POJ 3292, Semi-prime H-numbers

Time Limit: 1000MS  Memory Limit: 65536K
Total Submissions: 4166  Accepted: 1652

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

 

Source
Waterloo Local Contest, 2006.9.30

---

//  POJ3292.cpp : Defines the entry point for the console application.
//

#include  < iostream >
#include  < cmath >
using   namespace  std;

int  main( int  argc,  char *  argv[])
{
    unsigned  long   long   const  SIZE  =   1000005 ;
    unsigned  long   long  MAX  =  unsigned  long  (sqrt(SIZE * 1.0 ));
     int  H[SIZE];
    memset(H,  0 ,  sizeof (H));
    unsigned  long   long  num;
     for  (unsigned  long   long  i  =   5 ; i  <  MAX; i += 4 )
    {
         for (unsigned  long   long  j  =  i; (num  =  i  *  j ) <  SIZE; j += 4 )
        {
             if  (H[i]  !=   0   ||  H[j]  !=   0 ) 
                H[num]  =   - 1 ;
             else   if  (H[num] !=- 1 )
                H[num]  =   1 ;
        }
    }

     for  ( int  i  =   1 ; i  <  SIZE;  ++ i) 
    {
         if  (H[i]  ==   1 ) H[i]  =  H[i - 1 ]  +   1 ;
         else  H[i]  =  H[i - 1 ];
    }

     int  N;
     while (scanf( " %d " ,  & N) != EOF  &&  N != 0 )
        cout  <<  N  <<   "   "   <<  H[N]  <<  endl;

     return   0 ;
}