最笨的方法

解题思路:
首先输入结构体一变量的年月日，然后转到count函数先对闰年进行判断，取余为零的就是闰年。下一步到对月份的判断，在把每个月份的数累加

参考代码:

#include <stdio.h>

int count(int a,int b,int c);

struct date {

int year;

int month;

int day;

};

int main()

{

int number;

struct date day0;

scanf("%d %d %d", &day0.year, &day0.month, &day0.day);

printf("%d", count(day0.year, day0.month,day0.day));

return 0;

}

int count(int a,int b,int c)

{

int number=0;

if ((a % 4) == 0)

{

switch (b)

{

case 1:

number = c;

break;

case 2:

number = 31 + c;

break;

case 3:

number = 31 + 29 + c;

break;

case 4:

number = 31 + 29 + 31 + c;

break;

case 5:

number = 31 + 29 + 31 + 30 + c;

break;

case 6:

number = 31 + 29 + 31 + 30 + 31 +c;

break;

case 7:

number = 31 + 29 + 31 + 30 + 31 + 30 + c;

break;

case 8:

number = 31 + 29 + 31 + 30 + 31 + 30 + 31 + c;

break;

case 9:

number = 31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + c;

break;

case 10:

number = 31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + c;

break;

case 11:

number = 31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + c;

break;

case 12:

number = 31 + 29 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + c;

break;

}

}

else

{

switch (b)

{

case 1:

number = c;

break;

case 2:

number = 31 + c;

break;

case 3:

number = 31 + 28 + c;

break;

case 4:

number = 31 + 28 + 31 + c;

break;

case 5:

number = 31 + 28 + 31 + 30 + c;

break;

case 6:

number = 31 + 28 + 31 + 30 + 31 + c;

break;

case 7:

number = 31 + 28 + 31 + 30 + 31 + 30 + c;

break;

case 8:

number = 31 + 28 + 31 + 30 + 31 + 30 + 31 + c;

break;

case 9:

number = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + c;

break;

case 10:

number = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + c;

break;

case 11:

number = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + c;

break;

case 12:

number = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + c;

break;

}

}

return number;

}