POJ 3292, Semi-prime H-numbers

╰+哭是因爲堅強的太久メ 2021-12-22 06:45 155阅读 0赞

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 ;  
\}

转载于:https://www.cnblogs.com/asuran/archive/2009/10/22/1587802.html