Prime screening method -- SPOJ Problem 2 Prime Generator, -- spojprime

Prime screening method -- SPOJ Problem 2 Prime Generator, -- spojprime

Prime number is also called a prime number. Apart from 1 and itself, it cannot be divided into other natural numbers. In other words, this number has no other factors except 1 and itself; otherwise, it is called the sum. The minimum prime number is 2.

It is easy to judge whether an integer N is a prime number and whether it can be divisible by an integer between 2 and sqrt (N.

def isPrime(n):    if n%2==0:        return False    for i in xrange(3,int(math.sqrt(n)+1),2):        if n%i==0:            return False    return True

However, it is obviously not a good idea to identify all the prime numbers between 1 and N. Since the sum can be divided into a series of products of prime numbers, the sum between 1 and N is a multiple of a prime number between 1 and sqrt (N). Excluding the sum, the remainder is the prime number:

import mathimport timeitdef findPrime(n):    a=[True]*(n+1)    a[0]=False    a[1]=False    for i in xrange(2,int(math.sqrt(n)+1)):        if a[i]:            k=i*i            while k<=n:                a[k]=False                k=k+i      if __name__=='__main__':    t=timeit.Timer('findPrime(2000000)','from __main__ import findPrime')    print t.timeit(1)

The algorithm starts from 2 to determine whether it is a prime number and removes a multiple of the prime number. When both 2 and I are determined, it is clear whether I + 1 is a prime number.

SPOJ Problem 2 Prime Generator requires finding the Prime number between n and m, where 1 <= m <= n <= 1000000000, n-m <= 100000.

In this case, creating a 1000000000-length sequence is a waste of space. You need to first find the prime number between 1 and sqrt (n), and then eliminate the multiples of these prime numbers between n and m:

import mathdef findPrime(n):    a=[True]*(n+1)    a[0]=False    a[1]=False    for i in xrange(2,int(math.sqrt(n)+1)):        if a[i]:            k=i*i            while k<=n:                a[k]=False                k=k+i    for i in xrange(2,n+1):        if a[i]:            yield idef findPrimeBySeed(n,m):    if n==1:        n=2    seed=findPrime(int(math.sqrt(m)))    alist=[1]*(m-n+1)    for prime in seed:           if prime<n:            k=(prime-n%prime)%prime        else:            k=2*prime-n        while k<=m-n:            alist[k]=False                        k+=prime    for i in xrange(m-n+1):        if alist[i]:            print i+n                if __name__=='__main__':    line=int(raw_input())    for i in xrange(line):        n,m=raw_input().split()        findPrimeBySeed(int(n),int(m))        print