Add solutions to challenge 120

27 files changed, 1248 insertions, 0 deletions

diff --git a/challenge-120/paulo-custodio/ada/ch_1.adb b/challenge-120/paulo-custodio/ada/ch_1.adb
new file mode 100644
index 0000000000..8e7cf38977
--- /dev/null
+++ b/challenge-120/paulo-custodio/ada/ch_1.adb
@@ -0,0 +1,50 @@
+-- Challenge 120
+--
+-- TASK #1 - Swap Odd/Even bits
+-- Submitted by: Mohammad S Anwar
+-- You are given a positive integer $N less than or equal to 255.
+--
+-- Write a script to swap the odd positioned bit with even positioned bit and
+-- print the decimal equivalent of the new binary representation.
+--
+-- Example
+-- Input: $N = 101
+-- Output: 154
+--
+-- Binary representation of the given number is 01 10 01 01.
+-- The new binary representation after the odd/even swap is 10 01 10 10.
+-- The decimal equivalent of 10011010 is 154.
+--
+-- Input: $N = 18
+-- Output: 33
+--
+-- Binary representation of the given number is 00 01 00 10.
+-- The new binary representation after the odd/even swap is 00 10 00 01.
+-- The decimal equivalent of 100001 is 33.
+
+with Ada.Command_Line;
+with Ada.Text_IO; use Ada.Text_IO;
+
+procedure ch_1 is
+    package CL renames Ada.Command_Line;
+
+    function swapBits(n1: Integer) return Integer is
+        n: Integer := n1;
+        result: Integer := 0;
+        shift: Integer := 1;
+    begin
+        while n > 0 loop
+            if n mod 2 /= 0 then result := result + 2*shift; end if;
+            n := n / 2;
+            if n mod 2 /= 0 then result := result + 1*shift; end if;
+            n := n / 2;
+            shift := shift*4;
+        end loop;
+        return result;
+    end swapBits;
+
+    n : Integer := Integer'Value(CL.Argument(1));
+begin
+    n := swapBits(n);
+    Put_Line(Integer'Image(n));
+end ch_1;
diff --git a/challenge-120/paulo-custodio/ada/ch_2.adb b/challenge-120/paulo-custodio/ada/ch_2.adb
new file mode 100644
index 0000000000..35ae82ce1d
--- /dev/null
+++ b/challenge-120/paulo-custodio/ada/ch_2.adb
@@ -0,0 +1,64 @@
+-- Challenge 120
+--
+-- TASK #2 › Clock Angle
+-- Submitted by: Mohammad S Anwar
+-- You are given time $T in the format hh:mm.
+--
+-- Write a script to find the smaller angle formed by the hands of an analog
+-- clock at a given time.
+--
+-- HINT: A analog clock is divided up into 12 sectors. One sector represents 30
+-- degree (360/12 = 30).
+--
+-- Example
+-- Input: $T = '03:10'
+-- Output: 35 degree
+--
+-- The distance between the 2 and the 3 on the clock is 30 degree.
+-- For the 10 minutes i.e. 1/6 of an hour that have passed.
+-- The hour hand has also moved 1/6 of the distance between the 3 and the 4,
+-- which adds 5 degree (1/6 of 30).
+-- The total measure of the angle is 35 degree.
+--
+-- Input: $T = '04:00'
+-- Output: 120 degree
+
+with Ada.Command_Line;
+with Ada.Text_IO; use Ada.Text_IO;
+with Ada.Strings.Fixed; use Ada.Strings.Fixed;
+
+procedure ch_2 is
+    package CL renames Ada.Command_Line;
+
+    procedure parseTime(hhmm: String; hh, mm: out Integer) is
+        pos: Integer;
+    begin
+        pos := Index(hhmm, ":");
+        hh := Integer'Value(hhmm(hhmm'First..pos-1));
+        mm := Integer'Value(hhmm(pos+1..hhmm'Last));
+    end parseTime;
+
+    procedure clockAngles(hh, mm: Integer; hh_angle, mm_angle: out Integer) is
+    begin
+        mm_angle := mm*360/60;
+        hh_angle := (hh mod 12)*360/12 + mm_angle*(360/12)/360;
+    end clockAngles;
+
+    function clockAngle(hh, mm: Integer) return Integer is
+        hh_angle, mm_angle, angle: Integer;
+    begin
+        clockAngles(hh, mm, hh_angle, mm_angle);
+        angle := abs(hh_angle - mm_angle);
+        if angle > 180 then
+            angle := 360 - angle;
+        end if;
+        return angle;
+    end clockAngle;
+
+    hhmm : String := CL.Argument(1);
+    hh, mm, angle: Integer;
+begin
+    parseTime(hhmm, hh, mm);
+    angle := clockAngle(hh, mm);
+    Put_Line(Integer'Image(angle));
+end ch_2;
diff --git a/challenge-120/paulo-custodio/awk/ch-1.awk b/challenge-120/paulo-custodio/awk/ch-1.awk
new file mode 100644
index 0000000000..8c3eeb3cb3
--- /dev/null
+++ b/challenge-120/paulo-custodio/awk/ch-1.awk
@@ -0,0 +1,44 @@
+#!/usr/bin/env gawk
+
+# Challenge 120
+#
+# TASK #1 - Swap Odd/Even bits
+# Submitted by: Mohammad S Anwar
+# You are given a positive integer $N less than or equal to 255.
+#
+# Write a script to swap the odd positioned bit with even positioned bit and
+# print the decimal equivalent of the new binary representation.
+#
+# Example
+# Input: $N = 101
+# Output: 154
+#
+# Binary representation of the given number is 01 10 01 01.
+# The new binary representation after the odd/even swap is 10 01 10 10.
+# The decimal equivalent of 10011010 is 154.
+#
+# Input: $N = 18
+# Output: 33
+#
+# Binary representation of the given number is 00 01 00 10.
+# The new binary representation after the odd/even swap is 00 10 00 01.
+# The decimal equivalent of 100001 is 33.
+
+function swap_bits(n,  out, bits) {
+    out = 0
+    bits = 0
+    while (n > 0) {
+        if (and(n, 1) != 0) { out = or(out, lshift(2, bits)) }
+        if (and(n, 2) != 0) { out = or(out, lshift(1, bits)) }
+        n = rshift(n, 2)
+        bits = bits + 2;
+    }
+    return out
+}
+
+BEGIN {
+    n = ARGV[1]
+    n = swap_bits(n)
+    print n
+    exit
+}
diff --git a/challenge-120/paulo-custodio/awk/ch-2.awk b/challenge-120/paulo-custodio/awk/ch-2.awk
new file mode 100644
index 0000000000..26a0840386
--- /dev/null
+++ b/challenge-120/paulo-custodio/awk/ch-2.awk
@@ -0,0 +1,43 @@
+#!/usr/bin/env gawk
+
+# Challenge 120
+#
+# TASK #2 › Clock Angle
+# Submitted by: Mohammad S Anwar
+# You are given time $T in the format hh:mm.
+#
+# Write a script to find the smaller angle formed by the hands of an analog
+# clock at a given time.
+#
+# HINT: A analog clock is divided up into 12 sectors. One sector represents 30
+# degree (360/12 = 30).
+#
+# Example
+# Input: $T = '03:10'
+# Output: 35 degree
+#
+# The distance between the 2 and the 3 on the clock is 30 degree.
+# For the 10 minutes i.e. 1/6 of an hour that have passed.
+# The hour hand has also moved 1/6 of the distance between the 3 and the 4,
+# which adds 5 degree (1/6 of 30).
+# The total measure of the angle is 35 degree.
+#
+# Input: $T = '04:00'
+# Output: 120 degree
+
+function abs(v) {return v < 0 ? -v : v}
+
+function clock_angle(hh, mm,  hh_angle, mm_angle, angle) {
+    mm_angle = mm/60*360
+    hh_angle = (hh % 12)/12*360 + mm_angle/360*1/12*360
+    angle = abs(hh_angle - mm_angle)
+    if (angle > 180) { angle = 360 - angle }
+    return angle
+}
+
+BEGIN {
+    split(ARGV[1], a, ":")
+    angle = clock_angle(a[1], a[2])
+    print angle
+    exit
+}
diff --git a/challenge-120/paulo-custodio/basic/ch-1.bas b/challenge-120/paulo-custodio/basic/ch-1.bas
new file mode 100644
index 0000000000..615a05bf31
--- /dev/null
+++ b/challenge-120/paulo-custodio/basic/ch-1.bas
@@ -0,0 +1,43 @@
+' Challenge 120
+'
+' TASK #1 - Swap Odd/Even bits
+' Submitted by: Mohammad S Anwar
+' You are given a positive integer $N less than or equal to 255.
+'
+' Write a script to swap the odd positioned bit with even positioned bit and
+' print the decimal equivalent of the new binary representation.
+'
+' Example
+' Input: $N = 101
+' Output: 154
+'
+' Binary representation of the given number is 01 10 01 01.
+' The new binary representation after the odd/even swap is 10 01 10 10.
+' The decimal equivalent of 10011010 is 154.
+'
+' Input: $N = 18
+' Output: 33
+'
+' Binary representation of the given number is 00 01 00 10.
+' The new binary representation after the odd/even swap is 00 10 00 01.
+' The decimal equivalent of 100001 is 33.
+
+function swapBits(n as Integer) as Integer
+    dim result as Integer, shift as Integer
+
+    shift = 1
+    do while n > 0
+        if n mod 2 <> 0 then result = result + 2*shift
+        n = n \ 2
+        if n mod 2 <> 0 then result = result + 1*shift
+        n = n \ 2
+        shift = shift*4
+    loop
+    swapBits = result
+end function
+
+dim n as Integer
+
+n = val(Command(1))
+n = swapBits(n)
+print n
diff --git a/challenge-120/paulo-custodio/basic/ch-2.bas b/challenge-120/paulo-custodio/basic/ch-2.bas
new file mode 100644
index 0000000000..a9e158fca9
--- /dev/null
+++ b/challenge-120/paulo-custodio/basic/ch-2.bas
@@ -0,0 +1,53 @@
+' Challenge 120
+'
+' TASK #2 - Clock Angle
+' Submitted by: Mohammad S Anwar
+' You are given time $T in the format hh:mm.
+'
+' Write a script to find the smaller angle formed by the hands of an analog
+' clock at a given time.
+'
+' HINT: A analog clock is divided up into 12 sectors. One sector represents 30
+' degree (360/12 = 30).
+'
+' Example
+' Input: $T = '03:10'
+' Output: 35 degree
+'
+' The distance between the 2 and the 3 on the clock is 30 degree.
+' For the 10 minutes i.e. 1/6 of an hour that have passed.
+' The hour hand has also moved 1/6 of the distance between the 3 and the 4,
+' which adds 5 degree (1/6 of 30).
+' The total measure of the angle is 35 degree.
+'
+' Input: $T = '04:00'
+' Output: 120 degree
+
+sub parseTime(hhmm as String, byref hh as Integer, byref mm as Integer)
+    dim p as Integer
+
+    p = InStr(hhmm, ":")
+    hh = val(left(hhmm, p-1))
+    mm = val(mid(hhmm, p+1))
+end sub
+
+sub clockAngles(hh as Integer, mm as Integer, _
+                byref hh_angle as Integer, byref mm_angle as Integer)
+    mm_angle = mm*360\60
+    hh_angle = (hh mod 12)*360\12 + mm_angle*(360\12)\360
+end sub
+
+function clockAngle(hh as Integer, mm as Integer) as Integer
+    dim hh_angle as Integer, mm_angle as Integer, angle as Integer
+
+    clockAngles hh, mm, hh_angle, mm_angle
+    angle = abs(hh_angle - mm_angle)
+    if angle > 180 then angle = 360 - angle
+    clockAngle = angle
+end function
+
+dim hh as Integer, mm as Integer, angle as Integer
+
+parseTime Command(1), hh, mm
+angle = clockAngle(hh, mm)
+print angle
diff --git a/challenge-120/paulo-custodio/bc/ch-1.bc b/challenge-120/paulo-custodio/bc/ch-1.bc
new file mode 100644
index 0000000000..b9eb2f6f6b
--- /dev/null
+++ b/challenge-120/paulo-custodio/bc/ch-1.bc
@@ -0,0 +1,48 @@
+#!/usr/bin/bc -ql
+
+/*
+Challenge 120
+
+TASK #1 - Swap Odd/Even bits
+Submitted by: Mohammad S Anwar
+You are given a positive integer $N less than or equal to 255.
+
+Write a script to swap the odd positioned bit with even positioned bit and
+print the decimal equivalent of the new binary representation.
+
+Example
+Input: $N = 101
+Output: 154
+
+Binary representation of the given number is 01 10 01 01.
+The new binary representation after the odd/even swap is 10 01 10 10.
+The decimal equivalent of 10011010 is 154.
+
+Input: $N = 18
+Output: 33
+
+Binary representation of the given number is 00 01 00 10.
+The new binary representation after the odd/even swap is 00 10 00 01.
+The decimal equivalent of 100001 is 33.
+*/
+
+scale = 0
+n = read()
+
+define swap_bits(n) {
+    auto out, shift
+    out = 0
+    shift = 1
+    while (n > 0) {
+        if ((n % 2) != 0) { out = out + (2 * shift); }
+        n = n / 2
+        if ((n % 2) != 0) { out = out + (1 * shift); }
+        n = n / 2
+        shift = shift * 4;
+    }
+    return out;
+}
+
+out = swap_bits(n)
+out
+quit
diff --git a/challenge-120/paulo-custodio/bc/ch-2.bc b/challenge-120/paulo-custodio/bc/ch-2.bc
new file mode 100644
index 0000000000..bd7ae9179e
--- /dev/null
+++ b/challenge-120/paulo-custodio/bc/ch-2.bc
@@ -0,0 +1,42 @@
+#!/usr/bin/bc -ql
+
+/*
+Challenge 120
+
+TASK #2 › Clock Angle
+Submitted by: Mohammad S Anwar
+You are given time $T in the format hh:mm.
+
+Write a script to find the smaller angle formed by the hands of an analog
+clock at a given time.
+
+HINT: A analog clock is divided up into 12 sectors. One sector represents 30
+degree (360/12 = 30).
+
+Example
+Input: $T = '03:10'
+Output: 35 degree
+
+The distance between the 2 and the 3 on the clock is 30 degree.
+For the 10 minutes i.e. 1/6 of an hour that have passed.
+The hour hand has also moved 1/6 of the distance between the 3 and the 4,
+which adds 5 degree (1/6 of 30).
+The total measure of the angle is 35 degree.
+
+Input: $T = '04:00'
+Output: 120 degree
+*/
+
+scale = 0
+hh = read()
+mm = read()
+
+mm_angle = mm*360/60
+hh_angle = (hh % 12)*360/12 + mm_angle*(360/12)/360
+
+if (hh_angle > mm_angle) {  angle = (hh_angle - mm_angle)
+} else {                    angle = (mm_angle - hh_angle) }
+if (angle > 180) {          angle = 360 - angle }
+
+angle
+quit
diff --git a/challenge-120/paulo-custodio/c/ch-1.c b/challenge-120/paulo-custodio/c/ch-1.c
new file mode 100644
index 0000000000..0bb2545b9c
--- /dev/null
+++ b/challenge-120/paulo-custodio/c/ch-1.c
@@ -0,0 +1,46 @@
+/*
+Challenge 120
+
+TASK #1 - Swap Odd/Even bits
+Submitted by: Mohammad S Anwar
+You are given a positive integer $N less than or equal to 255.
+
+Write a script to swap the odd positioned bit with even positioned bit and
+print the decimal equivalent of the new binary representation.
+
+Example
+Input: $N = 101
+Output: 154
+
+Binary representation of the given number is 01 10 01 01.
+The new binary representation after the odd/even swap is 10 01 10 10.
+The decimal equivalent of 10011010 is 154.
+
+Input: $N = 18
+Output: 33
+
+Binary representation of the given number is 00 01 00 10.
+The new binary representation after the odd/even swap is 00 10 00 01.
+The decimal equivalent of 100001 is 33.
+*/
+
+#include <stdio.h>
+#include <stdlib.h>
+
+int swap_bits(int n) {
+    int out = 0;
+    int shift = 0;
+    while (n > 0) {
+        if ((n & 1) != 0) { out |= 2 << shift; }
+        if ((n & 2) != 0) { out |= 1 << shift; }
+        n >>= 2;
+        shift += 2;
+    }
+    return out;
+}
+
+int main(int argc, char* argv[]) {
+    int n = 0;
+    if (argc == 2) n = atoi(argv[1]);
+    printf("%d\n", swap_bits(n));
+}
diff --git a/challenge-120/paulo-custodio/c/ch-2.c b/challenge-120/paulo-custodio/c/ch-2.c
new file mode 100644
index 0000000000..b6d92b0cab
--- /dev/null
+++ b/challenge-120/paulo-custodio/c/ch-2.c
@@ -0,0 +1,52 @@
+/*
+Challenge 120
+
+TASK #2 › Clock Angle
+Submitted by: Mohammad S Anwar
+You are given time $T in the format hh:mm.
+
+Write a script to find the smaller angle formed by the hands of an analog
+clock at a given time.
+
+HINT: A analog clock is divided up into 12 sectors. One sector represents 30
+degree (360/12 = 30).
+
+Example
+Input: $T = '03:10'
+Output: 35 degree
+
+The distance between the 2 and the 3 on the clock is 30 degree.
+For the 10 minutes i.e. 1/6 of an hour that have passed.
+The hour hand has also moved 1/6 of the distance between the 3 and the 4,
+which adds 5 degree (1/6 of 30).
+The total measure of the angle is 35 degree.
+
+Input: $T = '04:00'
+Output: 120 degree
+*/
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <stdbool.h>
+
+bool parse_time(const char* hhmm, int* hh, int* mm) {
+    *hh = *mm = 0;
+    if (sscanf(hhmm, "%d:%d", hh, mm) != 2) return false;
+    return true;
+}
+
+int clock_angle(int hh, int mm) {
+    int mm_angle = mm*360/60;
+    int hh_angle = (hh % 12)*360/12 + mm_angle/12;
+    int angle = abs(hh_angle - mm_angle);
+    if (angle > 180) { angle = 360 - angle; }
+    return angle;
+}
+
+int main(int argc, char* argv[]) {
+    if (argc != 2) return EXIT_FAILURE;
+    int hh, mm;
+    if (!parse_time(argv[1], &hh, &mm)) return EXIT_FAILURE;
+    int angle = clock_angle(hh, mm);
+    printf("%d\n", angle);
+}
diff --git a/challenge-120/paulo-custodio/cpp/ch-1.cpp b/challenge-120/paulo-custodio/cpp/ch-1.cpp
new file mode 100644
index 0000000000..7d25a239ec
--- /dev/null
+++ b/challenge-120/paulo-custodio/cpp/ch-1.cpp
@@ -0,0 +1,46 @@
+/*
+Challenge 120
+
+TASK #1 - Swap Odd/Even bits
+Submitted by: Mohammad S Anwar
+You are given a positive integer $N less than or equal to 255.
+
+Write a script to swap the odd positioned bit with even positioned bit and
+print the decimal equivalent of the new binary representation.
+
+Example
+Input: $N = 101
+Output: 154
+
+Binary representation of the given number is 01 10 01 01.
+The new binary representation after the odd/even swap is 10 01 10 10.
+The decimal equivalent of 10011010 is 154.
+
+Input: $N = 18
+Output: 33
+
+Binary representation of the given number is 00 01 00 10.
+The new binary representation after the odd/even swap is 00 10 00 01.
+The decimal equivalent of 100001 is 33.
+*/
+
+#include <iostream>
+using namespace std;
+
+int swap_bits(int n) {
+    int out = 0;
+    int shift = 0;
+    while (n > 0) {
+        if ((n & 1) != 0) { out |= 2 << shift; }
+        if ((n & 2) != 0) { out |= 1 << shift; }
+        n >>= 2;
+        shift += 2;
+    }
+    return out;
+}
+
+int main(int argc, char* argv[]) {
+    int n = 0;
+    if (argc == 2) n = atoi(argv[1]);
+    cout << swap_bits(n) << endl;
+}
diff --git a/challenge-120/paulo-custodio/cpp/ch-2.cpp b/challenge-120/paulo-custodio/cpp/ch-2.cpp
new file mode 100644
index 0000000000..c3325269ed
--- /dev/null
+++ b/challenge-120/paulo-custodio/cpp/ch-2.cpp
@@ -0,0 +1,57 @@
+/*
+Challenge 120
+
+TASK #2 › Clock Angle
+Submitted by: Mohammad S Anwar
+You are given time $T in the format hh:mm.
+
+Write a script to find the smaller angle formed by the hands of an analog
+clock at a given time.
+
+HINT: A analog clock is divided up into 12 sectors. One sector represents 30
+degree (360/12 = 30).
+
+Example
+Input: $T = '03:10'
+Output: 35 degree
+
+The distance between the 2 and the 3 on the clock is 30 degree.
+For the 10 minutes i.e. 1/6 of an hour that have passed.
+The hour hand has also moved 1/6 of the distance between the 3 and the 4,
+which adds 5 degree (1/6 of 30).
+The total measure of the angle is 35 degree.
+
+Input: $T = '04:00'
+Output: 120 degree
+*/
+
+#include <iostream>
+#include <sstream>
+#include <string>
+using namespace std;
+
+bool parse_time(const string& hhmm, int* hh, int* mm) {
+    *hh = *mm = 0;
+    istringstream iss(hhmm);
+    char c;
+    if (!(iss >> *hh)) return false;
+    if (!(iss >> c) || c != ':') return false;
+    if (!(iss >> *mm)) return false;
+    return true;
+}
+
+int clock_angle(int hh, int mm) {
+    int mm_angle = mm*360/60;
+    int hh_angle = (hh % 12)*360/12 + mm_angle/12;
+    int angle = abs(hh_angle - mm_angle);
+    if (angle > 180) { angle = 360 - angle; }
+    return angle;
+}
+
+int main(int argc, char* argv[]) {
+    if (argc != 2) return EXIT_FAILURE;
+    int hh, mm;
+    if (!parse_time(argv[1], &hh, &mm)) return EXIT_FAILURE;
+    int angle = clock_angle(hh, mm);
+    cout << angle << endl;
+}
diff --git a/challenge-120/paulo-custodio/d/ch_1.d b/challenge-120/paulo-custodio/d/ch_1.d
new file mode 100644
index 0000000000..6960784c83
--- /dev/null
+++ b/challenge-120/paulo-custodio/d/ch_1.d
@@ -0,0 +1,46 @@
+/*
+Challenge 120
+
+TASK #1 - Swap Odd/Even bits
+Submitted by: Mohammad S Anwar
+You are given a positive integer $N less than or equal to 255.
+
+Write a script to swap the odd positioned bit with even positioned bit and
+print the decimal equivalent of the new binary representation.
+
+Example
+Input: $N = 101
+Output: 154
+
+Binary representation of the given number is 01 10 01 01.
+The new binary representation after the odd/even swap is 10 01 10 10.
+The decimal equivalent of 10011010 is 154.
+
+Input: $N = 18
+Output: 33
+
+Binary representation of the given number is 00 01 00 10.
+The new binary representation after the odd/even swap is 00 10 00 01.
+The decimal equivalent of 100001 is 33.
+*/
+
+import std.stdio;
+import std.conv;
+
+int swap_bits(int n) {
+    int result = 0;
+    int shift = 0;
+    while (n > 0) {
+        if ((n & 1) != 0) { result |= 2 << shift; }
+        if ((n & 2) != 0) { result |= 1 << shift; }
+        n >>= 2;
+        shift += 2;
+    }
+    return result;
+}
+
+void main(string[] args) {
+    int n = 0;
+    if (args.length == 2) n = to!int(args[1]);
+    writeln(swap_bits(n));
+}
diff --git a/challenge-120/paulo-custodio/d/ch_2.d b/challenge-120/paulo-custodio/d/ch_2.d
new file mode 100644
index 0000000000..faf4b0f163
--- /dev/null
+++ b/challenge-120/paulo-custodio/d/ch_2.d
@@ -0,0 +1,55 @@
+/*
+Challenge 120
+
+TASK #2 - Clock Angle
+Submitted by: Mohammad S Anwar
+You are given time $T in the format hh:mm.
+
+Write a script to find the smaller angle formed by the hands of an analog
+clock at a given time.
+
+HINT: A analog clock is divided up into 12 sectors. One sector represents 30
+degree (360/12 = 30).
+
+Example
+Input: $T = '03:10'
+Output: 35 degree
+
+The distance between the 2 and the 3 on the clock is 30 degree.
+For the 10 minutes i.e. 1/6 of an hour that have passed.
+The hour hand has also moved 1/6 of the distance between the 3 and the 4,
+which adds 5 degree (1/6 of 30).
+The total measure of the angle is 35 degree.
+
+Input: $T = '04:00'
+Output: 120 degree
+*/
+
+import std.stdio;
+import std.conv;
+import std.array : split;
+import std.math : abs;
+
+bool parse_time(string hhmm, int* hh, int* mm) {
+    string[] a = hhmm.split(":");
+    if (a.length != 2) return false;
+    *hh = to!int(a[0]);
+    *mm = to!int(a[1]);
+    return true;
+}
+
+int clock_angle(int hh, int mm) {
+    int mm_angle = mm*360/60;
+    int hh_angle = (hh % 12)*360/12 + mm_angle/12;
+    int angle = abs(hh_angle - mm_angle);
+    if (angle > 180) { angle = 360 - angle; }
+    return angle;
+}
+
+void main(string[] args) {
+    if (args.length != 2) return;
+    int hh, mm;
+    if (!parse_time(args[1], &hh, &mm)) return;
+    int angle = clock_angle(hh, mm);
+    writeln(angle);
+}
diff --git a/challenge-120/paulo-custodio/forth/ch-1.fs b/challenge-120/paulo-custodio/forth/ch-1.fs
new file mode 100644
index 0000000000..4bc5f6e65b
--- /dev/null
+++ b/challenge-120/paulo-custodio/forth/ch-1.fs
@@ -0,0 +1,38 @@
+#! /usr/bin/env gforth
+
+\ Challenge 120
+\
+\ TASK #1 - Swap Odd/Even bits
+\ Submitted by: Mohammad S Anwar
+\ You are given a positive integer $N less than or equal to 255.
+\
+\ Write a script to swap the odd positioned bit with even positioned bit and
+\ print the decimal equivalent of the new binary representation.
+\
+\ Example
+\ Input: $N = 101
+\ Output: 154
+\
+\ Binary representation of the given number is 01 10 01 01.
+\ The new binary representation after the odd/even swap is 10 01 10 10.
+\ The decimal equivalent of 10011010 is 154.
+\
+\ Input: $N = 18
+\ Output: 33
+\
+\ Binary representation of the given number is 00 01 00 10.
+\ The new binary representation after the odd/even swap is 00 10 00 01.
+\ The decimal equivalent of 100001 is 33.
+
+: swap_bits     { n -- n }
+    0 0         { out shift }
+    BEGIN n 0> WHILE
+        n 1 AND IF  2 shift LSHIFT out OR  TO out  THEN
+        n 2 AND IF  1 shift LSHIFT out OR  TO out  THEN
+        n 2 RSHIFT  TO n
+        shift 2 +  TO shift
+    REPEAT
+    out ;
+
+NEXT-ARG S>NUMBER? 0= THROW DROP
+swap_bits . CR BYE
diff --git a/challenge-120/paulo-custodio/forth/ch-2.fs b/challenge-120/paulo-custodio/forth/ch-2.fs
new file mode 100644
index 0000000000..078dbfbc19
--- /dev/null
+++ b/challenge-120/paulo-custodio/forth/ch-2.fs
@@ -0,0 +1,64 @@
+#! /usr/bin/env gforth
+
+\ Challenge 120
+\
+\ TASK #2 › Clock Angle
+\ Submitted by: Mohammad S Anwar
+\ You are given time $T in the format hh:mm.
+\
+\ Write a script to find the smaller angle formed by the hands of an analog
+\ clock at a given time.
+\
+\ HINT: A analog clock is divided up into 12 sectors. One sector represents 30
+\ degree (360/12 = 30).
+\
+\ Example
+\ Input: $T = '03:10'
+\ Output: 35 degree
+\
+\ The distance between the 2 and the 3 on the clock is 30 degree.
+\ For the 10 minutes i.e. 1/6 of an hour that have passed.
+\ The hour hand has also moved 1/6 of the distance between the 3 and the 4,
+\ which adds 5 degree (1/6 of 30).
+\ The total measure of the angle is 35 degree.
+\
+\ Input: $T = '04:00'
+\ Output: 120 degree
+
+: parse_time    ( addr len -- hh mm )
+    0. 2SWAP                ( 0  0 addr len )
+    >NUMBER                 ( hh 0 addr len )   \ convert hh
+    1 /STRING                                   \ skip ':'
+    0 -ROT                  ( hh 0 0 addr len )
+    >NUMBER                 ( hh mm 0 addr len )
+    2DROP DROP ;
+
+: clock_angle   { hh mm -- angle }
+    mm 360 60 */                            { mm_angle }
+    hh 12 MOD 360 12 */  mm_angle 12 /  +   { hh_angle }
+    hh_angle mm_angle - ABS                 ( angle )
+    DUP 180 > IF 360 SWAP - THEN ;
+
+NEXT-ARG parse_time clock_angle . CR BYE
+
+\ #include <stdio.h>
+\ #include <stdlib.h>
+\ #include <stdbool.h>
+\
+\
+\ int clock_angle(int hh, int mm) {
+\   int mm_angle = mm*360/60;
+\   int hh_angle = (hh % 12)*360/12 + mm_angle/12;
+\     int angle = abs(hh_angle - mm_angle);
+\     if (angle > 180) { angle = 360 - angle; }
+\     return angle;
+\ }
+\
+\ int main(int argc, char* argv[]) {
+\     if (argc != 2) return EXIT_FAILURE;
+\     int hh, mm;
+\     if (!parse_time(argv[1], &hh, &mm)) return EXIT_FAILURE;
+\     int angle = clock_angle(hh, mm);
+\     printf("%d\n", angle);
+\ }
+\
diff --git a/challenge-120/paulo-custodio/lua/ch-1.lua b/challenge-120/paulo-custodio/lua/ch-1.lua
new file mode 100644
index 0000000000..a4290510be
--- /dev/null
+++ b/challenge-120/paulo-custodio/lua/ch-1.lua
@@ -0,0 +1,44 @@
+#!/usr/bin/env lua
+
+--[[
+Challenge 120
+
+TASK #1 - Swap Odd/Even bits<