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-rw-r--r--challenge-101/paulo-custodio/ada/ch_1.adb30
-rw-r--r--challenge-101/paulo-custodio/ada/ch_2.adb22
-rw-r--r--challenge-101/paulo-custodio/basic/ch-1.bas28
-rw-r--r--challenge-101/paulo-custodio/basic/ch-2.bas8
-rw-r--r--challenge-101/paulo-custodio/c/ch-1.c8
-rw-r--r--challenge-101/paulo-custodio/cpp/ch-1.cpp26
-rw-r--r--challenge-101/paulo-custodio/lua/ch-1.lua18
-rw-r--r--challenge-101/paulo-custodio/lua/ch-2.lua10
-rw-r--r--challenge-101/paulo-custodio/perl/ch-1.pl18
-rw-r--r--challenge-101/paulo-custodio/perl/ch-2.pl10
-rw-r--r--challenge-101/paulo-custodio/python/ch-1.py24
-rw-r--r--challenge-101/paulo-custodio/python/ch-2.py10
-rw-r--r--challenge-101/paulo-custodio/t/test-2.yaml2
13 files changed, 107 insertions, 107 deletions
diff --git a/challenge-101/paulo-custodio/ada/ch_1.adb b/challenge-101/paulo-custodio/ada/ch_1.adb
index fd9f20df4c..f1c4071cdb 100644
--- a/challenge-101/paulo-custodio/ada/ch_1.adb
+++ b/challenge-101/paulo-custodio/ada/ch_1.adb
@@ -1,19 +1,19 @@
-- Challenge 101
---
+--
-- TASK #1 › Pack a Spiral
-- Submitted by: Stuart Little
---
+--
-- -- You are given an array @A of items (integers say, but they can be anything).
---
--- Your task is to pack that array into an MxN matrix spirally counterclockwise,
+--
+-- Your task is to pack that array into an MxN matrix spirally counterclockwise,
-- as tightly as possible.
---
--- ‘Tightly’ means the absolute value |M-N| of the difference has to be as small
+--
+-- ‘Tightly’ means the absolute value |M-N| of the difference has to be as small
-- as possible.
with Ada.Command_Line;
with Ada.Strings.Fixed; use Ada.Strings.Fixed;
-with Ada.Text_IO; use Ada.Text_IO;
+with Ada.Text_IO; use Ada.Text_IO;
procedure ch_1 is
-- command line arguments
@@ -27,7 +27,7 @@ procedure ch_1 is
number_rect : array(1 .. CL.Argument_Count, 1 .. CL.Argument_Count) of Integer;
width, height : Integer;
field_width : Integer := 1;
-
+
-- collect from command line
procedure collect_numbers is
begin
@@ -38,7 +38,7 @@ procedure ch_1 is
end if;
end loop;
end collect_numbers;
-
+
-- get smallest rectangle
procedure smallest_rect(width, height : out Integer) is
begin
@@ -63,7 +63,7 @@ procedure ch_1 is
number_rect(i, j) := -1;
end loop;
end loop;
-
+
idx := number_list'First;
row := height;
col := 1;
@@ -77,7 +77,7 @@ procedure ch_1 is
end loop;
col := col - 1;
row := row - 1;
-
+
-- go North
while row >= 1 loop
exit when number_rect(row, col) >= 0;
@@ -87,7 +87,7 @@ procedure ch_1 is
end loop;
row := row + 1;
col := col - 1;
-
+
-- go West
while col >= 1 loop
exit when number_rect(row, col) >= 0;
@@ -97,7 +97,7 @@ procedure ch_1 is
end loop;
col := col + 1;
row := row + 1;
-
+
-- go South
while row <= height loop
exit when number_rect(row, col) >= 0;
@@ -109,7 +109,7 @@ procedure ch_1 is
col := col + 1;
end loop;
end pack_numbers;
-
+
-- print the rectangle
procedure print_rect(width, heigth : Integer) is
begin
@@ -120,7 +120,7 @@ procedure ch_1 is
Put_Line("");
end loop;
end print_rect;
-
+
begin
collect_numbers;
smallest_rect(width, height);
diff --git a/challenge-101/paulo-custodio/ada/ch_2.adb b/challenge-101/paulo-custodio/ada/ch_2.adb
index f52e3faa33..be2268e7bc 100644
--- a/challenge-101/paulo-custodio/ada/ch_2.adb
+++ b/challenge-101/paulo-custodio/ada/ch_2.adb
@@ -2,17 +2,17 @@
--
-- TASK #2 › Origin-containing Triangle
-- Submitted by: Stuart Little
--- You are given three points in the plane, as a list of six co-ordinates:
+-- You are given three points in the plane, as a list of six co-ordinates:
-- A=(x1,y1), B=(x2,y2) and C=(x3,y3).
---
--- Write a script to find out if the triangle formed by the given three
+--
+-- Write a script to find out if the triangle formed by the given three
-- co-ordinates contain origin (0,0).
---
+--
-- Print 1 if found otherwise 0.
with Ada.Command_Line;
with Ada.Strings.Fixed; use Ada.Strings.Fixed;
-with Ada.Text_IO; use Ada.Text_IO;
+with Ada.Text_IO; use Ada.Text_IO;
procedure ch_2 is
-- command line arguments
@@ -24,21 +24,21 @@ procedure ch_2 is
type point is record
x, y : Float;
end record;
-
+
P : array (1 .. 3) of point;
P0 : point;
function point_in_triangle(P0, P1, P2, P3 : Point) return Integer is
-
+
function sign(P1, P2, P3 : Point) return Float is
begin
- return (P1.x - P3.x) * (P2.y - P3.y)
+ return (P1.x - P3.x) * (P2.y - P3.y)
- (P2.x - P3.x) * (P1.y - P3.y);
end sign;
-
+
d : array(1..3) of Float;
has_neg, has_pos : Boolean;
-
+
begin
d(1) := sign(P0, P1, P2);
d(2) := sign(P0, P2, P3);
@@ -53,7 +53,7 @@ procedure ch_2 is
return 0;
end if;
end point_in_triangle;
-
+
begin
-- read points
for i in 1 .. 3 loop
diff --git a/challenge-101/paulo-custodio/basic/ch-1.bas b/challenge-101/paulo-custodio/basic/ch-1.bas
index f958245d3e..e68f212358 100644
--- a/challenge-101/paulo-custodio/basic/ch-1.bas
+++ b/challenge-101/paulo-custodio/basic/ch-1.bas
@@ -1,14 +1,14 @@
' Challenge 101
-'
+'
' TASK #1 › Pack a Spiral
' Submitted by: Stuart Little
-'
+'
' You are given an array @A of items (integers say, but they can be anything).
-'
-' Your task is to pack that array into an MxN matrix spirally counterclockwise,
+'
+' Your task is to pack that array into an MxN matrix spirally counterclockwise,
' as tightly as possible.
-'
-' ‘Tightly’ means the absolute value |M-N| of the difference has to be as small
+'
+' ‘Tightly’ means the absolute value |M-N| of the difference has to be as small
' as possible.
const max_numbers as integer = 100
@@ -19,12 +19,12 @@ dim shared num_width as integer, rect_width as integer, rect_height as integer
dim shared num_numbers as integer
' collect from command line
-sub collect_numbers()
+sub collect_numbers()
dim i as integer
i = 1
do while command(i)<>""
number_list(i) = val(command(i))
- if len(command(i))>num_width then
+ if len(command(i))>num_width then
num_width = len(command(i))
end if
num_numbers = i
@@ -56,13 +56,13 @@ sub pack_numbers(rect_width as integer, rect_height as Integer)
number_rect(row, col) = -1
next
next
-
+
idx = 1
row = rect_height
col = 1
do while idx <= num_numbers
' go East
- do while col <= rect_width
+ do while col <= rect_width
if number_rect(row, col) >= 0 then exit do
number_rect(row, col) = number_list(idx)
idx = idx + 1
@@ -70,7 +70,7 @@ sub pack_numbers(rect_width as integer, rect_height as Integer)
loop
col = col - 1
row = row - 1
-
+
' go North
do while row >= 1
if number_rect(row, col) >= 0 then exit do
@@ -80,7 +80,7 @@ sub pack_numbers(rect_width as integer, rect_height as Integer)
loop
row = row + 1
col = col - 1
-
+
' go West
do while col >= 1
if number_rect(row, col) >= 0 then exit do
@@ -90,9 +90,9 @@ sub pack_numbers(rect_width as integer, rect_height as Integer)
loop
col = col + 1
row = row + 1
-
+
' go South
- do while row <= rect_height
+ do while row <= rect_height
if number_rect(row, col) >= 0 then exit do
number_rect(row, col) = number_list(idx)
idx = idx + 1
diff --git a/challenge-101/paulo-custodio/basic/ch-2.bas b/challenge-101/paulo-custodio/basic/ch-2.bas
index 30c62f9fb6..68b0ad627b 100644
--- a/challenge-101/paulo-custodio/basic/ch-2.bas
+++ b/challenge-101/paulo-custodio/basic/ch-2.bas
@@ -2,12 +2,12 @@
'
' TASK #2 › Origin-containing Triangle
' Submitted by: Stuart Little
-' You are given three points in the plane, as a list of six co-ordinates:
+' You are given three points in the plane, as a list of six co-ordinates:
' A=(x1,y1), B=(x2,y2) and C=(x3,y3).
-'
-' Write a script to find out if the triangle formed by the given three
+'
+' Write a script to find out if the triangle formed by the given three
' co-ordinates contain origin (0,0).
-'
+'
' Print 1 if found otherwise 0.
type Point
diff --git a/challenge-101/paulo-custodio/c/ch-1.c b/challenge-101/paulo-custodio/c/ch-1.c
index 066f1b628b..b2a166312b 100644
--- a/challenge-101/paulo-custodio/c/ch-1.c
+++ b/challenge-101/paulo-custodio/c/ch-1.c
@@ -6,10 +6,10 @@ Submitted by: Stuart Little
You are given an array @A of items (integers say, but they can be anything).
-Your task is to pack that array into an MxN matrix spirally counterclockwise,
+Your task is to pack that array into an MxN matrix spirally counterclockwise,
as tightly as possible.
-‘Tightly’ means the absolute value |M-N| of the difference has to be as small
+‘Tightly’ means the absolute value |M-N| of the difference has to be as small
as possible.
*/
@@ -60,7 +60,7 @@ int** make_empty_rect(int width, int height) {
}
void free_rect(int**rect, int height) {
- for (int r = 0; r < height; r++)
+ for (int r = 0; r < height; r++)
free(rect[r]);
free(rect);
}
@@ -124,6 +124,6 @@ int main(int argc, char* argv[]) {
fputs("Usage: ch-1 N...", stderr);
exit(EXIT_FAILURE);
}
- else
+ else
spiral(--argc, ++argv);
}
diff --git a/challenge-101/paulo-custodio/cpp/ch-1.cpp b/challenge-101/paulo-custodio/cpp/ch-1.cpp
index 9d81d1916e..c16fae2ad2 100644
--- a/challenge-101/paulo-custodio/cpp/ch-1.cpp
+++ b/challenge-101/paulo-custodio/cpp/ch-1.cpp
@@ -6,10 +6,10 @@ Submitted by: Stuart Little
You are given an array @A of items (integers say, but they can be anything).
-Your task is to pack that array into an MxN matrix spirally counterclockwise,
+Your task is to pack that array into an MxN matrix spirally counterclockwise,
as tightly as possible.
-‘Tightly’ means the absolute value |M-N| of the difference has to be as small
+‘Tightly’ means the absolute value |M-N| of the difference has to be as small
as possible.
*/
@@ -20,7 +20,7 @@ as possible.
int numbers_width;
std::vector<int> collect_numbers(int argc, char* argv[]) {
- std::vector<int> numbers;
+ std::vector<int> numbers;
for (int i = 0; i < argc; i++) {
numbers.push_back(atoi(argv[i]));
if (numbers[i] >= 1000 && numbers_width < 4) numbers_width = 4;
@@ -43,20 +43,20 @@ void smallest_rect(int n, int& width, int& height) {
}
std::vector<std::vector<int>> make_empty_rect(int width, int height) {
- std::vector<std::vector<int>> rect;
+ std::vector<std::vector<int>> rect;
for (int r = 0; r < height; r++) {
- rect.push_back({});
+ rect.push_back({});
for (int c = 0; c < width; c++)
rect[r].push_back(-1);
}
return rect;
}
-void fill_spiral(std::vector<std::vector<int>>& rect,
- std::vector<int>& numbers) {
- int width = static_cast<int>(rect[0].size());
- int height = static_cast<int>(rect.size());
- int count = static_cast<int>(numbers.size());
+void fill_spiral(std::vector<std::vector<int>>& rect,
+ std::vector<int>& numbers) {
+ int width = static_cast<int>(rect[0].size());
+ int height = static_cast<int>(rect.size());
+ int count = static_cast<int>(numbers.size());
int i = 0;
int r = height - 1;
int c = 0;
@@ -92,8 +92,8 @@ void fill_spiral(std::vector<std::vector<int>>& rect,
}
void print_spiral(std::vector<std::vector<int>>& rect) {
- int width = static_cast<int>(rect[0].size());
- int height = static_cast<int>(rect.size());
+ int width = static_cast<int>(rect[0].size());
+ int height = static_cast<int>(rect.size());
for (int r = 0; r < height; r++) {
for (int c = 0; c < width; c++)
std::cout << std::setw(numbers_width + 1) << rect[r][c];
@@ -115,6 +115,6 @@ int main(int argc, char* argv[]) {
std::cerr << "Usage: ch-1 N..." << std::endl;
exit(EXIT_FAILURE);
}
- else
+ else
spiral(--argc, ++argv);
}
diff --git a/challenge-101/paulo-custodio/lua/ch-1.lua b/challenge-101/paulo-custodio/lua/ch-1.lua
index 7eac6f60f5..602143d1ee 100644
--- a/challenge-101/paulo-custodio/lua/ch-1.lua
+++ b/challenge-101/paulo-custodio/lua/ch-1.lua
@@ -8,10 +8,10 @@ Submitted by: Stuart Little
You are given an array @A of items (integers say, but they can be anything).
-Your task is to pack that array into an MxN matrix spirally counterclockwise,
+Your task is to pack that array into an MxN matrix spirally counterclockwise,
as tightly as possible.
-‘Tightly’ means the absolute value |M-N| of the difference has to be as small
+‘Tightly’ means the absolute value |M-N| of the difference has to be as small
as possible.
--]]
@@ -44,7 +44,7 @@ function build_empty_rectangle(m, n)
local rect = {}
for r=1, m do
rect[r] = {}
- for c=1, n do
+ for c=1, n do
rect[r][c] = ""
end
end
@@ -67,7 +67,7 @@ function spiral(numbers)
i = i + 1
c = c + 1
end
- c = c - 1
+ c = c - 1
r = r - 1
-- go North
while r >= 1 do
@@ -77,7 +77,7 @@ function spiral(numbers)
i = i + 1
r = r - 1
end
- r = r + 1
+ r = r + 1
c = c - 1
-- go West
while c >= 1 do
@@ -87,7 +87,7 @@ function spiral(numbers)
i = i + 1
c = c - 1
end
- c = c + 1
+ c = c + 1
r = r + 1
-- go South
while r <= m do
@@ -97,13 +97,13 @@ function spiral(numbers)
i = i + 1
r = r + 1
end
- r = r - 1
+ r = r - 1
c = c + 1
end
-
+
-- print result
for r=1, m do
- for c=1, n do
+ for c=1, n do
io.write(rect[r][c])
end
io.write("\n")
diff --git a/challenge-101/paulo-custodio/lua/ch-2.lua b/challenge-101/paulo-custodio/lua/ch-2.lua
index 47a12b657e..54b02aa59d 100644
--- a/challenge-101/paulo-custodio/lua/ch-2.lua
+++ b/challenge-101/paulo-custodio/lua/ch-2.lua
@@ -5,10 +5,10 @@ Challenge 101
TASK #2 › Origin-containing Triangle
Submitted by: Stuart Little
-You are given three points in the plane, as a list of six co-ordinates:
+You are given three points in the plane, as a list of six co-ordinates:
A=(x1,y1), B=(x2,y2) and C=(x3,y3).
-Write a script to find out if the triangle formed by the given three
+Write a script to find out if the triangle formed by the given three
co-ordinates contain origin (0,0).
Print 1 if found otherwise 0.
@@ -24,12 +24,12 @@ function point_in_triangle(xp,yp, x1,y1,x2,y2,x3,y3)
local d3 = sign(xp,yp, x3,y3, x1,y1)
local has_neg
- if (d1 < 0) or (d2 < 0) or (d3 < 0) then
+ if (d1 < 0) or (d2 < 0) or (d3 < 0) then
has_neg = true
else
has_neg = false
end
-
+
local has_neg = (d1 < 0) or (d2 < 0) or (d3 < 0)
local has_pos = (d1 > 0) or (d2 > 0) or (d3 > 0)
@@ -40,7 +40,7 @@ function point_in_triangle(xp,yp, x1,y1,x2,y2,x3,y3)
end
end
-io.write(point_in_triangle(0,0,
+io.write(point_in_triangle(0,0,
tonumber(arg[1]), tonumber(arg[2]),
tonumber(arg[3]), tonumber(arg[4]),
tonumber(arg[5]), tonumber(arg[6])), "\n")
diff --git a/challenge-101/paulo-custodio/perl/ch-1.pl b/challenge-101/paulo-custodio/perl/ch-1.pl
index b203d06266..f6b8f3097e 100644
--- a/challenge-101/paulo-custodio/perl/ch-1.pl
+++ b/challenge-101/paulo-custodio/perl/ch-1.pl
@@ -1,16 +1,16 @@
#!/usr/bin/perl
# Challenge 101
-#
+#
# TASK #1 › Pack a Spiral
# Submitted by: Stuart Little
-#
+#
# You are given an array @A of items (integers say, but they can be anything).
-#
-# Your task is to pack that array into an MxN matrix spirally counterclockwise,
+#
+# Your task is to pack that array into an MxN matrix spirally counterclockwise,
# as tightly as possible.
-#
-# ‘Tightly’ means the absolute value |M-N| of the difference has to be as small
+#
+# ‘Tightly’ means the absolute value |M-N| of the difference has to be as small
# as possible.
use strict;
@@ -44,13 +44,13 @@ sub smallest_rect {
sub spiral {
my($m, $n, @list) = @_;
-
+
# find max width of elements
my $width = 1;
for (@list) {
$width = length($_) if length($_) > $width;
}
-
+
# build rectangle
my @rect;
my $r = $m; my $c = 1;
@@ -80,6 +80,6 @@ sub spiral {
}
$r--; $c++;
}
-
+
return @rect;
}
diff --git a/challenge-101/paulo-custodio/perl/ch-2.pl b/challenge-101/paulo-custodio/perl/ch-2.pl
index b5e7a71f64..d9efb89857 100644
--- a/challenge-101/paulo-custodio/perl/ch-2.pl
+++ b/challenge-101/paulo-custodio/perl/ch-2.pl
@@ -1,15 +1,15 @@
#!/usr/bin/perl
# Challenge 101
-#
+#
# TASK #2 › Origin-containing Triangle
# Submitted by: Stuart Little
-# You are given three points in the plane, as a list of six co-ordinates:
+# You are given three points in the plane, as a list of six co-ordinates:
# A=(x1,y1), B=(x2,y2) and C=(x3,y3).
-#
-# Write a script to find out if the triangle formed by the given three
+#
+# Write a script to find out if the triangle formed by the given three
# co-ordinates contain origin (0,0).
-#
+#
# Print 1 if found otherwise 0.
use strict;
diff --git a/challenge-101/paulo-custodio/python/ch-1.py b/challenge-101/paulo-custodio/python/ch-1.py
index 57d08d0847..3f5fd5f79e 100644
--- a/challenge-101/paulo-custodio/python/ch-1.py
+++ b/challenge-101/paulo-custodio/python/ch-1.py
@@ -1,23 +1,23 @@
#!/usr/bin/env python
# Challenge 101
-#
+#
# TASK #1 > Pack a Spiral
# Submitted by: Stuart Little
-#
+#
# You are given an array @A of items (integers say, but they can be anything).
-#
-# Your task is to pack that array into an MxN matrix spirally counterclockwise,
+#
+# Your task is to pack that array into an MxN matrix spirally counterclockwise,
# as tightly as possible.
-#
-# 'Tightly' means the absolute value |M-N| of the difference has to be as small
+#
+# 'Tightly' means the absolute value |M-N| of the difference has to be as small
# as possible.
import sys
import math
def spiral(numbers):
-
+
# find max width of elements, convert to int
def max_width(numbers):
num_width = 1
@@ -46,7 +46,7 @@ def spiral(numbers):
# build spiral rectangle
num_width = max_width(numbers)
m, n = build_rect(numbers)
-
+
r, c = m, 1
i = 0
while (i < len(numbers)):
@@ -57,7 +57,7 @@ def spiral(numbers):
rect[r][c] = ("{:"+str(num_width+1)+"d}").format(numbers[i])
i += 1
c += 1
- c -= 1
+ c -= 1
r -= 1
# go North
while (r >= 1):
@@ -66,7 +66,7 @@ def spiral(numbers):
rect[r][c] = ("{:"+str(num_width+1)+"d}").format(numbers[i])
i += 1
r -= 1
- r += 1
+ r += 1
c -= 1
# go West
while (c >= 1):
@@ -75,7 +75,7 @@ def spiral(numbers):
rect[r][c] = ("{:"+str(num_width+1)+"d}").format(numbers[i])
i += 1
c -= 1
- c += 1
+ c += 1
r += 1
# go South
while (r <= m):
@@ -84,7 +84,7 @@ def spiral(numbers):
rect[r][c] = ("{:"+str(num_width+1)+"d}").format(numbers[i])
i += 1
r += 1
- r -= 1
+ r -= 1
c += 1
for r in range (1, m+1):
diff --git a/challenge-101/paulo-custodio/python/ch-2.py b/challenge-101/paulo-custodio/python/ch-2.py
index 0e510d9eca..b706fb9636 100644
--- a/challenge-101/paulo-custodio/python/ch-2.py
+++ b/challenge-101/paulo-custodio/python/ch-2.py
@@ -1,15 +1,15 @@
#! /usr/bin/env python
# Challenge 101
-#
+#
# TASK #2 > Origin-containing Triangle
# Submitted by: Stuart Little
-# You are given three points in the plane, as a list of six co-ordinates:
+# You are given three points in the plane, as a list of six co-ordinates:
# A=(x1,y1), B=(x2,y2) and C=(x3,y3).
-#
-# Write a script to find out if the triangle formed by the given three
+#
+# Write a script to find out if the triangle formed by the given three
# co-ordinates contain origin (0,0).
-#
+#
# Print 1 if found otherwise 0.
import sys
diff --git a/challenge-101/paulo-custodio/t/test-2.yaml b/challenge-101/paulo-custodio/t/test-2.yaml
index a90cb7c6ef..c4e3e91391 100644
--- a/challenge-101/paulo-custodio/t/test-2.yaml
+++ b/challenge-101/paulo-custodio/t/test-2.yaml
@@ -1,6 +1,6 @@
- setup:
cleanup:
- args: 0 1 1 0 2 2
+ args: 0 1 1 0 2 2
input:
output: 0
- setup:
generated by cgit (git 2.34.1) at 2026-07-03 20:06:47 +0000