Çözüldü Kombinasyon - Basamak Analizi - Programlama

1'den 10^19'a kadar doğal sayılardan kaç tanesinin rakamları toplamı 2'dir?
https://www.facebook.com/photo.php?fbid=168677627597083&set=gm.1639546842849519&type=3&theater (başka bir çözüm)
(Sorunun gönderildiği Facebook grubu 25 Eylül 2022 tarihinde "Private" duruma getirildiği için aslını ve varsa diğer çözümleri ancak üyeleri görebilir.)

C(19, 2) + 19 = 190

www.mathful.com'daki yapay zekânın 463 saniyede verdiği 21 sayfalık incelemesiyle çözümü (ektedir) ve sonundaki özeti:

https://i.ibb.co/39dkRvWT/Mathful-AI-Basamaklar-Toplami.png

Bilgisayar programlamayla ilgilenen öğrenciler için yararlı olacağı düşüncesiyle www.workik.com sitesindeki yapay zekâya yaptırdığım (çok küçük bir hatasını düzelttim) Fortran programı:

https://i.ibb.co/RpT9cVpS/rakamlar-toplami-Fortran.png

Program:

Kod:

! Program generated by the AI at www.workik.com
! I have slightly modified and corrected one mistake by placing the
! variable d at the proper place.

! Explanation:
! We use a dynamic programming approach to count the number of numbers of length
! max_digits=19 whose digits sum to 2.
! dp(pos, s) stores how many ways to form a number with exactly pos digits and the digit
! sum equal to s.
! We allow leading zeros to handle all numbers from 0 up to 10^19 - 1 uniformly.
! Since the problem counts natural numbers from 1 to 10^19 and 0 is excluded but has
! digit sum 0, no subtraction is necessary.
! The upper limit 10^19 is represented as a 20-digit number, but since we only use 19
! digits from 0 to 10^19-1, it's consistent.
! The count of numbers with digit sum 2 is the answer.
! The program should execute fast because dp is very small (only 20 x 3 in size).
! The integer kind(8) (64-bit integers) is used for holding the result as it can be quite large.
program count_sum_digits_two
  implicit none
  integer, parameter :: max_digit_sum = 2
  integer, parameter :: max_digits = 19
  integer(8) :: result
  integer :: i, sum_digit, d
  integer(kind=8), dimension(0:max_digits,0:max_digit_sum) :: dp = 0_8

  ! dp(pos, s) = number of ways to form a number with 'pos' digits summing to 's'

   dp(0,0) = 1_8

  ! For each digit position
  do i = 1, max_digits
    do sum_digit = 0, max_digit_sum
       do d = 0, 9
        if (sum_digit - d >= 0) then
          dp(i,sum_digit) = dp(i,sum_digit) + dp(i-1,sum_digit - d)
        end if
      end do
    end do
  end do

  ! We want numbers from 1 to 10^19 with digit sum = 2
  ! dp(max_digits, 2) includes numbers with leading zeros, i.e., numbers with fewer digits padded with zeros
  ! This counts numbers from 0 to 10^19-1 with digit sum 2.
  ! Since 0 has digit sum 0, no correction is required.
  ! Also, 10^19 itself is a 1 followed by 19 zeros, digit sum=1, so it does not count anyway.

  result = dp(max_digits, max_digit_sum)

  write(6,10)"Number of natural numbers from 1 to 10^19 with digit sum 2: ", result

10 format (a,i3)

end program count_sum_digits_two