• Linear Sieve Method for Prime Numbers


    Problem description:When we calculate for prime numbers with a sieve method,we delete so many numbers which is not necessary repeatly.For instance,there is a number which consists of 3x7x17x23,and we delete it when we delete the multiples of 3 as we delete the same number when we delete the multiples of 7,17,and 23.Please write a program that will not do these jobs more than once.
       
    Thinking: There is a factorization theorem:every composite number could be decomposed into the multiplication of some primer numbers.Hence,the number can be decomposed in the form of  (both of p and q are prime numbers and p < q).Therefore,what we need to remove is:,,...and,i=1,2,3.....The value of p and q is the numbers which are not removed currently and in a sequence from small to large.It is easy to write the program.

     

     1 #include <stdio.h>
     2  #define MAX 1000
     3  #define null1 0
     4  #define NEXT(x)  x=next[x]
     5  #define REMOVE(x) {   previous[next[x]]=previous[x];   \
     6                        next[previous[x]]=next[x];       \
     7                    }
     8  
     9  #define INITIAL(n)  { unsigned long i;                    \
    10                        for(i=2;i<=n;i++)                   \
    11                            previous[i]=i-1,next[i]=i+1;    \
    12                        previous[2]=next[n]=null1;           \
    13                      }
    14  
    15  int main()
    16  {
    17      unsigned long previous[MAX+1]={0};
    18      unsigned long next[MAX+1]={0};
    19      unsigned long prime,fact,i,mult;
    20      unsigned long n;
    21      unsigned long count=0;
    22      
    23      scanf("%lu",&n);
    24  
    25      INITIAL(n); //initial the array
    26  
    27      for(prime=2;prime*prime<=n;NEXT(prime))
    28      {
    29          for(fact=prime;prime*fact<=n;NEXT(fact)) 
    30          {
    31              for(mult=prime*fact;mult<=n;mult*=prime) 
    32                  REMOVE(mult);
    33          }
    34      }
    35      for(i=2;i!=null1;NEXT(i))
    36          printf("%lu ",i),count++;
    37      printf("\nThe sum of the prime numbers is %lu\n",count);
    38  }

    Reference material: C语言名题精选百则技巧篇 in Chinese.

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  • 原文地址:https://www.cnblogs.com/NeilHappy/p/2843047.html
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