Prime number "template"

Ordinary prime number judgment

int isprime (int N)      //Note #include<cmath>{    if (N <= 1)  return 0;    int i;    for (i = 2; I <= sqrt (n*1.0); ++i)        if (n%i = = 0)            return 0;    return 1;}

Sieve method for primes [1,n]

const int MAXN = 1000000;bool prime[maxn+100];   Prime[i] = = True means I is a prime number void IsPrime () {for    (int i = 2; I <= maxn; ++i)        prime[i] = true;    for (int i = 2; I <= MAXN, ++i)        if (Prime[i]) for            (int j = i+i; J <= Maxn; j+=i)                prime[j] = false;}

Two-time sieve method for primes [l,r]

BOOL prime[50010];    Deposit 50000 inside prime judgment result int primer[1000010];   The prime number of bool prime1[1000010] between the storage interval [l,r];    Determines whether the number in the interval [l,r] is a prime int isprime ()//The prime in the first sieve 50000 {int num = 0;    for (int i = 2; I <= 50000; i++) prime[i] = true;            for (int i = 2; I <= 50000; i++) {if (Prime[i]) {primer[num++] = i;        for (int j = i+i; J <= 50000; j+=i) prime[j] = false;     }} return num; Num is a number of elements in the range of 50000}int solve (__int64 A,__int64 b)/* On the basis of the first sieve prime, using the prime number within 50000, the number of pixels between the range "a", the remainder is the prime */{int num =    IsPrime ();    memset (prime1,true,sizeof (Prime1)); The Prime1 array is used to hold the primality of the range "A, B" to determine if (a = = 1)//Note here that 1 is not prime prime1[0] = 0; This indicates that 0+1 is not a prime number for (__int64 i = 0; i < num && primer[i] * Primer[i] <= b; i++) {__int64 begin =        A/primer[i] + (a%primer[i]! = 0);  The upper A/primer calculates how many times a is a prime primer[i] (a%primer[i]!=0) to start the sieve from a/primer[i] times of primer[i], or a/primer[i]+1 times sieve if (begin = = 1)//If the result is 1 time times the number of sieved prime, the sieve begin++ starts from twice times the prime number;    for (begin = begin*primer[i]; begin <= B; begin + = Primer[i]) prime1[begin-a] = false;    }//The Primer array is re-used here to hold the prime number between the interval "a, b", Num is the prime count of Memset (primer,0,sizeof (Primer));    num = 0;    for (__int64 i = A; I <= b; i++) if (prime1[i-a]==1) primer[num++] = i-a;     return num; Num is the number of characters in the interval [a, b]}

Test method for Miller prime number

#include <stdio.h> #include <stdlib.h> #include <time.h> #include <math.h> #define Max_val (POW (    2.0,60)) #define LL __int64/*ll Mod_mul (ll x,ll y,ll mo)//calculate x * y% mo{LL t;    x%= mo;    for (t = 0; y; x = (x<<1)%mo,y>>=1) if (Y & 1) t = (t+x)%mo; return t;}    */ll Mod_mul (LL x,ll y,ll mo)//calculate x * y% mo{LL T,t,a,b,c,d,e,f,g,h,v,ans;    T = (LL) (sqrt (double (MO) +0.5));    t = T*t-mo;    A = x/t;    b = x% T;    c = y/t;    d = y% T;    e = a*c/t;    f = a*c% T;    v = ((a*d+b*c)%mo + e*t)% Mo;    g = v/t;    h = v% T;    Ans = (((f+g) *t%mo + b*d)% mo + h*t)%mo;    while (ans < 0) ans + = mo; return ans;}    ll Mod_exp (ll num,ll t,ll mo)//calculate num^t% mo{LL ret = 1, temp = num% mo;    for (; t; t >>=1,temp=mod_mul (Temp,temp,mo)) if (T & 1) ret = Mod_mul (RET,TEMP,MO); return ret;}    BOOL Miller_rabbin (LL N)//miller_rabbin primality test {if (n = = 2) return true; if (N < 2|| !     (n&1))    return false;    int t = 0;    LL a,x,y,u = n-1;        while ((u & 1) = = 0) {t++;    U >>= 1;        } for (int i = 0; i < i++) {a = rand ()% (n-1) +1;        x = Mod_exp (a,u,n);            for (int j = 0; J < T; j + +) {y = Mod_mul (x,x,n);            if (y = = 1 && x! = 1 && x! = n-1) return false;        x = y;    } if (x! = 1) return false; } return true;

Prime number "template"