楼上有点小错误
被100整除的年份必须要被400整除才是闰年,像2100年就不是闰年

有两个类似的程序：

输入年和月，输出该月有多少天？（我们的作业题，嘿嘿） 在TURBO C V2.01 下调试通过。

第一个：
#include "stdio.h"
#include "conio.h"
main ()
{
int year,month;
clrscr ();
printf ("please input number like this: [year],[month]\n\n");
scanf ("[%d],[%d]",&year,&month);
printf("\n");
if (year%4==0&&year%100!=0||year%400==0)
{
printf ("year=%d;month=%d;",year,month);
switch (month)
{
case 1:
case 3:
case 5:
case 7:
case 8:
case 10:
case 12:printf ("day=31\n\n"); break;
case 2:printf ("day=29\n\n"); break;
case 4:
case 6:
case 9:
case 11:printf ("day=30\n\n"); break;
default:printf ("the number is error!\n\n"); break;
}

}
else
{
printf ("year=%d;month=%d;",year,month);
switch (month)
{
case 1:
case 3:
case 5:
case 7:
case 8:
case 10:
case 12:printf ("day=31\n\n"); break;
case 2:printf ("day=28\n\n"); break;
case 4:
case 6:
case 9:
case 11:printf ("day=30\n\n"); break;
default:printf ("the number is error!\n\n"); break;
}
}
}

第二个：

#include "stdio.h"
#include "conio.h"
main()
{
int year,month;
clrscr();
printf("please input number like this: [year],[month]\n");
printf("\n\n");
scanf("[%d],[%d]", &year, &month);
if(year%4==0&&year%100!=0||year%400==0)
if(1==month||3==month||5==month||7==month||8==month||10==month||12==month)
printf ("day=31\n");
else if (2==month)
printf ("day=29\n");
else printf("day=30\n");
else
if (1==month||3==month||5==month||7==month||8==month||10==month||12==month)
printf("day=31\n");
else if (2==month) printf("day=28\n");
else printf("day=30\n");
}

main()
{
int day, month, year, sum, leap;
printf("\nplease input year, month, day\n");
scanf("%d, %d, %d", &year, &month, &day);
switch(month) /* 先计算某月以前月份的总天数 */
{
case 1: sum=0; break;
case 2: sum=31; break;
case 3: sum=59; break;
case 4: sum=90; break;
case 5: sum=120; break;
case 6: sum=151; break;
case 7: sum=181; break;
case 8: sum=212; break;
case 9: sum=243; break;
case 10: sum=273; break;
case 11: sum=304; break;
case 12: sum=334; break;
default: printf("data error");break;
}
sum = sum + day; /* 再加上某天的天数 */
if(year %400 == 0 || (year % 4 == 0 && year % 100 != 0)) /* 判断是不是闰年 */
leap = 1;
else
leap = 0;
if(leap == 1 && month > 2) /* 如果是闰年且月份大于2, 总天数应该加一天 */
sum++;
printf("It is the %dth day.", sum);
}
这个不知道可以不可以，你看看吧

#include <stdio.h>
#include <stdlib.h>
int isLeapYear(int year);

void main()
{
int date[2][13]={{0,30,28,31,30,31,30,31,31,30,31,30,31},//定义二维数组，存放每个月的天数！
{0,30,29,31,30,31,30,31,31,30,31,30,31}};
int y,m;
printf("请输入年份:");
scanf("%d",&y);
printf("请输入月份:");
scanf("%d",&m);
if(m<1||m>12)//判断用户输入的月份是否合理
{
printf("月份输入错误!程序退出!\n");
exit(0);
}
printf("%d年的%d月有%d天!\n",y,m,date[isLeapYear(y)][m]);
}

int isLeapYear(int year)//闰年的判断
{
if (year%400==0||(year%4==0&&year%100!=0))
{
return 1;
}
else
{
return 0;
}
}

都是书上的题目

怎样能这样实现
输入：1985年1月20日
     （或2006/2/19）
     
输出：20
      71