Mathematics:Semi-prime H-numbers(POJ 3292)

　　　　　　　　　　　　　　

　　　　　　　　　　　   Semi-prime H-numbers

　　题目大意，令4n+1的数叫H数，H数素数x的定义是只能被x=1*h（h是H数）,其他都叫合数，特别的，当一个数只能被两个H素数乘积得到时，叫H-semi数

　　做法，筛法暴力打表，记得要打表不然会TLE

　　

#include <iostream>
#include <functional>
#include <algorithm>
#define MAX_N 1000100

using namespace std;

static int H_Semi[MAX_N], H_Semi_Flag[MAX_N], H_Semi_Sum[MAX_N];

void Inivilize(void);

int main(void)
{
    int range;
    Inivilize();
    while (~scanf("%d", &range))
    {
        if (range == 0) break;
        printf("%d %d
", range, H_Semi_Sum[range]);
    }
    return 0;
}

void Inivilize(void)
{
    //-1不是H数 ,0表示是H_P,1表示H_C,2表示H_s
    int j, i, ans = 0;
    memset(H_Semi_Flag, -1, sizeof(H_Semi_Flag));
    for (i = 1; 4 * i + 1 <= 1000001; i++)
        H_Semi_Flag[4 * i + 1] = 0;
    for (i = 1; 4 * i + 1 <= 1000001; i++)
    {
        for (j = 1; (4 * i + 1) * (4 * j + 1) <= 1000001 && j <= i; j++)
        {
            if (H_Semi_Flag[(4 * j + 1)] == 0)
            {
                if (H_Semi_Flag[(4 * i + 1)] == 0)
                {
                    if (H_Semi_Flag[(4 * i + 1) * (4 * j + 1)] == 0)
                        H_Semi_Flag[(4 * i + 1) * (4 * j + 1)] = 2;
                    else if (H_Semi_Flag[(4 * i + 1) * (4 * j + 1)] == 2)
                        continue;
                }
                else if (H_Semi_Flag[(4 * i + 1) * (4 * j + 1)] == 2)
                    H_Semi_Flag[(4 * i + 1) * (4 * j + 1)] = 1;
            }
            else if (H_Semi_Flag[(4 * i + 1) * (4 * j + 1)] == 0 || H_Semi_Flag[(4 * i + 1) * (4 * j + 1)] == 2)
                H_Semi_Flag[(4 * i + 1) * (4 * j + 1)] = 1;
        }
    }
    for (int i = 1; i <= 1000001; i++)
    {
        if (H_Semi_Flag[i] == 2)
            ans++;
        H_Semi_Sum[i] = ans;
    }
}