POJ3292(素数筛选)

Semi-prime H-numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8706		Accepted: 3809

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 
85
789
0

Sample Output

21 0
85 5
789 62
思路：注意H数的域是模4余1的数。如9是H数，虽然在自然数范围内9不是素数但是在H数域内9是H-prime。

#include <iostream>
#include <string.h>
using namespace std;
const int MAXN=1000005;
bool isHprime[MAXN];
int Hprime[MAXN],top;
int h[MAXN];
void sieve()
{
    memset(isHprime,true,sizeof(isHprime));
    for(int i=5;i<MAXN;i+=4)
    {
        if(isHprime[i])
        {
            Hprime[top++]=i;
            for(int j=i+i;j<MAXN;j+=i)
            {
                if((j-1)%4==0)
                {
                    isHprime[j]=false;
                }
            }
        }
    }
    for(int i=0;i<top;i++)
    {
        if(Hprime[i]>10000)    break;
        for(int j=i;j<top;j++)
        {
            int mul=Hprime[i]*Hprime[j];
            if(mul>=MAXN)
            {
                break;
            }
            h[mul]=1;
        }
    }
    for(int i=1;i<MAXN;i++)
    {
        h[i]+=h[i-1];
    }
}
int main()
{
    sieve();
    int n;
    while(cin>>n&&n!=0)
    {
        cout<<n<<" "<<h[n]<<endl;
    }
    return 0;
}