poj 3292 Semi-prime H-numbers

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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 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

21 857890

Sample Output

21 085 5789 62

题目大意：定义一种数叫H-numbers，它是所有能除以四余一的数。
在H-numbers中分三种数：
1、H-primes，这种数只能被1和它本身整除，不能被其他的H-number整除，例如9是一个H-number，能被1，3，9整除，但3不是H-number，所以他是H-primes。
2、H-semi-primes是由两个H-primes相乘得出的。
3、剩下的是H-composite
问给一个数，求1到这个数之间有多少个H-semi-primes。

ps：根据要求进行筛选即可

<span style="font-size:18px;">#include <iostream>#include<cstring>#define N 1000100using namespace std;int vis[N+10],a[N+10];int main(){    for(int i=5;i<=N;i=i+4)    {        for(int j=5;j<=N;j=j+4)        {            if(i*j>N)break;            if(!vis[i]&&!vis[j])            {                vis[i*j]=1;            }            else            {                vis[i*j]=2;            }        }    }    int num=0;    for(int i=1;i<=N;i++)    {        if(vis[i]==1)        {            num++;        }        a[i]=num;    }    int n;    while(cin>>n,n)    {        cout<<n<<" "<<a[n]<<endl;    }    return 0;}</span>