呵呵,动脑筋的算法请教了(C语言)

只看该作者 · 2009-8-6 09:32

给出一个总的分钟数 unsigned long  tminute; ,  如tminute 是65230;
想求得这个分钟数是**年**月**日**时**分00秒.

这个日期的时间是以2000年01月01日00时00分00秒为起始算的.
//当然秒是不考虑的.

最后的日期,存储在TIME_BCD[0]; //年
                     TIME_BCD[1]; //月
                     TIME_BCD[2]; //日
                     TIME_BCD[3]; //时
                     TIME_BCD[4];//分
                     TIME_BCD[5]=0x00;//秒
/////
void get_bcdtime(unsigned long tminute)
      {
      }

只看该作者 · 2009-8-6 10:08

2楼朋友,关键还要考虑闰年的问题

只看该作者 · 2009-8-6 12:57

闲着无事，用VC进行了简单的编程。
#include "stdafx.h"

#define YYearMinute (366*24*60)
#define NYearminute (365*24*60)
#define Month31 (31*24*60)
#define Month30 (30*24*60)
#define Month28 (28*24*60)
#define Month29 (29*24*60)
#define Day (24*60)
#define Hour 60

unsigned long tminute = 65230;
unsigned char TIME_BCD[6];

void get_bcdtime()
{
int i = 1;
int left=tminute;
char flag ='N';

//get year
while ( left >= NYearminute ){
      //it is better using a function to check if the year is a leap year or not
if ( (TIME_BCD[0]==0x00) || (TIME_BCD[0]==0x04) || (TIME_BCD[0]==0x08) ){
flag ='Y';

if (left>=YYearMinute){
         left = left - YYearMinute;
}
else break;
}
else{
         left = left - NYearminute;
   flag ='N';
}

TIME_BCD[0]=TIME_BCD[0]+0x01;

}
//get month
for (;i<=12;i++){
if ( (left>=Month30) || (left>=Month31) ){
   if ( (i==1) || (i==3) || (i==5) || (i==7) || (i==8) || (i==10) || (i==12) ){
         left = left - Month31;
   }
   else if ( i==2 ){
   flag=='Y' ? (left=left-Month29):(left=left-Month28);
   }
   else{
   left = left - Month30;
   }

   TIME_BCD[1]=TIME_BCD[1]+0x01;
}
else break;
}

//get day
while(left>=Day){
      left = left - Day;
TIME_BCD[2]=TIME_BCD[2]+0x01;
}

//get hour
while(left>=Hour){
      left = left - Hour;
TIME_BCD[3]=TIME_BCD[3]+0x01;
}

//get minute
TIME_BCD[4]=left;

}
void main(){

TIME_BCD[0]=0x00;
TIME_BCD[1]=0x01;
TIME_BCD[2]=0x01;
TIME_BCD[3]=0x00;
TIME_BCD[4]=0x00;
TIME_BCD[5]=0x00;

get_bcdtime();

printf("20%02d-%02d-%02d %02d:%02d:00", TIME_BCD[0],TIME_BCD[1],TIME_BCD[2],TIME_BCD[3],TIME_BCD[4]);

}