下面是题目复述：

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 are H-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

789

Output

21 0

85 5

789 62

下面是解题思路：

题目大意：H 是除4余1的数字。

H——prime就是指不可以被其他H数字相除得到的数字。可以类似于素数。

H——semi-primes.是指两个H——prime相乘得到的数字。

题目要求是要求求出输入数字内有多少H——-semi-primes.

分为三步。

和求质数类似，H数相当于从5开始每次自增4，然后按照标准筛法筛选即可（这里按照的是埃氏筛法）。
然后把两个H——prime相乘，得到H——-semi-primes，存入数组。
保存每个数字内对应的有几个H——-semi-primes。

需要注意的是，变量较多，写的时候注意变量名称。

下面是AC代码：

#include <iostream>
#include <cstdio>

const int N=1e6+10;
using namespace std;
int prime[N],num=0;
bool str[N],flag[N];//str为false表示质数，flag true表示满足满足semi-primes
void in()//找出来h质数
{
    for(int i=5;i<=N;i+=4)
    {
        if(!str[i])
        {
            prime[num++]=i;
            for(int j=i+i;j<=N;j+=i) str[j]=true;
        }
    }
}
void on()//找出来结果要的数字类型
{
    for(int i=5;i<=N;i+=4)
    {
        for(int j=5;j<=N;j+=4)
        {
            if(i*j>N) break;

            if(!str[i]&&!str[j]) flag[i*j]=true;
        }
    }

}

int sum[N],ans=0;
void under ()
{
    for(int i=1;i<=N;i++)
    {
        if(flag[i])ans++;
        sum[i]=ans;
        //printf("i= %d,ans=%d\n",i,ans);
    }

}
int main()
{
    in();
    on();
    under();
    while(1)
    {
        int num;
        scanf("%d",&num);
        if(num==0) break;

        printf("%d %d\n",num,sum[num]);
    }
    return 0;
}

【题解】 Semi-prime H-numbers

下面是题目复述：

下面是解题思路：

下面是AC代码：