输入一个日期，输出该日期是星期几

假定输入日期合法正确。

先找一个参考日期，找星期天的日期为最好。我一时没想到就选了今天，星期一，也不错。然后求出输入日期与参考日期之间间隔的天数n，n为负时则表示输入日期在参考日期之前，n为正时则表示输入日期在参考日期之后。因为星期为循环星期1到星期天，又根据补码的原理，可知n = （（n % 7） +  8） % 7， 此时的n为几则是星期几。（PS:星期天用0来表示，因为我选的是参考日期是星期一所以是+8，如果选的是星期天则是+7）。

 1 //给定一个日期，求这个日期是星期几？
 2 #include <stdio.h>
 3 #include <stdlib.h>
 4
 5 int is_leap_year(unsigned y);
 6 unsigned days_in_year(unsigned y, unsigned m, unsigned d);
 7 void swap(unsigned *m, unsigned *n);
 8 int days(unsigned y1, unsigned m1, unsigned d1);
 9
10 int main(int argc, char *argv[])
11 {
12     unsigned y, m, d;
13     int n, res;
14
15     while (fflush(stdin),(res = scanf("%d%d%d", &y, &m, &d)) != EOF)
16     {
17         n = days(y, m, d);
18         n %= 7;
19         n = (n + 8) % 7 ;
20         printf("%d年%d月%d日是星期%d\n", y, m, d, n);
21     }
22     system("pause");
23     return 0;
24 }
25
26 int is_leap_year(unsigned y)
27 {
28     if((y % 4 == 0 && y % 100 != 0) || y % 400 == 0)
29         return 1;
30     else
31         return 0;
32 }
33
34 unsigned days_in_year(unsigned y, unsigned m, unsigned d)
35 {
36     unsigned n = 0, i;
37     unsigned months[13] = {0,31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
38
39     if(is_leap_year(y))
40         months[2] = 29;
41     else
42         months[2] = 28;
43     for ( i = 1; i < m; i++)
44         n += months[i];
45     n += d;
46     return n;
47 }
48
49 void swap(unsigned *m, unsigned *n)
50 {
51     unsigned tmp;
52     tmp = *m;
53     *m = *n;
54     *n = tmp;
55 }
56
57 int days(unsigned y2, unsigned m2, unsigned d2)
58 {
59     unsigned y1, m1, d1, i, n = 0;
60     int flag = 0;
61
62     y1 = 2015;
63     m1 = 3;
64     d1 = 9;
65     if ( y1 > y2 || (y1 == y2 && m1 > m2) || (y1 == y2 && m1 == m2 && d1 > d2))
66     {
67         flag = 1;
68         swap(&y1, &y2);
69         swap(&m1, &m2);
70         swap(&d1, &d2);
71     }
72     if (y1 != y2)
73     {
74         if (is_leap_year(y1))
75             n += 366 - days_in_year(y1, m1, d1);
76         else
77             n += 365 - days_in_year(y1, m1, d1);
78         for ( i = y1 + 1; i < y2; i++)
79         {
80             if( is_leap_year(i) )
81                 n += 366;
82             else
83                 n += 365;
84         }
85         n += days_in_year(y2, m2, d2);
86     }
87     else
88         n = days_in_year(y2, m2, d2) - days_in_year(y1, m1, d1);
89     if (flag) n = -n;
90     return n;
91 }