#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>

using namespace std;

int	days[][12] =
{
	{ 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31, },
	{ 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31, },
};

int is_leap(int year);

int
main(int argc, char *argv[])
{
	int	n;
	cin >> n;
	
	while (--n >= 0)
	{
		int	hour, min, secs;
		int	day, month, year;
		char	tmp;
		
		scanf("%d:%d:%d %d.%d.%d", &hour, &min, &secs, &day, &month, &year);
		
		int	total = 0;
		total += 365*(year - 2000) + (year - 2000)/4 - (year - 2000)/100 + (year - 2000)/400;
		total += !is_leap(year);
		
		for (int i = 0; i < month-1; ++i)
			total += days[is_leap(year)][i];
		
		total += day - 1;
		
		day = total % 100 + 1;
		month = total/100 % 10 + 1;
		year = total/1000;
		
		total = 0;
		total += hour * 60 * 60;
		total += min * 60;
		total += secs;
		
		total *= 125;
		total = total/108;
		
		secs = total % 100;
		min = total/100 % 100;
		hour = total / 10000;
		
		printf("%d:%d:%d %d.%d.%d\n", hour, min, secs, day, month, year);
	}
	
	return 0;
}

int
is_leap(int year)
{
	return (year%4 == 0) && year%100 || (year%400 == 0);
}

• Re:这题关键是闰年的处理, 我把代码贴出来给你参考 (48) qywyh_scut 2006-03-19 14:07:49
  • 相当于total = total * 100000 / 86400; (56) frkstyc 2006-03-19 14:09:46
    • 注意:特殊日历1天有100000秒(这里的秒跟正常的秒不一样),而正常日历有86400秒 (78) zsc09_leaf 2010-08-05 17:36:33