POJ 3292 Semi-prime H-numbers（类素数筛法）_of that. an h-number is a positive number which is-优快云博客

本文介绍了一种特殊的数——H数，及其子集H质数和H半质数的概念，并提供了一个C++实现的例子，用于计算指定范围内H半质数的数量。

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Description

This problem is based on an exercise of David Hilbert, who pedagogically suggested that one study the theory of 4n+1 numbers. Here, we do only a bit of that.

An H-number is a positive number which is one more than a multiple of four: 1, 5, 9, 13, 17, 21,... are the H-numbers. For this problem we pretend that these are the only numbers. The H-numbers are closed under multiplication.

As with regular integers, we partition the H-numbers into units, H-primes, and H-composites. 1 is the only unit. An H-number h is H-prime if it is not the unit, and is the product of two H-numbers in only one way: 1 × h. The rest of the numbers are H-composite.

For examples, the first few H-composites are: 5 × 5 = 25, 5 × 9 = 45, 5 × 13 = 65, 9 × 9 = 81, 5 × 17 = 85.

Your task is to count the number of H-semi-primes. An H-semi-prime is an H-number which is the product of exactly two H-primes. The two H-primes may be equal or different. In the example above, all five numbers areH-semi-primes. 125 = 5 × 5 × 5 is not an H-semi-prime, because it's the product of three H-primes.

Input

Each line of input contains an H-number ≤ 1,000,001. The last line of input contains 0 and this line should not be processed.

Output

For each inputted H-number h, print a line stating h and the number of H-semi-primes between 1 and h inclusive, separated by one space in the format shown in the sample.

Sample Input

Sample Output

21 0
85 5
789 62

H-numbers是一类4n+1的数（在这个问题里，只有这类数字）

H-primes是一类因子只有1和它本身的H-numbers（类似于平常我们见到的质数）

H-semi-primes是一类数是两个H-primes的乘积

打个表，然后输出就可以

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;
int H_number[1000011];
void fun()
{
    int i,j;
    memset(H_number,0,sizeof(H_number));
    for(i=5;i<=1000001;i+=4)
    {
        for(j=5;j<=1000001;j+=4)
        {
            int t = i*j;
            if(t > 1000001)
                break;
            if(H_number[i] == 0 && H_number[j] == 0)
                H_number[t] = 1;
            else
                H_number[t] = -1;
        }
    }
    int cnt = 0;
    for(i=1;i<=1000001;i++)
    {
        if(H_number[i] == 1)
        {
            cout << i << endl;
            cnt++;
        }

        H_number[i] = cnt;
    }
}
int main(void)
{
    int n;
    fun();
    while(scanf("%d",&n)&&n)
    {
        printf("%d %d\n",n,H_number[n]);
    }

    return 0;
}