97 lines
2.7 KiB
C
97 lines
2.7 KiB
C
/* You are given the following information, but you may prefer to do some research for yourself.
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*
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* 1 Jan 1900 was a Monday.
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* Thirty days has September,
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* April, June and November.
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* All the rest have thirty-one,
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* Saving February alone,
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* Which has twenty-eight, rain or shine.
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* And on leap years, twenty-nine.
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* A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.
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*
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* How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?*/
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#define _POSIX_C_SOURCE 199309L
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#include <stdio.h>
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#include <stdlib.h>
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#include <time.h>
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typedef enum {jan, feb, mar, apr, may, jun, jul, aug, sep, oct, nov, dec} months;
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typedef enum {mon, tue, wed, thu, fri, sat, sun} days;
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int main(int argc, char **argv)
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{
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int year, i, limit, count = 0;
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double elapsed;
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struct timespec start, end;
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months month;
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days day;
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clock_gettime(CLOCK_MONOTONIC, &start);
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day = mon;
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month = jan;
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year = 1900;
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while(year < 2001)
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{
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/* February has 29 days on leap years, otherwise 28. Leap years are those
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* divisible by 4, but not if they're divisible by 100, except when they're
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* divisible by 400.*/
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if(month == feb)
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{
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if(year % 400 == 0 || (year % 4 == 0 && year % 100 != 0))
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{
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limit = 29;
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}
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else
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{
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limit = 28;
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}
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}
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/* April, June, September and November have 30 days.*/
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else if(month == apr || month == jun || month == sep || month == nov)
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{
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limit = 30;
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}
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/* All other months have 31 days.*/
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else
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{
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limit = 31;
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}
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/* Loop on every day of the month.*/
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for(i = 1; i <= limit; i++)
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{
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/* If it's the first day of the month and it's Sunday, increase
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* counter, except if year=1900 (we need to count Sundays from
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* 1901 to 2000.*/
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if(year > 1900 && i == 1 && day == sun)
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{
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count++;
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}
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/* Change day of the week.*/
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day = (day + 1) % 7;
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}
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/* At the end of the month, go to next month.*/
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month = (month + 1) % 12;
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/* If we're back to january, increase the year.*/
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if(month == jan)
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{
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year++;
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}
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}
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clock_gettime(CLOCK_MONOTONIC, &end);
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elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;
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printf("Project Euler, Problem 19\n");
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printf("Answer: %d\n", count);
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printf("Elapsed time: %.9lf seconds\n", elapsed);
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return 0;
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}
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