Add Fortran solution to challenge 119

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diff --git a/challenge-119/paulo-custodio/fortran/ch-1.f90 b/challenge-119/paulo-custodio/fortran/ch-1.f90
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+! Challenge 119
+!
+! TASK #1 - Swap Nibbles
+! Submitted by: Mohammad S Anwar
+! You are given a positive integer $N.
+!
+! Write a script to swap the two nibbles of the binary representation of the
+! given number and print the decimal number of the new binary representation.
+!
+! A nibble is a four-bit aggregation, or half an octet.
+!
+! To keep the task simple, we only allow integer less than or equal to 255.
+!
+! Example
+! Input: $N = 101
+! Output: 86
+!
+! Binary representation of decimal 101 is 1100101 or as 2 nibbles (0110)(0101).
+! The swapped nibbles would be (0101)(0110) same as decimal 86.
+!
+! Input: $N = 18
+! Output: 33
+!
+! Binary representation of decimal 18 is 10010 or as 2 nibbles (0001)(0010).
+! The swapped nibbles would be (0010)(0001) same as decimal 33.
+
+program ch1
+    implicit none
+
+    integer :: n, stat
+    character(len=50) :: arg
+
+    call get_command_argument(1, arg)
+    if (len_trim(arg) /= 0) then
+        read(arg,*,iostat=stat) n
+        n = ((mod((n / 16), 16)) + (mod(n, 16) * 16))
+        print *, n
+    end if
+end program ch1
diff --git a/challenge-119/paulo-custodio/fortran/ch-2.f90 b/challenge-119/paulo-custodio/fortran/ch-2.f90
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+! Challenge 119
+!
+! TASK #2 - Sequence without 1-on-1
+! Submitted by: Cheok-Yin Fung
+! Write a script to generate sequence starting at 1. Consider the increasing
+! sequence of integers which contain only 1's, 2's and 3's, and do not have any
+! doublets of 1's like below. Please accept a positive integer $N and print the
+! $Nth term in the generated sequence.
+!
+! 1, 2, 3, 12, 13, 21, 22, 23, 31, 32, 33, 121, 122, 123, 131, ...
+!
+! Example
+! Input: $N = 5
+! Output: 13
+!
+! Input: $N = 10
+! Output: 32
+!
+! Input: $N = 60
+! Output: 2223
+
+program ch2
+    implicit none
+
+    integer :: i, num, n, stat
+    character(len=50) :: arg
+
+    call get_command_argument(1, arg)
+    if (len_trim(arg) /= 0) then
+        read(arg,*,iostat=stat) num
+
+        n = 0
+        do i = 1, num
+            n = next_seq(n)
+        end do
+
+        print *, n
+    end if
+
+contains
+    function num_ok(n1)
+        logical :: num_ok
+        integer, intent(in) :: n1
+        integer :: n, digit, last_digit
+        logical :: failed
+
+        n = n1
+        failed = .false.
+        if (n <= 0) then
+            failed = .true.
+        end if
+
+        digit = 0
+        do while (n > 0)
+            last_digit = digit
+            digit = mod(n, 10)
+            n = n / 10
+            if (digit < 1 .or. digit > 3 .or. (digit == 1 .and. last_digit == 1)) then
+                failed = .true.
+            end if
+        end do
+
+        num_ok = .not. failed
+    end function num_ok
+
+    function next_seq(n1)
+        integer :: next_seq
+        integer, intent(in) :: n1
+        integer :: n
+
+        n = n1
+        n = n+1
+        do while (.true.)
+            if (num_ok(n)) then
+                exit
+            else
+                n = n+1
+            end if
+        end do
+
+        next_seq = n
+    end function next_seq
+end program ch2