历法类的题目，纯模拟，处理2月份的代码可能不是很精练，不过思路还是很清晰的
#include <stdio.h>
int judgeleap(int year)
{
if ( ( !(year%4) && year%100 ) || !(year%400) )
return 1;
else return 0;
}
int leapmon[]={ 0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
int mon[]={ 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
int n;
int main()
{
scanf("%d", &n);
for ( int l = 1; l <= n; l++ )
{
int year, day, month, inday, inmon, num, flag;
scanf("%d/%d/%d%d/%d", &month, &day, &year, &inmon, &inday );
printf("%d ", l);
if ( month == inmon )
{
num= day-inday;
num=num>0?num:-num;
day>inday?(flag=0):(flag=1);
}
else if ( 1 == month && 12 == inmon )
{
num=31-inday+day;
year--;
flag= 0;
}
else if ( 1 == inmon && 12 == month )
{
num=31-day+inday;
year++;
flag= 1;
}
else if ( ( month < inmon-1 ) || ( inmon < month-1 ) )
{
num=8;
}
else if ( 2 == month )
{
if ( month < inmon )
{
if ( judgeleap(year) )
{
num=29-day+inday;
flag= 1;
}
else
{
num=28-day+inday;
flag= 1;
}
}
else
{
num=31-inday+day;
flag= 0;
}
}
else if ( 2 == inmon )
{
if ( month < inmon )
{
num=31-day+inday;
flag= 1;
}
else
{
if ( judgeleap(year) )
{
num=29-inday+day;
}
else
{
num=28-inday+day;
}
flag= 0;
}
}
else if ( month < inmon )
{
if ( judgeleap(year) )
{
num=leapmon[month]-day+inday;
flag= 1;
}
else
{
num=mon[month]-day+inday;
flag= 0;
}
}
else if ( inmon < month )
{
if ( judgeleap(year) )
{
num=leapmon[month]-inday+day;
flag= 0;
}
else
{
num=mon[month]-inday+day;
flag= 1;
}
}
if ( 0 == num )
printf("SAME DAY\n");
else if ( 1 == num )
{
printf("%d/%d/%d IS %d DAY ", inmon, inday, year, num);
if (flag) printf("AFTER\n");
else printf("PRIOR\n");
}
else if ( 8 > num)
{
printf("%d/%d/%d IS %d DAYS ", inmon, inday, year, num);
if (flag) printf("AFTER\n");
else printf("PRIOR\n");
}
else
printf("OUT OF RANGE\n");
}
return 0;
}