素数-原创手记-慕课网

埃氏筛法

#define exp 2.7182818285

#define Y 0.57721566490153286060651209

#define ln2 0.69314718055995

#define ln3 1.0986122886681

#define lg2 0.301029995663981

#define lg5 0.698970004336018

#define ln5 1.6094379124341

int prime[N];

void pre()

{

memset(prime,1,sizeof(prime));prime[1]=0;

for(int i=2;i<N;i++)

{

if(prime[i])

for(int j=2*i;j<N;j+=i)prime[j]=0;

}

线性筛

int p[MAX],cnt;

bool isprime[MAX];

void prime()

{

cnt=1;

memset(isprime,1,sizeof(isprime));//初始化认为所有数都为素数

isprime[0]=isprime[1]=0;//0和1不是素数

for(int i=2;i<MAX;i++)

{

if(isprime[i])

p[cnt++]=i;//保存素数i

for(int j=1;j<cnt&&p[j]*i<MAX;j++)

{

isprime[p[j]*i]=0;

If(i%isprimep[j]=0)break;

//筛掉小于等于i的素数和i的积构成的合数

}

Meisell-Lehmer

//求n以内的素数个数

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;

const int N = 5*1e6 + 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(cin >> n) {

cout << lehmer_pi(n) << endl;

}

Min_25 prime<n

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#define M 5333333

using namespace std;

typedef long long ll;

ll A[M], B[M];

double inv[M];

int main() {

ll n;

scanf("%lld", &n);

int lim = sqrt(n) + 1;

if(n <= 10) return printf("%d\n", (n >= 2) + (n >= 3) + (n >= 5) + (n >= 7)), 0;

for(int i = 1; i <= lim; i++) inv[i] = 1.0/ (i * 2 - 1), A[i] = (n / (i * 2 - 1) + 1) / 2, B[i] = (i + 1) / 2 - (i == 1);

for(int i = 3; i <= lim; i += 2)if(B[i] != B[i - 1]) {

unsigned l = B[i - 1];

for(int j = 1; i * (j * 2 - 1) <= lim; j++) A[j] -= A[i * j - (i - 1) / 2] - l;

double tmp = n / i;

for(int j = (lim / i + 1) / 2 + 1, s = min(n / i / i, (ll) lim); j * 2 - 1 <= s; j++) A[j] -= B[(int) (tmp * inv[j] + 1e-7)] - l;

for(int j = lim / i; j >= i; j--)

for(int k = i - 1; k >= 0; k--) B[j * i + k] -= B[j] - l;

}

printf("%lld\n", A[1]);

return 0;

}

Miller-rabin

//判断n是否为素数

#include <iostream>

#include <cstdio>

#include <algorithm>

#include <cmath>

#include <cstring>

#include <map>

using namespace std;

const int times = 20;

int number = 0;

//map<long long, int>m;

long long Random( long long n ) //生成[ 0 , n ]的随机数

{

return ((double)rand( ) / RAND_MAX*n + 0.5);

}

long long q_mul( long long a, long long b, long long mod ) //快速计算 (a*b) % mod

{

long long ans = 0;

while(b)

{

if(b & 1)

{

b--;

ans =(ans+ a)%mod;

}

b /= 2;

a = (a + a) % mod;

}

return ans;

}

long long q_pow( long long a, long long b, long long mod ) //快速计算 (a^b) % mod

{

long long ans = 1;

while(b)

{

if(b & 1)

{

ans = q_mul( ans, a, mod );

}

b /= 2;

a = q_mul( a, a, mod );

}

return ans;

}

bool witness( long long a, long long n )//miller_rabin算法的精华

{//用检验算子a来检验n是不是素数

long long tem = n - 1;

int j = 0;

while(tem % 2 == 0)

{

tem /= 2;

j++;

}

//将n-1拆分为a^r * s

long long x = q_pow( a, tem, n ); //得到a^r mod n

if(x == 1 || x == n - 1) return true; //余数为1则为素数

while(j--) //否则试验条件2看是否有满足的 j

{

x = q_mul( x, x, n );

if(x == n - 1) return true;

}

return false;

}

bool miller_rabin( long long n ) //检验n是否是素数

{

if(n == 2)

return true;

if(n < 2 || n % 2 == 0)

return false; //如果是2则是素数，如果<2或者是>2的偶数则不是素数

for(int i = 1; i <= times; i++) //做times次随机检验

{

long long a = Random( n - 2 ) + 1; //得到随机检验算子 a

if(!witness( a, n )) //用a检验n是否是素数

return false;

}

return true;

}

main( )

{

int t;cin>>t;

while(t--)

{

long long n;scanf("%lld",&n);

if(miller_rabin(n))

{

puts("Prime");continue;

}

else puts("NO");

}