Black Friday's verdict

A simple C language programming exercise, judging black Friday, prompted to enter a year, and then give the year all the existing black Friday months, and give the number of black Friday. Judge the week by using the Caille formula.

/* * author: nonkey  * time  : 2015-08-02 sunday *  black friday judge  *  Requirements: Enter a year  ,  determine if there is a black Friday &nbsp in this year, and if so, the output has a total of  *        several, and the output month.  *  idea: Because only need to determine whether there is black Friday, so, do not have to every month every day of the calculation, Black star  *   period five of course only appears in 13th, so, as long as the number of years to determine the input of the year 13th whether there is a star  *   Five on the line, judging the week with Caille formula.  * */#include  <stdio.h>int main (void) {int year , moon ,day  ,week ,y_in;int i ,count = 0 ;d ay = 13 ;//always judged number 13th, and the rest no matter printf (" pls input year :  ")  ;  scanf ("%d ", &y_in)  ;for (i = 1  ;i <= 12 ;i++) {if (i == 1 | | &NBSP;I&NBSP;==&NBSP;2) {moon = i + 12 ;year = y_in - 1 ;} else{moon = i ;year = y_in ;} if ((year < 1752) &Nbsp;| |  (year == 1752 && moon < 9)  | |   (year == 1752 && moon == 9 && day <  3) week =  (day + 2 * moon + 3 *  (moon + 1)) &NBSP;/&NBSP;5&NBSP;+&NBSP;YEAR&NBSP;+&NBSP;YEAR&NBSP;/&NBSP;4&NBSP;+&NBSP;5)  % 7 ;elseweek  =  (day + 2 * moon + 3 *  (moon + 1)  / 5  + year + year / 4 - year / 100 + year /  400)  % 7 ;week++ ;//except  7  to get the number of  0--6 , need  + 1  To revise to our customary  1 --7  weekday notation  . if (week == 5) {count++ ;p rintf ("moon : %d\n", i)  ;}} if (count == 0) {printf ("This year don ' t have black friday.\n")  ;} else printf ("all black friday&Nbsp;is %d:\n ", count)  ;return 0 ;} 

This article is from the "Hao Shui Mo Negative Blog" blog, please be sure to keep this source http://nonkey.blog.51cto.com/6923493/1681410

Black Friday's verdict