各位高手帮帮忙，新手求问 （100分求助）

[direren]

关于时间的一个（C/C++）程序：
假设一天的时间是24小时的模式，-hh：mm：ss ，当问““现在是几点了？”产生一个具有人性化的回答。
例子：
00：01：12 => it's twelve at night
13:38:49=>it's one-forty pm

更高的期待：
00：01：12 => it's midnight
13:38:49=>it's about twenty till two in the afternoon (还有二十分钟到下午俩点）
还可以这么回答 “the little hand is on the one, the big hand is almost on the eight" (时针在1那，分针快到8了）


各位高手谢谢了，如果回答的好，我还可以在加分的。

[direren]

有人会LISP语言吗？这道题好像是用LISP解得，我就说没那么简单。。。。呜呜。。。

[hbvanguard]

分析:
1.不需要考虑秒数，所以忽略秒，只考虑时钟和分钟。
2.以24时计时法 hh(时钟):0--23, mm(分钟):0--59 为取值范围
3.假设18:00---00:00为晚上(night) ，1--11为AM,12为Midnoon,13--17为PM
4.为了输出更精确，规定分钟 >=55 的时候为下一时刻，否则为50分，例08:57 输出为9:00 ，08:52输出为8:50
以下为程序:

#include "stdio.h"
#include "stdlib.h"
#include "string.h"
typedef enum TimeKind
{
  None = -1,
  Night,
  AM,
  PM,
  MidNoon
}Kind;
//得到时钟对应的英文时间
void GetHourStr(int hour,char *getHour,Kind *timeKind)
{
  memset(getHour,0,sizeof(getHour));
  switch(hour)
	{
	  case 0: strcpy(getHour," twelve");*timeKind = Night; break;
	  case 1: strcpy(getHour," one"); *timeKind = AM;  break;
	  case 2: strcpy(getHour," two"); *timeKind = AM;  break;
	  case 3: strcpy(getHour," three");*timeKind = AM; break;
	  case 4: strcpy(getHour," four");*timeKind = AM;  break;
	  case 5: strcpy(getHour," five");*timeKind = AM;  break;
	  case 6: strcpy(getHour," six");*timeKind = AM;   break;
	  case 7: strcpy(getHour," seven");*timeKind = AM; break;
	  case 8: strcpy(getHour," eight");*timeKind = AM; break;
	  case 9: strcpy(getHour," nine");*timeKind = AM;  break;
	  case 10: strcpy(getHour, " ten");*timeKind = AM;  break;
	  case 11: strcpy(getHour," eleven");*timeKind = AM;break;
	  case 12: strcpy(getHour," twelve");*timeKind = MidNoon;break;
	  case 13: strcpy(getHour," one");*timeKind = PM;  break;
	  case 14: strcpy(getHour," two");*timeKind = PM;  break;
	  case 15: strcpy(getHour," three");*timeKind = PM;break;
	  case 16: strcpy(getHour," four");*timeKind = PM; break;
	  case 17: strcpy(getHour," five");*timeKind = PM; break;
	  case 18: strcpy(getHour," six");*timeKind = Night;  break;
	  case 19: strcpy(getHour," seven");*timeKind = Night;break;
	  case 20: strcpy(getHour," eight");*timeKind = Night;break;
	  case 21: strcpy(getHour," nine");*timeKind = Night; break;
	  case 22: strcpy(getHour," ten");*timeKind = Night;  break;
	  case 23: strcpy(getHour," eleven");*timeKind = Night;break;
	  default:
		 break;
	}
   return;
}
char * GetTime(const char *strTime )
{
	char pHour[3] = {0};
	char pMinute[3] = {0};
	char hourBuff[16] = {0};
	char *DesStr = NULL;
	int iHour = 0;
	int iMinute = 0;
	int tmp = 0;
	Kind timeKind = None;
	DesStr = (char *)malloc(sizeof(char)*64);
        strncpy(pHour,strTime,2);      //时钟
	strncpy(pMinute,strTime + 3,2);//分钟
        iHour = atoi(pHour);
	iMinute = atoi(pMinute);
        strcpy(DesStr,"it's");
	//得到时钟对应的英文时间
	GetHourStr(iHour,hourBuff,&timeKind);
	strcat(DesStr,hourBuff);
	tmp = iMinute%10;
	//得到模糊分钟
	switch(iMinute/10)
	{
		case 0: break;
		case 1: (tmp>=5)?strcat(DesStr,"-twenty "):strcat(DesStr,"-ten ");break;
		case 2: (tmp>=5)?strcat(DesStr,"-thirty "):strcat(DesStr,"-twenty ");break;
		case 3: (tmp>=5)?strcat(DesStr,"-forty "):strcat(DesStr,"-thirty ");break;
		case 4:(tmp>=5)?strcat(DesStr,"-fifty "):strcat(DesStr,"-forty ");break;
		case 5:
			if(tmp>5)
                        {
                          GetHourStr(iHour + 1,hourBuff,&timeKind); 
			   memset(DesStr,0,sizeof(DesStr));
			   strcpy(DesStr,"it's");
			   strcat(DesStr,hourBuff);
			   break;
			 }
			else
			{strcat(DesStr,"-fifty ");break;}
	}
	switch(timeKind)
	{
	case Night:strcat(DesStr," at night "); break;
	case AM:strcat(DesStr," am "); break;
	case PM:strcat(DesStr," pm "); break;
	case MidNoon:strcat(DesStr," at Midnoon "); break;
	default:
		 break;
	}
      return DesStr;
}
int main(int argc, char* argv[])
  { 
     char *p = "20:46:46";
     char *pTimeStr = GetTime(p);
     printf("%s",pTimeStr);
     if(pTimeStr != NULL)
       {
	  free(pTimeStr);
       }
     getchar();
     return 0;
   } 
//vs2005 运行结果:
//it's eight-fifty  at night

[selooloo]

LZ可以根据需要再添加些answer

#include <stdio.h>

typedef struct mytime{
        int hour;
        int minute;
        }MT;
void answer1(MT n)
{
     char hour[12][12]={"one","two","three","four","five","six","seven","eight","nine","ten","eleven","twelve"};
     char minute[60][12]={"one","two","three","four","five","six","seven","eight","nine","ten","eleven","twelve",
                           "thirtheen","fourtheen","fifteen","sixtheen","seventheen","eighteen","nintheen"};
     char m[10][12]={"twenty","thirty","fourty","fifty","fifty","sixty"}; 
     printf("it's ");
     if(n.hour>12)
     {
       printf("%s  ",hour[n.hour-12-1]);
       if(n.minute>=20)
       {
                       if(n.minute%10==0)
                           printf("%s-%s pm.\n",m[n.minute/10-2]);
                       else
                           printf("%s-%s pm.\n",m[n.minute/10-2],minute[n.minute%10-1]);
       }
       else
       printf("%s pm.\n",minute[n.minute]);
     }
     else
     {
       printf("%s ",hour[n.hour-1]);
       if(n.minute>20)
       printf("%s-%s am.\n",m[n.minute/10-2],minute[n.minute%10-1]);
       else
       printf("%s am.\n",minute[n.minute]);
     }
}
void answer2(MT n)
{
     char big[12][12]={"one","two","three","four","five","six","seven","eight","nine","ten","eleven","twelve"};
     char little[12][12]={"one","two","three","four","five","six","seven","eight","nine","ten","eleven","twelve"};
     int min;
     printf("the little hand is on the %s,",little[n.hour%12-1]);
     printf("the big hand is almost on the %s\n",big[n.minute/5]);
     
}
void asktime(void)
{
     char time[10]={};
     MT now;
     puts("现在是几点了？");
     sprintf(time,"%s",__TIME__);
     puts(time);
     sscanf(time,"%d:%d",&now.hour,&now.minute);
     answer1(now);
     answer2(now);
}

int main(void)
{
    asktime();
    getchar();
    return 0;
}

[daidodo]

mark

[MasterLuo]

我觉得应该就是一个把时间转换为数字的过程。当然你要回答的话是你先保存起来的，然后用时间来替换原来的时间，回答方式的不同，就使数字的表达方式不同。
00：01：12 => it's midnight
如果你想选择这种回答的话，那么，it's是不变的，那么你只需要判断时间是哪个时间段，再用相应的单词来替代（预先定制）。
这个还是很简单的，你可以对每一个转换方式写一个函数，还可以随机选机回答方式，我觉得这样做比较好。

[yjqidai]

例如时间为 13:38:49

if(时针==?)
{
Switch(60-分钟)
{
case ...;
break;
.
.
default:....;
}
else
.
.
.
if(时针==13)
{
Switch(60-38)
{
case 1:printf("....");
break;
.
.
.
.
case 12:printf("it's about twenty till two in the afternoon ");
.
.
.
default:printf("...");
}
.
.
.


---------------------
先声明,我是个新手.可能有错,望高手指错.谢谢

[yjqidai]

PS: 我的程序完全是:汇编脑袋.上面那程序要写24*60条以上的程序.

[liuchaotao]

用模糊数学的方法先把一天分成几个段（早上、上午、中午、下午、晚上、午夜等），并把每个小时也用模糊的方法进行分段。然后根据具体的时间通过查表取得回答的词语，按照格式回答就可以了。

[小魔菇]

switch
case

[direren]

有没有更简单的方法呢？就是不要把全部的时间和语句都switch case 一遍， 这样就没有意义了。能不能给个具体的例子呢，像是怎么录入之类的

[晴天1999]

新手哈!多多指教哟!

[a4626458]

switch 语句啊

[cattycat]

定义与12小时对应的字符数组"one","two",...,"twelve"，然后三个区间属性的字符数组"morning", "afternoon","night". 对分也定义模糊的区间，10，20，30，40，50对应的字符串也差不多了。

然后根据hh-mm取得的值，对应到数组下标，用一个格式化输出的函数就行了，这个函数根据这些数组的下标格式化输出It's about xx(h) xx(m) in the xx(morning,afternoon,evening)。
类似这个样子