Prime number Screening method (to find the prime number of 1~n)

1, the following is the general method to find the prime number of 1~n:

The general method of finding the prime number of 1~n
#include <iostream>
using namespace std;
int main ()
{
   int n,i,j,k=1;
   bool Bo;
   cin>>n;cout<< "2";
     for (i=3;i<=n;i+=2)
	{
	   bo=true;
           for (j=3;j*j<=i;j+=2)
            if (i%j==0)
	     {
	       bo=false;
	       break;
	       if (bo)
		 {
	           if (k%10==0) cout<<endl;
		    cout<<i<< ';
		      k++
	         }
	}
      cout<<endl<<k<<endl;
  return 0;
}

2, for the number of prime screening method, the specific method is not elaborated, Baidu can be found, simply say is to find a prime number of its multiples marked as a non prime.

Reference code:

The prime
#include <iostream>
using namespace std
of 1~n were obtained by the method of number selection. #define N 1000000000
  bool Isprime[n];
  int prime[100000];
int main ()
{
	int n,i,j,k=1;
	memset (isprime,true,sizeof (IsPrime));
	   cin>>n;
	     int num=0;
		   prime[num++]=2;
     	        for (i=3;i<=n;i+=2)
		   {
		      if (Isprime[i])
			   {
			      prime[num++]=i;
				for (j=i+i;j<=n;j+=i)  
				   isprime[j]=false;
			   }
		   }
		   for (i=0;i<num;i++)
		   {
	             cout<<prime[i]<< ';
		       if ((i+1)%10==0)

         		  cout<<endl;
		   }
      cout<<endl<<num<<endl;
  return 0;
}

3, the following is my small optimization of the algorithm:

is to change the for (j=i+i; j<=n j+=i) to for (J=i*i; j<=n; j+=2*i) because 2*i is a multiple of 2, 3*i is a multiple of 3, and 5*i is a multiple of 5 ~~~~~~, so J directly Starting from I*i, and every time J 2*i,

This will eliminate the even number. After modification, you can reduce the amount of duplicate assignments.

Complete code:

#include <iostream>
using namespace std;
#define N 1000000000
  bool Isprime[n];
  int prime[100000];
int main ()
{
	int n,i,j,k=1;
	memset (isprime,true,sizeof (IsPrime));
	   cin>>n;
	     int num=0;
	       prime[num++]=2;
     	   for (i=3;i<=n;i+=2)
	     {
	       if (Isprime[i])
	         {
		    prime[num++]=i;
		      for (j=i*i;j<=n;j+=2*i)
			 isprime[j]=false;
	         }
	     }
	  for (i=0;i<num;i++)
	    {
	      cout<<prime[i]<< ';
	         if ((i+1)%10==0)
	           cout<<endl;
	    }
      cout<<endl<<num<<endl;
  return 0;
}