Add C solution

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+/*
+Challenge 178
+
+Task 2: Business Date
+Submitted by: Mohammad S Anwar
+
+You are given $timestamp (date with time) and $duration in hours.
+
+Write a script to find the time that occurs $duration business hours after
+$timestamp. For the sake of this task, let us assume the working hours is 9am
+to 6pm, Monday to Friday. Please ignore timezone too.
+
+For example,
+
+Suppose the given timestamp is 2022-08-01 10:30 and the duration is 4 hours.
+Then the next business date would be 2022-08-01 14:30.
+
+Similar if the given timestamp is 2022-08-01 17:00 and the duration is 3.5 hours.
+Then the next business date would be 2022-08-02 11:30.
+*/
+
+#include <assert.h>
+#include <stdio.h>
+#include <stdlib.h>
+#include <stdbool.h>
+
+#define DAY_START   9
+#define DAY_END     18
+
+enum { Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday };
+
+int day_of_week(int y, int m, int d) {
+    return (d += m < 3 ? y-- : y - 2, 23*m/9 + d + 4 + y/4- y/100 + y/400)%7;
+}
+
+// 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;
+}
+
+void jd_to_gregorian_date(long jd, int* yp, int* mp, int* dp) {
+    int y, m, d;
+    for (y = jd / 366 - 4715; gregorian_date_to_jd(y + 1, 1, 1) <= jd; y++);
+    for (m = 1; gregorian_date_to_jd(y, m + 1, 1) <= jd; m++);
+    for (d = 1; gregorian_date_to_jd(y, m, d + 1) <= jd; d++);
+    *yp = y; *mp = m; *dp = d;
+}
+
+void next_business_date(int* y, int* m, int* d, int* hh, int* mm, int minutes) {
+    while (true) {
+        int now = *hh * 60 + *mm;
+        int end_day = DAY_END * 60;
+        if (now + minutes < end_day) {
+            *mm += minutes % 60;
+            *hh += minutes / 60;
+            return;
+        }
+        minutes -= end_day - now;
+        long jd = gregorian_date_to_jd(*y, *m, *d);
+        int dow;
+        do {
+            jd++;
+            jd_to_gregorian_date(jd, y, m, d);
+            dow = day_of_week(*y, *m, *d);
+        } while (dow == Saturday || dow == Sunday);
+        *hh = DAY_START;
+        *mm = 0;
+    }
+}
+
+int main(int argc, char* argv[]) {
+    if (argc != 4) {
+        fputs("usage: ch-2 yyyy-mm-dd HH:MM hours", stderr);
+        return EXIT_FAILURE;
+    }
+
+    int y, m, d;
+    if (sscanf(argv[1], "%d-%d-%d", &y, &m, &d) != 3) {
+        fputs("invalid date", stderr);
+        return EXIT_FAILURE;
+    }
+
+    int hh, mm;
+    if (sscanf(argv[2], "%d:%d", &hh, &mm) != 2) {
+        fputs("invalid hour", stderr);
+        return EXIT_FAILURE;
+    }
+
+    float hours;
+    if (sscanf(argv[3], "%f", &hours) != 1) {
+        fputs("invalid hours", stderr);
+        return EXIT_FAILURE;
+    }
+
+    next_business_date(&y, &m, &d, &hh, &mm, (int)(hours*60));
+
+    printf("%04d-%02d-%02d %02d:%02d\n", y, m, d, hh, mm);
+}