Several variants of Sieve method

We know that the Sieve method is generally open up a large array, and then the cycle can be 2,3,5, such as multiple culling, through yesterday's ACM training, I learned several variants, very interesting.

1): General Screening Method:

	BOOL Prime[max];
    memset (prime,true,sizeof (Prime));
    for (k=4;k<max;k+=2)
        prime[k]=false;
    for (I=3;i<=sqrtmax;++i)
    {
        if (Prime[i])
        {for
            (int j=i*i;j<max;j+=2*i)
                prime[j]=false;
        }
    }

J Starting from the I*i, it is clear that the number of smaller than i*i has been removed, but this method note that the size of I may lead to i*i out of bounds, this time to open a wide range or to be judged divided into i*i and i+i two cases of the first twice times the number of elimination, then can j+=2*i.

Attached: HDU1397

#include <iostream>
#include <cmath>
#include <cstring>
using namespace std;
const int max=32770;
const int sqrtmax=181;
const int submax=10000;
BOOL Prime[max];
int Subprime[submax];
void Creatprime (void)
{
	int i,k;
    memset (prime,true,sizeof (Prime));
    for (k=4;k<max;k+=2)
        prime[k]=false;
    for (I=3;i<=sqrtmax;++i)
    {
        if (Prime[i])
        {for
            (int j=i*i;j<max;j+=2*i)
                prime[j]= false;
        }
    }
    subprime[0]=0;
	for (I=2;i<max;++i)
		if (Prime[i])
			subprime[++subprime[0]]=i;
}
int main (void)
{
    int n,count;
	Creatprime ();
    cin>>n;
	while (n!=0)
    {
        count=0;
        if (Prime[n-2])
            count++;
        for (int i=1;i<=subprime[0]&&subprime[i]<=n/2;i++)
            if (prime[subprime[i]]&&prime[ N-subprime[i]])
                count++;
        cout<<count<<endl;
        cin>>n;
    }
	return 0;
}

2): Solve the sum of a number of factors: on the basis of the general sieve slightly deformed, prime[i] accumulation factor can be.

   memset (prime,0,sizeof (Prime));
    for (I=1;i<=max/2;++i)
	{for
        (j=i+i;j<max;j+=i)
            prime[j]+=i;
	}

Attached: HDU1215

#include <stdio.h>
#include <string.h>
#define MAX 500001
int Prime[max];
void Creatprime (void)
{
	int i,j;
    memset (prime,0,sizeof (Prime));
    for (I=1;i<=max/2;++i)
	{for
        (j=i+i;j<max;j+=i)
            prime[j]+=i;
	}
}
int main ()
{
	int n,ncase;
	Creatprime ();
	scanf ("%d", &ncase);
	while (ncase--)
	{
		scanf ("%d", &n);
		printf ("%d\n", Prime[n]);
	}
	return 0;
}

3): Solve a number of the maximum mass factor (not more than he itself in the quality table of the first few) such as 9 of the maximum factor 3 ranked second, 15 of the largest factor 5 ranked third:

   memset (prime,0,sizeof (Prime));
    for (I=2;i<=max;++i)
    {
        if (prime[i]==0)
        {
			count++;
            for (j=i;j<max;j+=i)
                prime[j]=count;
        }
    }

Attached: HDU2136

#include <stdio.h>
#include <string.h>
#define MAX 1000000
int Prime[max];
void Creatprime (void)
{
	int i,j,count=0;
    memset (prime,0,sizeof (Prime));
    for (I=2;i<=max;++i)
    {
        if (prime[i]==0)
        {
			count++;
            for (j=i;j<max;j+=i)
                prime[j]=count;
}}} int main ()
{
	int n;
	Creatprime ();
	while (scanf ("%d", &n)!=eof)
	{
		printf ("%d\n", Prime[n]);
	}
	return 0;
}

4): solves the maximum mass factor of a number (if it is prime, it is itself)

memset (prime,0,sizeof (Prime));
for (int i=2;i<max;++i)
{
	if (prime[i]==0)
	{for
		(int j=i;j<max;j+=i)
			prime[j]=i;
	}
}

Yesterday's ACM training learned something, let us learn to think, but the key is to practice more thinking, continue to work hard.