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-rw-r--r--challenge-243/paulo-custodio/Makefile2
-rw-r--r--challenge-243/paulo-custodio/c/ch-1.c64
-rw-r--r--challenge-243/paulo-custodio/c/ch-2.c66
-rw-r--r--challenge-243/paulo-custodio/c/utarray.h252
-rw-r--r--challenge-243/paulo-custodio/cpp/ch-1.cpp61
-rw-r--r--challenge-243/paulo-custodio/cpp/ch-2.cpp63
-rw-r--r--challenge-243/paulo-custodio/perl/ch-1.pl45
-rw-r--r--challenge-243/paulo-custodio/perl/ch-2.pl48
-rw-r--r--challenge-243/paulo-custodio/t/test-1.yaml10
-rw-r--r--challenge-243/paulo-custodio/t/test-2.yaml10
10 files changed, 621 insertions, 0 deletions
diff --git a/challenge-243/paulo-custodio/Makefile b/challenge-243/paulo-custodio/Makefile
new file mode 100644
index 0000000000..c3c762d746
--- /dev/null
+++ b/challenge-243/paulo-custodio/Makefile
@@ -0,0 +1,2 @@
+all:
+ perl ../../challenge-001/paulo-custodio/test.pl
diff --git a/challenge-243/paulo-custodio/c/ch-1.c b/challenge-243/paulo-custodio/c/ch-1.c
new file mode 100644
index 0000000000..39ffc3dc91
--- /dev/null
+++ b/challenge-243/paulo-custodio/c/ch-1.c
@@ -0,0 +1,64 @@
+/*
+Challenge 243
+
+Task 1: Reverse Pairs
+Submitted by: Mohammad S Anwar
+
+You are given an array of integers.
+
+Write a script to return the number of reverse pairs in the given array.
+
+A reverse pair is a pair (i, j) where: a) 0 <= i < j < nums.length and b)
+nums[i] > 2 * nums[j].
+Example 1
+
+Input: @nums = (1, 3, 2, 3, 1)
+Output: 2
+
+(1, 4) => nums[1] = 3, nums[4] = 1, 3 > 2 * 1
+(3, 4) => nums[3] = 3, nums[4] = 1, 3 > 2 * 1
+
+Example 2
+
+Input: @nums = (2, 4, 3, 5, 1)
+Output: 3
+
+(1, 4) => nums[1] = 4, nums[4] = 1, 4 > 2 * 1
+(2, 4) => nums[2] = 3, nums[4] = 1, 3 > 2 * 1
+(3, 4) => nums[3] = 5, nums[4] = 1, 5 > 2 * 1
+*/
+
+#include "utarray.h"
+#include <stdio.h>
+#include <stdlib.h>
+
+int num_reverse_pairs(UT_array* nums) {
+ int count = 0;
+ for (size_t i = 0; i < utarray_len(nums)-1; i++) {
+ for (size_t j = i+1; j < utarray_len(nums); j++) {
+ if (*(int*)utarray_eltptr(nums, i) > 2 * *(int*)utarray_eltptr(nums, j))
+ count++;
+ }
+ }
+ return count;
+}
+
+int main(int argc, char* argv[]) {
+ if (argc < 3) {
+ fputs("Usage: ch-1 n n n ...\n", stderr);
+ exit(EXIT_FAILURE);
+ }
+
+ UT_array* nums;
+ utarray_new(nums, &ut_int_icd);
+
+ for (int i = 1; i < argc; i++) {
+ int n = atoi(argv[i]);
+ utarray_push_back(nums, &n);
+ }
+
+ int count = num_reverse_pairs(nums);
+ printf("%d\n", count);
+
+ utarray_free(nums);
+}
diff --git a/challenge-243/paulo-custodio/c/ch-2.c b/challenge-243/paulo-custodio/c/ch-2.c
new file mode 100644
index 0000000000..47974d633c
--- /dev/null
+++ b/challenge-243/paulo-custodio/c/ch-2.c
@@ -0,0 +1,66 @@
+/*
+Challenge 243
+
+Task 2: Floor Sum
+Submitted by: Mohammad S Anwar
+
+You are given an array of positive integers (>=1).
+
+Write a script to return the sum of floor(nums[i] / nums[j]) where
+0 <= i,j < nums.length. The floor() function returns the integer part of the
+division.
+
+Example 1
+
+Input: @nums = (2, 5, 9)
+Output: 10
+
+floor(2 / 5) = 0
+floor(2 / 9) = 0
+floor(5 / 9) = 0
+floor(2 / 2) = 1
+floor(5 / 5) = 1
+floor(9 / 9) = 1
+floor(5 / 2) = 2
+floor(9 / 2) = 4
+floor(9 / 5) = 1
+
+Example 2
+
+Input: @nums = (7, 7, 7, 7, 7, 7, 7)
+Output: 49
+*/
+
+#include "utarray.h"
+#include <stdio.h>
+#include <stdlib.h>
+
+int sum_floor(UT_array* nums) {
+ int sum = 0;
+ for (size_t i = 0; i < utarray_len(nums); i++) {
+ for (size_t j = 0; j < utarray_len(nums); j++) {
+ sum += *(int*)utarray_eltptr(nums, i) / *(int*)utarray_eltptr(nums, j);
+ }
+ }
+ return sum;
+}
+
+int main(int argc, char* argv[]) {
+ if (argc < 3) {
+ fputs("Usage: ch-2 n n n ...\n", stderr);
+ exit(EXIT_FAILURE);
+ }
+
+ UT_array* nums;
+ utarray_new(nums, &ut_int_icd);
+
+ for (int i = 1; i < argc; i++) {
+ int n = atoi(argv[i]);
+ utarray_push_back(nums, &n);
+ }
+
+ int sum = sum_floor(nums);
+ printf("%d\n", sum);
+
+ utarray_free(nums);
+}
diff --git a/challenge-243/paulo-custodio/c/utarray.h b/challenge-243/paulo-custodio/c/utarray.h
new file mode 100644
index 0000000000..1fe8bc1c74
--- /dev/null
+++ b/challenge-243/paulo-custodio/c/utarray.h
@@ -0,0 +1,252 @@
+/*
+Copyright (c) 2008-2022, Troy D. Hanson https://troydhanson.github.io/uthash/
+All rights reserved.
+
+Redistribution and use in source and binary forms, with or without
+modification, are permitted provided that the following conditions are met:
+
+ * Redistributions of source code must retain the above copyright
+ notice, this list of conditions and the following disclaimer.
+
+THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS
+IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED
+TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A
+PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT OWNER
+OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL,
+EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO,
+PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR
+PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF
+LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING
+NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE OF THIS
+SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
+*/
+
+/* a dynamic array implementation using macros
+ */
+#ifndef UTARRAY_H
+#define UTARRAY_H
+
+#define UTARRAY_VERSION 2.3.0
+
+#include <stddef.h> /* size_t */
+#include <string.h> /* memset, etc */
+#include <stdlib.h> /* exit */
+
+#ifdef __GNUC__
+#define UTARRAY_UNUSED __attribute__((__unused__))
+#else
+#define UTARRAY_UNUSED
+#endif
+
+#ifndef utarray_oom
+#define utarray_oom() exit(-1)
+#endif
+
+typedef void (ctor_f)(void *dst, const void *src);
+typedef void (dtor_f)(void *elt);
+typedef void (init_f)(void *elt);
+typedef struct {
+ size_t sz;
+ init_f *init;
+ ctor_f *copy;
+ dtor_f *dtor;
+} UT_icd;
+
+typedef struct {
+ unsigned i,n;/* i: index of next available slot, n: num slots */
+ UT_icd icd; /* initializer, copy and destructor functions */
+ char *d; /* n slots of size icd->sz*/
+} UT_array;
+
+#define utarray_init(a,_icd) do { \
+ memset(a,0,sizeof(UT_array)); \
+ (a)->icd = *(_icd); \
+} while(0)
+
+#define utarray_done(a) do { \
+ if ((a)->n) { \
+ if ((a)->icd.dtor) { \
+ unsigned _ut_i; \
+ for(_ut_i=0; _ut_i < (a)->i; _ut_i++) { \
+ (a)->icd.dtor(utarray_eltptr(a,_ut_i)); \
+ } \
+ } \
+ free((a)->d); \
+ } \
+ (a)->n=0; \
+} while(0)
+
+#define utarray_new(a,_icd) do { \
+ (a) = (UT_array*)malloc(sizeof(UT_array)); \
+ if ((a) == NULL) { \
+ utarray_oom(); \
+ } \
+ utarray_init(a,_icd); \
+} while(0)
+
+#define utarray_free(a) do { \
+ utarray_done(a); \
+ free(a); \
+} while(0)
+
+#define utarray_reserve(a,by) do { \
+ if (((a)->i+(by)) > (a)->n) { \
+ char *utarray_tmp; \
+ while (((a)->i+(by)) > (a)->n) { (a)->n = ((a)->n ? (2*(a)->n) : 8); } \
+ utarray_tmp=(char*)realloc((a)->d, (a)->n*(a)->icd.sz); \
+ if (utarray_tmp == NULL) { \
+ utarray_oom(); \
+ } \
+ (a)->d=utarray_tmp; \
+ } \
+} while(0)
+
+#define utarray_push_back(a,p) do { \
+ utarray_reserve(a,1); \
+ if ((a)->icd.copy) { (a)->icd.copy( _utarray_eltptr(a,(a)->i++), p); } \
+ else { memcpy(_utarray_eltptr(a,(a)->i++), p, (a)->icd.sz); }; \
+} while(0)
+
+#define utarray_pop_back(a) do { \
+ if ((a)->icd.dtor) { (a)->icd.dtor( _utarray_eltptr(a,--((a)->i))); } \
+ else { (a)->i--; } \
+} while(0)
+
+#define utarray_extend_back(a) do { \
+ utarray_reserve(a,1); \
+ if ((a)->icd.init) { (a)->icd.init(_utarray_eltptr(a,(a)->i)); } \
+ else { memset(_utarray_eltptr(a,(a)->i),0,(a)->icd.sz); } \
+ (a)->i++; \
+} while(0)
+
+#define utarray_len(a) ((a)->i)
+
+#define utarray_eltptr(a,j) (((j) < (a)->i) ? _utarray_eltptr(a,j) : NULL)
+#define _utarray_eltptr(a,j) ((void*)((a)->d + ((a)->icd.sz * (j))))
+
+#define utarray_insert(a,p,j) do { \
+ if ((j) > (a)->i) utarray_resize(a,j); \
+ utarray_reserve(a,1); \
+ if ((j) < (a)->i) { \
+ memmove( _utarray_eltptr(a,(j)+1), _utarray_eltptr(a,j), \
+ ((a)->i - (j))*((a)->icd.sz)); \
+ } \
+ if ((a)->icd.copy) { (a)->icd.copy( _utarray_eltptr(a,j), p); } \
+ else { memcpy(_utarray_eltptr(a,j), p, (a)->icd.sz); }; \
+ (a)->i++; \
+} while(0)
+
+#define utarray_inserta(a,w,j) do { \
+ if (utarray_len(w) == 0) break; \
+ if ((j) > (a)->i) utarray_resize(a,j); \
+ utarray_reserve(a,utarray_len(w)); \
+ if ((j) < (a)->i) { \
+ memmove(_utarray_eltptr(a,(j)+utarray_len(w)), \
+ _utarray_eltptr(a,j), \
+ ((a)->i - (j))*((a)->icd.sz)); \
+ } \
+ if ((a)->icd.copy) { \
+ unsigned _ut_i; \
+ for(_ut_i=0;_ut_i<(w)->i;_ut_i++) { \
+ (a)->icd.copy(_utarray_eltptr(a, (j) + _ut_i), _utarray_eltptr(w, _ut_i)); \
+ } \
+ } else { \
+ memcpy(_utarray_eltptr(a,j), _utarray_eltptr(w,0), \
+ utarray_len(w)*((a)->icd.sz)); \
+ } \
+ (a)->i += utarray_len(w); \
+} while(0)
+
+#define utarray_resize(dst,num) do { \
+ unsigned _ut_i; \
+ if ((dst)->i > (unsigned)(num)) { \
+ if ((dst)->icd.dtor) { \
+ for (_ut_i = (num); _ut_i < (dst)->i; ++_ut_i) { \
+ (dst)->icd.dtor(_utarray_eltptr(dst, _ut_i)); \
+ } \
+ } \
+ } else if ((dst)->i < (unsigned)(num)) { \
+ utarray_reserve(dst, (num) - (dst)->i); \
+ if ((dst)->icd.init) { \
+ for (_ut_i = (dst)->i; _ut_i < (unsigned)(num); ++_ut_i) { \
+ (dst)->icd.init(_utarray_eltptr(dst, _ut_i)); \
+ } \
+ } else { \
+ memset(_utarray_eltptr(dst, (dst)->i), 0, (dst)->icd.sz*((num) - (dst)->i)); \
+ } \
+ } \
+ (dst)->i = (num); \
+} while(0)
+
+#define utarray_concat(dst,src) do { \
+ utarray_inserta(dst, src, utarray_len(dst)); \
+} while(0)
+
+#define utarray_erase(a,pos,len) do { \
+ if ((a)->icd.dtor) { \
+ unsigned _ut_i; \
+ for (_ut_i = 0; _ut_i < (len); _ut_i++) { \
+ (a)->icd.dtor(utarray_eltptr(a, (pos) + _ut_i)); \
+ } \
+ } \
+ if ((a)->i > ((pos) + (len))) { \
+ memmove(_utarray_eltptr(a, pos), _utarray_eltptr(a, (pos) + (len)), \
+ ((a)->i - ((pos) + (len))) * (a)->icd.sz); \
+ } \
+ (a)->i -= (len); \
+} while(0)
+
+#define utarray_renew(a,u) do { \
+ if (a) utarray_clear(a); \
+ else utarray_new(a, u); \
+} while(0)
+
+#define utarray_clear(a) do { \
+ if ((a)->i > 0) { \
+ if ((a)->icd.dtor) { \
+ unsigned _ut_i; \
+ for(_ut_i=0; _ut_i < (a)->i; _ut_i++) { \
+ (a)->icd.dtor(_utarray_eltptr(a, _ut_i)); \
+ } \
+ } \
+ (a)->i = 0; \
+ } \
+} while(0)
+
+#define utarray_sort(a,cmp) do { \
+ qsort((a)->d, (a)->i, (a)->icd.sz, cmp); \
+} while(0)
+
+#define utarray_find(a,v,cmp) bsearch((v),(a)->d,(a)->i,(a)->icd.sz,cmp)
+
+#define utarray_front(a) (((a)->i) ? (_utarray_eltptr(a,0)) : NULL)
+#define utarray_next(a,e) (((e)==NULL) ? utarray_front(a) : (((a)->i != utarray_eltidx(a,e)+1) ? _utarray_eltptr(a,utarray_eltidx(a,e)+1) : NULL))
+#define utarray_prev(a,e) (((e)==NULL) ? utarray_back(a) : ((utarray_eltidx(a,e) != 0) ? _utarray_eltptr(a,utarray_eltidx(a,e)-1) : NULL))
+#define utarray_back(a) (((a)->i) ? (_utarray_eltptr(a,(a)->i-1)) : NULL)
+#define utarray_eltidx(a,e) (((char*)(e) - (a)->d) / (a)->icd.sz)
+
+/* last we pre-define a few icd for common utarrays of ints and strings */
+static void utarray_str_cpy(void *dst, const void *src) {
+ char *const *srcc = (char *const *)src;
+ char **dstc = (char**)dst;
+ if (*srcc == NULL) {
+ *dstc = NULL;
+ } else {
+ *dstc = (char*)malloc(strlen(*srcc) + 1);
+ if (*dstc == NULL) {
+ utarray_oom();
+ } else {
+ strcpy(*dstc, *srcc);
+ }
+ }
+}
+static void utarray_str_dtor(void *elt) {
+ char **eltc = (char**)elt;
+ if (*eltc != NULL) free(*eltc);
+}
+static const UT_icd ut_str_icd UTARRAY_UNUSED = {sizeof(char*),NULL,utarray_str_cpy,utarray_str_dtor};
+static const UT_icd ut_int_icd UTARRAY_UNUSED = {sizeof(int),NULL,NULL,NULL};
+static const UT_icd ut_ptr_icd UTARRAY_UNUSED = {sizeof(void*),NULL,NULL,NULL};
+
+
+#endif /* UTARRAY_H */
diff --git a/challenge-243/paulo-custodio/cpp/ch-1.cpp b/challenge-243/paulo-custodio/cpp/ch-1.cpp
new file mode 100644
index 0000000000..ce52b9b8f2
--- /dev/null
+++ b/challenge-243/paulo-custodio/cpp/ch-1.cpp
@@ -0,0 +1,61 @@
+/*
+Challenge 243
+
+Task 1: Reverse Pairs
+Submitted by: Mohammad S Anwar
+
+You are given an array of integers.
+
+Write a script to return the number of reverse pairs in the given array.
+
+A reverse pair is a pair (i, j) where: a) 0 <= i < j < nums.length and b)
+nums[i] > 2 * nums[j].
+Example 1
+
+Input: @nums = (1, 3, 2, 3, 1)
+Output: 2
+
+(1, 4) => nums[1] = 3, nums[4] = 1, 3 > 2 * 1
+(3, 4) => nums[3] = 3, nums[4] = 1, 3 > 2 * 1
+
+Example 2
+
+Input: @nums = (2, 4, 3, 5, 1)
+Output: 3
+
+(1, 4) => nums[1] = 4, nums[4] = 1, 4 > 2 * 1
+(2, 4) => nums[2] = 3, nums[4] = 1, 3 > 2 * 1
+(3, 4) => nums[3] = 5, nums[4] = 1, 5 > 2 * 1
+*/
+
+#include <iostream>
+#include <vector>
+using namespace std;
+
+int num_reverse_pairs(const vector<int>& nums) {
+ int count = 0;
+ for (size_t i = 0; i < nums.size()-1; i++) {
+ for (size_t j = i+1; j < nums.size(); j++) {
+ if (nums[i] > 2 * nums[j])
+ count++;
+ }
+ }
+ return count;
+}
+
+int main(int argc, char* argv[]) {
+ if (argc < 3) {
+ cerr << "Usage: ch-1 n n n ..." << endl;
+ exit(EXIT_FAILURE);
+ }
+
+ vector<int> nums;
+
+ for (int i = 1; i < argc; i++) {
+ int n = atoi(argv[i]);
+ nums.push_back(n);
+ }
+
+ int count = num_reverse_pairs(nums);
+ cout << count << endl;
+}
diff --git a/challenge-243/paulo-custodio/cpp/ch-2.cpp b/challenge-243/paulo-custodio/cpp/ch-2.cpp
new file mode 100644
index 0000000000..8bcad05d7c
--- /dev/null
+++ b/challenge-243/paulo-custodio/cpp/ch-2.cpp
@@ -0,0 +1,63 @@
+/*
+Challenge 243
+
+Task 2: Floor Sum
+Submitted by: Mohammad S Anwar
+
+You are given an array of positive integers (>=1).
+
+Write a script to return the sum of floor(nums[i] / nums[j]) where
+0 <= i,j < nums.length. The floor() function returns the integer part of the
+division.
+
+Example 1
+
+Input: @nums = (2, 5, 9)
+Output: 10
+
+floor(2 / 5) = 0
+floor(2 / 9) = 0
+floor(5 / 9) = 0
+floor(2 / 2) = 1
+floor(5 / 5) = 1
+floor(9 / 9) = 1
+floor(5 / 2) = 2
+floor(9 / 2) = 4
+floor(9 / 5) = 1
+
+Example 2
+
+Input: @nums = (7, 7, 7, 7, 7, 7, 7)
+Output: 49
+*/
+
+#include <iostream>
+#include <vector>
+using namespace std;
+
+int sum_floor(const vector<int>& nums) {
+ int sum = 0;
+ for (size_t i = 0; i < nums.size(); i++) {
+ for (size_t j = 0; j < nums.size(); j++) {
+ sum += nums[i] / nums[j];
+ }
+ }
+ return sum;
+}
+
+int main(int argc, char* argv[]) {
+ if (argc < 3) {
+ cerr << "Usage: ch-1 n n n ..." << endl;
+ exit(EXIT_FAILURE);
+ }
+
+ vector<int> nums;
+
+ for (int i = 1; i < argc; i++) {
+ int n = atoi(argv[i]);
+ nums.push_back(n);
+ }
+
+ int sum = sum_floor(nums);
+ cout << sum << endl;
+}
diff --git a/challenge-243/paulo-custodio/perl/ch-1.pl b/challenge-243/paulo-custodio/perl/ch-1.pl
new file mode 100644
index 0000000000..157e5c0828
--- /dev/null
+++ b/challenge-243/paulo-custodio/perl/ch-1.pl
@@ -0,0 +1,45 @@
+#!/usr/bin/env perl
+
+# Challenge 243
+#
+# Task 1: Reverse Pairs
+# Submitted by: Mohammad S Anwar
+#
+# You are given an array of integers.
+#
+# Write a script to return the number of reverse pairs in the given array.
+#
+# A reverse pair is a pair (i, j) where: a) 0 <= i < j < nums.length and b)
+# nums[i] > 2 * nums[j].
+# Example 1
+#
+# Input: @nums = (1, 3, 2, 3, 1)
+# Output: 2
+#
+# (1, 4) => nums[1] = 3, nums[4] = 1, 3 > 2 * 1
+# (3, 4) => nums[3] = 3, nums[4] = 1, 3 > 2 * 1
+#
+# Example 2
+#
+# Input: @nums = (2, 4, 3, 5, 1)
+# Output: 3
+#
+# (1, 4) => nums[1] = 4, nums[4] = 1, 4 > 2 * 1
+# (2, 4) => nums[2] = 3, nums[4] = 1, 3 > 2 * 1
+# (3, 4) => nums[3] = 5, nums[4] = 1, 5 > 2 * 1
+
+use Modern::Perl;
+
+@ARGV>0 or die "Usage: ch-1.pl n n ...\n";
+say num_reverse_pairs(@ARGV);
+
+sub num_reverse_pairs {
+ my(@n) = @_;
+ my $count = 0;
+ for my $i (0..$#n-1) {
+ for my $j ($i+1..$#n) {
+ $count++ if $n[$i] > 2* $n[$j];
+ }
+ }
+ return $count;
+}
diff --git a/challenge-243/paulo-custodio/perl/ch-2.pl b/challenge-243/paulo-custodio/perl/ch-2.pl
new file mode 100644
index 0000000000..e8fd020119
--- /dev/null
+++ b/challenge-243/paulo-custodio/perl/ch-2.pl
@@ -0,0 +1,48 @@
+#!/usr/bin/env perl
+
+# Challenge 243
+#
+# Task 2: Floor Sum
+# Submitted by: Mohammad S Anwar
+#
+# You are given an array of positive integers (>=1).
+#
+# Write a script to return the sum of floor(nums[i] / nums[j]) where
+# 0 <= i,j < nums.length. The floor() function returns the integer part of the
+# division.
+#
+# Example 1
+#
+# Input: @nums = (2, 5, 9)
+# Output: 10
+#
+# floor(2 / 5) = 0
+# floor(2 / 9) = 0
+# floor(5 / 9) = 0
+# floor(2 / 2) = 1
+# floor(5 / 5) = 1
+# floor(9 / 9) = 1
+# floor(5 / 2) = 2
+# floor(9 / 2) = 4
+# floor(9 / 5) = 1
+#
+# Example 2
+#
+# Input: @nums = (7, 7, 7, 7, 7, 7, 7)
+# Output: 49
+
+use Modern::Perl;
+
+@ARGV>0 or die "Usage: ch-2.pl n n ...\n";
+say sum_floor(@ARGV);
+
+sub sum_floor {
+ my(@n) = @_;
+ my $sum = 0;
+ for my $i (0..$#n) {
+ for my $j (0..$#n) {
+ $sum += int($n[$i] / $n[$j]);
+ }
+ }
+ return $sum;
+}
diff --git a/challenge-243/paulo-custodio/t/test-1.yaml b/challenge-243/paulo-custodio/t/test-1.yaml
new file mode 100644
index 0000000000..2156a4274d
--- /dev/null
+++ b/challenge-243/paulo-custodio/t/test-1.yaml
@@ -0,0 +1,10 @@
+- setup:
+ cleanup:
+ args: 1 3 2 3 1
+ input:
+ output: 2
+- setup:
+ cleanup:
+ args: 2 4 3 5 1
+ input:
+ output: 3
diff --git a/challenge-243/paulo-custodio/t/test-2.yaml b/challenge-243/paulo-custodio/t/test-2.yaml
new file mode 100644
index 0000000000..fdbcf42adf
--- /dev/null
+++ b/challenge-243/paulo-custodio/t/test-2.yaml
@@ -0,0 +1,10 @@
+- setup:
+ cleanup:
+ args: 2 5 9
+ input:
+ output: 10
+- setup:
+ cleanup:
+ args: 7 7 7 7 7 7 7
+ input:
+ output: 49
generated by cgit (git 2.34.1) at 2026-04-17 14:28:34 +0000