考虑闰年计算一个月天数的通用公式

Question

有人知道这样的数学公式来计算一个月中的天数吗

28 + (x + Math.floor(x/8)) % 2 + 2 % x + 2 * Math.floor(1/x);

但哪个也考虑闰年？还应考虑到公历每 400 年省略 3 个闰日，这是其闰周期的长度。

Answer 1

在表达式中添加一个术语 (m == 2) * leapyear(yyyy) 来确定闰年二月的正确天数并不难。此 C 代码显示了一种方法：

#include <stdio.h>
#include <stdbool.h>

static inline bool leapyear(int yy)
{
    if (yy %   4 != 0) return false;
    if (yy % 100 != 0) return true;
    if (yy % 400 != 0) return false;
    return true;
}

static inline int old_dim(int mm)
{
    return (28 + (mm + (mm/8)) % 2 + 2 % mm + 2 * (1/mm));
}

static inline int new_dim(int mm, int yyyy)
{
    return (28 + (mm + (mm/8)) % 2 + 2 % mm + 2 * (1/mm) + ((mm == 2) * leapyear(yyyy)));
}

int main(void)
{
    /*28 + (x + Math.floor(x/8)) % 2 + 2 % x + 2 * Math.floor(1/x);*/
    for (int mm = 1; mm <= 12; mm++)
        printf("mm = %2d, DIM = %2d\n", mm, old_dim(mm));

    for (int yyyy = 1900; yyyy < 2101; yyyy += 5)
    {
        for (int mm = 1; mm <= 12; mm++)
            printf("yyyy = %4d, mm = %2d: DIM = %2d\n", yyyy, mm, new_dim(mm, yyyy));
    }

    return 0;
}

mm = 2 的输出（从完整输出中过滤）是：

yyyy = 1900, mm =  2: DIM = 28
yyyy = 1905, mm =  2: DIM = 28
yyyy = 1910, mm =  2: DIM = 28
yyyy = 1915, mm =  2: DIM = 28
yyyy = 1920, mm =  2: DIM = 29
yyyy = 1925, mm =  2: DIM = 28
yyyy = 1930, mm =  2: DIM = 28
yyyy = 1935, mm =  2: DIM = 28
yyyy = 1940, mm =  2: DIM = 29
yyyy = 1945, mm =  2: DIM = 28
yyyy = 1950, mm =  2: DIM = 28
yyyy = 1955, mm =  2: DIM = 28
yyyy = 1960, mm =  2: DIM = 29
yyyy = 1965, mm =  2: DIM = 28
yyyy = 1970, mm =  2: DIM = 28
yyyy = 1975, mm =  2: DIM = 28
yyyy = 1980, mm =  2: DIM = 29
yyyy = 1985, mm =  2: DIM = 28
yyyy = 1990, mm =  2: DIM = 28
yyyy = 1995, mm =  2: DIM = 28
yyyy = 2000, mm =  2: DIM = 29
yyyy = 2005, mm =  2: DIM = 28
yyyy = 2010, mm =  2: DIM = 28
yyyy = 2015, mm =  2: DIM = 28
yyyy = 2020, mm =  2: DIM = 29
yyyy = 2025, mm =  2: DIM = 28
yyyy = 2030, mm =  2: DIM = 28
yyyy = 2035, mm =  2: DIM = 28
yyyy = 2040, mm =  2: DIM = 29
yyyy = 2045, mm =  2: DIM = 28
yyyy = 2050, mm =  2: DIM = 28
yyyy = 2055, mm =  2: DIM = 28
yyyy = 2060, mm =  2: DIM = 29
yyyy = 2065, mm =  2: DIM = 28
yyyy = 2070, mm =  2: DIM = 28
yyyy = 2075, mm =  2: DIM = 28
yyyy = 2080, mm =  2: DIM = 29
yyyy = 2085, mm =  2: DIM = 28
yyyy = 2090, mm =  2: DIM = 28
yyyy = 2095, mm =  2: DIM = 28
yyyy = 2100, mm =  2: DIM = 28

这正确地将 1900 年和 2100 年视为非闰年，但将 2000 年视为闰年。

yyyy = 1900, mm =  1: DIM = 31
yyyy = 1900, mm =  2: DIM = 28
yyyy = 1900, mm =  3: DIM = 31
yyyy = 1900, mm =  4: DIM = 30
yyyy = 1900, mm =  5: DIM = 31
yyyy = 1900, mm =  6: DIM = 30
yyyy = 1900, mm =  7: DIM = 31
yyyy = 1900, mm =  8: DIM = 31
yyyy = 1900, mm =  9: DIM = 30
yyyy = 1900, mm = 10: DIM = 31
yyyy = 1900, mm = 11: DIM = 30
yyyy = 1900, mm = 12: DIM = 31

…

yyyy = 2000, mm =  1: DIM = 31
yyyy = 2000, mm =  2: DIM = 29
yyyy = 2000, mm =  3: DIM = 31
yyyy = 2000, mm =  4: DIM = 30
yyyy = 2000, mm =  5: DIM = 31
yyyy = 2000, mm =  6: DIM = 30
yyyy = 2000, mm =  7: DIM = 31
yyyy = 2000, mm =  8: DIM = 31
yyyy = 2000, mm =  9: DIM = 30
yyyy = 2000, mm = 10: DIM = 31
yyyy = 2000, mm = 11: DIM = 30
yyyy = 2000, mm = 12: DIM = 31

考虑闰年计算一个月天数的通用公式

Universal formula to calculate the number of days in a month taking into account leap years

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