int main(){
pcf::init();
long long n, res;
cin>>n;
printf("%lld\n",solve(n));
return 0;
}