Determining day of the week using Zeller's Congruence

烂漫一生 提交于 2019-12-01 06:22:06
congusbongus

Here's a working version:

#include <stdio.h>
#include <math.h>
int main()
{
  int h,q,m,k,j,day,month,year;
  printf("Enter the date (dd/mm/yyyy)\n");
  scanf("%i/%i/%i",&day,&month,&year);
  if(month == 1)
  {
    month = 13;
    year--;
  }
  if (month == 2)
  {
    month = 14;
    year--;
  }
  q = day;
  m = month;
  k = year % 100;
  j = year / 100;
  h = q + 13*(m+1)/5 + k + k/4 + j/4 + 5*j;
  h = h % 7;
  switch(h)
  {
    case 0 : printf("Saturday.\n"); break;
    case 1 : printf("Sunday.\n"); break;
    case 2 : printf("Monday. \n"); break;
    case 3 : printf("Tuesday. \n"); break;
    case 4 : printf("Wednesday. \n"); break;
    case 5 : printf("Thurday. \n"); break;
    case 6 : printf("Friday. \n"); break;
  }
  return 0;
}

Live demo.

The key is this in your h formula: 13/5*(m+1). This is using integer division, which calculates 13/5 first, so the result is equivalent to 2*(m+1). Swap the 5 and (m+1) around and the result will be correct.

By the way you do need the year decrements if Jan/Feb, as the wiki article explains.

Why do you include "h= year % 100" and "j= year / 100"?????

#include <stdio.h>
#include <math.h>
int main()
{
  int h,q,m,k,j,day,month,year;
  printf("Enter the date (dd/mm/yyyy)\n");
  scanf("%i/%i/%i",&day,&month,&year);
  if(month == 1)
  {
    month = 13;
    year--;
  }
  if (month == 2)
  {
    month = 14;
    year--;
  }
  q = day;
  m = month;
  k = year % 100;
  j = year / 100;
  h = q + 13*(m+1)/5 + k + k/4 + j/4 + 5*j;
  h = h % 7;
  switch(h)
  {
    case 0 : printf("Saturday.\n"); break;
    case 1 : printf("Sunday.\n"); break;
    case 2 : printf("Monday. \n"); break;
    case 3 : printf("Tuesday. \n"); break;
    case 4 : printf("Wednesday. \n"); break;
    case 5 : printf("Thurday. \n"); break;
    case 6 : printf("Friday. \n"); break;
  }
  return 0;
}
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