From abdbde519e16ab23d4a2858ea1f7345e2773d23f Mon Sep 17 00:00:00 2001
From: Paulo Custodio <pauloscustodio@gmail.com>
Date: Sun, 2 Apr 2023 01:33:11 +0100
Subject: Add C solution

---
 challenge-187/paulo-custodio/c/ch-1.c | 105 ++++++++++++++++++++++++++++++++++
 1 file changed, 105 insertions(+)
 create mode 100644 challenge-187/paulo-custodio/c/ch-1.c

diff --git a/challenge-187/paulo-custodio/c/ch-1.c b/challenge-187/paulo-custodio/c/ch-1.c
new file mode 100644
index 0000000000..4816fabe9d
--- /dev/null
+++ b/challenge-187/paulo-custodio/c/ch-1.c
@@ -0,0 +1,105 @@
+/*
+Challenge 187
+
+Task 1: Days Together
+Submitted by: Mohammad S Anwar
+
+Two friends, Foo and Bar gone on holidays seperately to the same city. You are
+given their schedule i.e. start date and end date.
+
+To keep the task simple, the date is in the form DD-MM and all dates belong to
+the same calendar year i.e. between 01-01 and 31-12. Also the year is non-leap
+year and both dates are inclusive.
+
+Write a script to find out for the given schedule, how many days they spent
+together in the city, if at all.
+Example 1
+
+Input: Foo => SD: '12-01' ED: '20-01'
+       Bar => SD: '15-01' ED: '18-01'
+
+Output: 4 days
+
+Example 2
+
+Input: Foo => SD: '02-03' ED: '12-03'
+       Bar => SD: '13-03' ED: '14-03'
+
+Output: 0 day
+
+Example 3
+
+Input: Foo => SD: '02-03' ED: '12-03'
+       Bar => SD: '11-03' ED: '15-03'
+
+Output: 2 days
+
+Example 4
+
+Input: Foo => SD: '30-03' ED: '05-04'
+       Bar => SD: '28-03' ED: '02-04'
+
+Output: 4 days
+*/
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+#include <stdbool.h>
+
+#define YEAR 2023   // any non-leap year
+
+#define MIN(a, b)   ((a)<(b)?(a):(b))
+#define MAX(a, b)   ((a)>(b)?(a):(b))
+
+// https://pdc.ro.nu/jd-code.html
+long gregorian_date_to_jd(int y, int m, int d) {
+    y += 8000;
+    if (m < 3) { y--; m += 12; }
+    return (y * 365) + (y / 4) - (y / 100) + (y / 400) - 1200820
+        + (m * 153 + 3) / 5 - 92 + d - 1;
+}
+
+int days_together(long as, long ae, long bs, long be) {
+    if (bs > ae || as > be) return 0;
+    long os = MAX(as, bs);
+    long oe = MIN(ae, be);
+    long days = oe - os + 1;
+    return (int)days;
+}
+
+int main(int argc, char* argv[]) {
+    if (argc != 5) {
+        fputs("usage: ch-2 dd-mm dd-mm dd-mm dd-mm", stderr);
+        return EXIT_FAILURE;
+    }
+
+    int d, m;
+    if (sscanf(argv[1], "%d-%d", &d, &m) != 2) {
+        fputs("invalid date", stderr);
+        return EXIT_FAILURE;
+    }
+    long as = gregorian_date_to_jd(YEAR, m, d);
+
+    if (sscanf(argv[2], "%d-%d", &d, &m) != 2) {
+        fputs("invalid date", stderr);
+        return EXIT_FAILURE;
+    }
+    long ae = gregorian_date_to_jd(YEAR, m, d);
+
+    if (sscanf(argv[3], "%d-%d", &d, &m) != 2) {
+        fputs("invalid date", stderr);
+        return EXIT_FAILURE;
+    }
+    long bs = gregorian_date_to_jd(YEAR, m, d);
+
+    if (sscanf(argv[4], "%d-%d", &d, &m) != 2) {
+        fputs("invalid date", stderr);
+        return EXIT_FAILURE;
+    }
+    long be = gregorian_date_to_jd(YEAR, m, d);
+
+    int days = days_together(as, ae, bs, be);
+
+    printf("%d\n", days);
+}
-- 
cgit