Add C solution

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+/*
+Challenge 183
+
+Task 2: Date Difference
+Submitted by: Mohammad S Anwar
+
+You are given two dates, $date1 and $date2 in the format YYYY-MM-DD.
+
+Write a script to find the difference between the given dates in terms on years
+and days only.
+Example 1
+
+Input: $date1 = '2019-02-10'
+       $date2 = '2022-11-01'
+Output: 3 years 264 days
+
+Example 2
+
+Input: $date1 = '2020-09-15'
+       $date2 = '2022-03-29'
+Output: 1 year 195 days
+
+Example 3
+
+Input: $date1 = '2019-12-31'
+       $date2 = '2020-01-01'
+Output: 1 day
+
+Example 4
+
+Input: $date1 = '2019-12-01'
+       $date2 = '2019-12-31'
+Output: 30 days
+
+Example 5
+
+Input: $date1 = '2019-12-31'
+       $date2 = '2020-12-31'
+Output: 1 year
+
+Example 6
+
+Input: $date1 = '2019-12-31'
+       $date2 = '2021-12-31'
+Output: 2 years
+
+Example 7
+
+Input: $date1 = '2020-09-15'
+       $date2 = '2021-09-16'
+Output: 1 year 1 day
+
+Example 8
+
+Input: $date1 = '2019-09-15'
+       $date2 = '2021-09-16'
+Output: 2 years 1 day
+*/
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+#include <stdbool.h>
+
+// 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;
+}
+
+int main(int argc, char* argv[]) {
+    if (argc != 3) {
+        fputs("usage: ch-2 yyyy-mm-dd yyyy-mm-dd", stderr);
+        return EXIT_FAILURE;
+    }
+
+    int y1, m1, d1;
+    if (sscanf(argv[1], "%d-%d-%d", &y1, &m1, &d1) != 3) {
+        fputs("invalid date", stderr);
+        return EXIT_FAILURE;
+    }
+    long jd1 = gregorian_date_to_jd(y1, m1, d1);
+
+    int y2, m2, d2;
+    if (sscanf(argv[2], "%d-%d-%d", &y2, &m2, &d2) != 3) {
+        fputs("invalid date", stderr);
+        return EXIT_FAILURE;
+    }
+    long jd2 = gregorian_date_to_jd(y2, m2, d2);
+
+    int years = 0;
+    while (gregorian_date_to_jd(y1 + years + 1, m1, d1) <= jd2)
+        years++;
+    jd1 = gregorian_date_to_jd(y1 + years, m1, d1);
+
+    if (years != 0)
+        printf("%d year%s ", years, years == 1 ? "" : "s");
+
+    int days = jd2 - jd1;
+    if (days != 0)
+        printf("%d day%s ", days, days == 1 ? "" : "s");
+    printf("\n");
+}