From c751359c3fd4c743d2e99ef22e0fff7bb586e26a Mon Sep 17 00:00:00 2001
From: Krishna Vedala <krishna.vedala@ieee.org>
Date: Wed, 1 Apr 2020 12:10:59 -0400
Subject: [PATCH] Project Euler / Problem 19

---
 project_euler/Problem 19/sol1.c | 120 ++++++++++++++++++++++++++++++++
 1 file changed, 120 insertions(+)
 create mode 100644 project_euler/Problem 19/sol1.c

diff --git a/project_euler/Problem 19/sol1.c b/project_euler/Problem 19/sol1.c
new file mode 100644
index 00000000..70fb7f16
--- /dev/null
+++ b/project_euler/Problem 19/sol1.c	
@@ -0,0 +1,120 @@
+#include <stdio.h>
+
+/**
+ * returns number of days in a month. 
+ * Month is identified by an integer -
+ * 0 = Jan and 11 = December 
+ * For February, adjust for leap year outside the function.
+ **/
+char get_month_days(short month)
+{
+    if (month == 1) /* February has 28 days. Adjust leap year in the loop */
+        return 28;
+    else if (month <= 6) /* odd months till July have 30 days - Jan = 0 (even)*/
+    {
+        if (month & 0x01)
+            return 30;
+        else
+            return 31;
+    } else if (month >= 7) /* odd months after July have 31 days*/
+    {
+        if (month & 0x01)
+            return 31;
+        else
+            return 30;
+    }
+    /* should never reach here! */
+    perror("Should never have reached this point!\n");
+    return -1;
+}
+
+/**
+ * return 1 if input year is a leap year
+ * otherwise, return 0
+ **/
+char is_leap_year(short year)
+{
+    if ((year % 400 == 0) || ((year % 4 == 0) && (year % 100 != 0)))
+        return 1;
+    
+    return 0;
+}
+
+#ifdef DEBUG
+const char* day_string(int day)
+{
+    switch(day)
+    {
+        case 0:
+            return "Sunday";
+        case 1:
+            return "Monday";
+        case 2:
+            return "Tuesday";
+        case 3:
+            return "Wednesday";
+        case 4:
+            return "Thursday";
+        case 5:
+            return "Friday";
+        case 6:
+            return "Saturday";
+        default:
+            return "Shouldnt see this!";
+    }
+}
+#endif
+
+
+int main(int argc, char **argv)
+{
+    int count_sundays = 0;
+    const short start_year = 1901;
+    const short end_year = 2000;
+
+    /**
+     * Let us identify days i.e., Sunday thru Saturday with integers - 0 thru 6 respectively
+     *  Jan 1 1901 was a Tuesday
+     **/
+    char start_day = 2;
+
+    for (int year = start_year; year <= end_year; year++)
+    {
+        char is_leap = is_leap_year(year);
+        for (char month = 0; month < 12; month ++)
+        {
+            /**
+             * These two for-loops count the start of day for the next month. Hence, 
+             * we have to skip the last December count */
+            if (year == end_year && month == 11)
+                continue;
+
+            int days = get_month_days(month);
+    
+            if (is_leap && month == 1)  /* for a leap year february, add a day */
+                days ++;
+            
+            #ifdef DEBUG
+            if (year == end_year)
+            {
+                printf("Year: %d\t Month: %d\t Days: %d\t First of day: %s\n", year, month, days, day_string(start_day));
+            }
+            #endif
+
+            /** Main Algorithm:
+             * every week has 7 days hence, the start of next day would be modulo 7
+             * add to this, the current start date and ensure the result is still
+             * modulo 7!
+             **/
+            start_day = ((days % 7) + start_day) % 7;
+
+            /* If start-day is a Sunday, increment counter */
+            if (start_day == 0)
+                count_sundays ++;
+        }
+    }
+
+    printf("Total number of Sundays that happened on the 1st of a month in the last century: %d\n", count_sundays);
+
+    return 0;
+}
\ No newline at end of file