HDU 2136 largest prime factor three methods for prime number table

Three methods for prime number table are summarized.

First:

void intline(){    memset(prime,true,sizeof(prime));    prime[1]=false;    prime[0]=false;    for(int i=2; i*i<MAXN; i++)    {        for(int j=2; j*i<MAXN; j++)            prime[j*i]=false;    }}

Second: similar to the first principle, it is also the method used for this question, but this is an array opened with an int, prime [I] = J indicates that the maximum prime factor of I is the prime number J, and 2 is the first prime number). Therefore, this method has many variants.

void intline(){    int num=0;   memset(prime,0,sizeof(prime));    for(int i=2;i<MAXN;i++)    {        if(!prime[i])        {            num++;            for(int j=i;j<MAXN;j+=i)                prime[j]=num;        }    }}

The third method is the most common method.

int isprime(int n){    for(int i=2; i*i<=n; i++)        if(n%i==0)  return 0;    return 1;}

Fourth type: in fact, like the second type, it is the first type of deformation. This prime [I] = J indicates that the prime number of I is J.

void intline(){    memset(num,0,sizeof(num));    num[0]=num[1]=1;    int cnt=0;    for(int i=2; i*i<MAXN; i++)    {        if(!num[i])        {            prime[cnt++]=i;            for(int j=2; j*i<MAXN; j++)                num[j*i]=1;        }    }}

Code for this question:

#include<cstdio>#include<cstring>#include<iostream>#define MAXN  1000001using namespace std;int prime[MAXN];void intline(){    int num=0;    for(int i=2;i<MAXN;i++)    {        if(!prime[i])        {            num++;            for(int j=i;j<MAXN;j+=i)                prime[j]=num;        }    }}int main(){    //freopen("in.txt","r",stdin);    intline();    int n;    while(scanf("%d",&n)!=EOF)    {        printf("%d\n",prime[n]);    }    return 0;}