解题报告_18.5.23_POJ_3292_0

转载自：https://blog.youkuaiyun.com/lyy289065406/article/details/6648537

#include<iostream>  
using namespace std;  
  
const int size=1000001;  
  
int H_Number[size+1];  
  
/*筛法打表*/  
void Table(void)  
{  
    memset(H_Number,0,sizeof(H_Number));  //H_Number[i]=0 表示 i为H-prime  
  
    for(int i=5;i<=size;i+=4)  
    {  
        for(int j=5;j<=size;j+=4)  
        {  
            int multiply=i*j;  
            if(multiply>size)  
                break;  
  
            if(H_Number[i]==0 && H_Number[j]==0)  //i与j均为H-prime  
                H_Number[multiply]=1;  //multiply为H-semi-primes  
            else  
                H_Number[multiply]=-1; //multiply为H-composite  
        }  
    }  
  
    int Pcount=0; //H-prime计数器  
    for(int k=1;k<=size;k++)  
    {  
        if(H_Number[k]==1)  
            Pcount++;  
        H_Number[k]=Pcount;   //从1到k有Pcount个H-semi-primes  
    }  
    return;  
}  
  
int main(void)  
{  
    Table();  
    int h;  
    while(cin>>h && h)  
        cout<<h<<' '<<H_Number[h]<<endl;  
  
    return 0;  
}