一年的第一天公式导致结果不一致

First Day of the year Formula resulting in inconsistent results

objective 是找到一年的第一天,然后据此计算日历和闰年。我 运行 关注的问题是;虽然 class 给出了一个公式(无法使用数组或函数),但我能够在大多数给定年份得到正确的结果,除了计算 saturdays(模数后的余数为 0。)

我的公式有问题吗?

我会link分解公式的页面,当然也提供我的。

**这是包含公式的页面:**

Math Forum - Formula for the First Day of a Year

这是我的公式和我的代码:

 //Calculating Dates:
//N = d + 2m + [3(m+1)/5] + y + [y/4] - [y/100] + [y/400] + 2
//N = 1 + 2(13) + [3(13+1)/5] + year_int + [year_int/4] - [year_int/100] + 
//[year_int/400] + 2
// I'm using a hard-coded "13" to represent January in the place of m
 day_one =
        1 + 2*(13) + 3*(13 + 1)/5 + year_int + (year_int / 4) - (year_int / 100) + (year_int / 400) +
        2;
day_one = day_one % 7;


if (day_one < 0 || day_one >= 7) {
printf("Invalid Day\n");
} else {
switch (day_one) {


    case 1:
        weekday = "Sunday";
        printf("%s", weekday);
        break;
    case 2:
        weekday = "Monday";
        printf("%s", weekday);
        break;
    case 3:
        weekday = "Tuesday";
        printf("%s", weekday);
        break;
    case 4:
        weekday = "Wednesday";
        printf("%s", weekday);
        break;
    case 5:
        weekday = "Thursday";
        printf("%s", weekday);
        break;
    case 6:
        weekday = "Friday";
        printf("%s", weekday);
        break;
    case 0:
        weekday = "Saturday";
        printf("%s", weekday);
        break;
    default: printf("Error");
    break;
}
}

非常感谢任何帮助,我只是在寻求指导。

谢谢!

我发现要找到年份的第一天应该包含去年的整数。然后我回头看了看阅读材料,发现它是用纯文本说的。

脸掌

感谢您检查您是否在这里!

我希望这更清楚。请查看带有 !!! 的评论!在函数内部

/* 
ZELLER'S Algorithm
https://en.wikipedia.org/wiki/Zeller's_congruence
Gregorian calendar
*/
 #include <stdio.h>

 int day_of_week(int d,int m,int y);
 int main(void)
 {
  int day,month,year;
  printf ("Day, Month, Year\n");
  scanf ("%d %d %d",&day,&month,&year); 
  printf("%d",day_of_week(day,month,year)); 
  return 0;
 }

 int day_of_week(int d, int m, int y)
 //This function returns the ISO number of the day for a given day,month,year
 {
  int day;  
  if (d<1 || d>31 || m<1 || m>12 || y<1583 ) return 0; // In that case we have major error!
  if ((m==1) || (m==2)) 
    {m=m+12;--y;} // You have to do this correction !!!!
  day=d+(m+1)*26/10+y+y/4+6*(y/100)+y/400;
  day=day % 7; // 0=Saturday 1=Sunday ...
  day=((day+5) % 7)+1; // Convertion to ISO date: 1=Monday, 2=Tuesday ..
  return (day);
 }