How to resolve the algorithm Increasing gaps between consecutive Niven numbers step by step in the Fortran programming language

Published on 12 May 2024 09:40 PM
#Fortran

How to resolve the algorithm Increasing gaps between consecutive Niven numbers step by step in the Fortran programming language

Table of Contents

Problem Statement

Note:   Niven   numbers are also called   Harshad   numbers.

Niven numbers are positive integers which are evenly divisible by the sum of its digits   (expressed in base ten). Evenly divisible   means   divisible with no remainder.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Increasing gaps between consecutive Niven numbers step by step in the Fortran programming language

Source code in the fortran programming language

       program nivengaps      
       implicit none
       integer*8 prev /1/, gap /0/, sum /0/
       integer*8 nividx /0/, niven /1/
       integer gapidx /1/
       
       character*13 idxfmt
       character*14 nivfmt
       write (*,*) 'Gap no  Gap   Niven index    Niven number '
       write (*,*) '------  ---  -------------  --------------'
       
 10    call divsum(niven, sum)
       if (mod(niven, sum) .EQ. 0) then
          if (niven .GT. prev + gap) then
             gap = niven - prev
             call fmtint(nividx,13,idxfmt)
             call fmtint(prev,14,nivfmt)
             write (*,20) gapidx,gap,idxfmt,nivfmt
             gapidx = gapidx + 1
          end if
          prev = niven
          nividx = nividx + 1
       end if          
       niven = niven + 1
       if (gapidx .LE. 32) go to 10
       
       stop
 20    format (i7,'  ',i3,'  ',a13,'  ',a14)       
       end program

C      Sum of divisors of NN, given the sum of divisors of NN-1       
       subroutine divsum(nn,sum)
          implicit none
          integer*8 n,nn,sum
          n = nn          
          sum = sum + 1
 30       if (n.GT.0 .AND. mod(n,10).EQ.0) then
             sum = sum - 9
             n = n / 10
             go to 30
          end if
       end subroutine  
 
       integer*8 function mod(a,b)
          implicit none
          integer*8 a,b
          mod = a - a/b * b
       end function

C      Format a positive integer with ',' as the thousands separator.
       subroutine fmtint(num, len, str)
          implicit none 
          integer*8 n, num
          integer pos, len, th
          character(*) str
          n=num
          pos=len
          th=2
 40       if (pos.GT.0) then
             if (n.EQ.0) then
                str(pos:pos) = ' '
             else
                str(pos:pos) = achar(mod(n,10) + iachar('0'))
                if (th.EQ.0 .AND. n.GE.10 .AND. pos.GT.1) then
                    th = 2
                    pos = pos-1
                    str(pos:pos) = ','
                else
                    th = th-1
                end if
            end if
            pos = pos - 1
            n = n/10
            go to 40
          end if    
       end subroutine

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