C语言作业如何把 输入的123456 转换成 12:34:56 和 12小时制时间 还有计算距离到晚上十二点的时间还有多久

2025-03-16 16:11:23
推荐回答(3个)
回答1:

C程序:

#include "stdio.h"

void main()

{

long num;

        int hour, minute, second, hour1;

        char time1[9], time2[11];

        int i;

        int isAm = 1;

        double diff = 0;

        printf("input time : ");

scanf("%ld", &num);

        hour = num / 10000;

        minute = num / 100 % 100;

second = num % 100;

        /* time for 24 hour */

        i = 0;

        if (hour < 10)

        {

            time1[i++] = '0';

            time1[i++] = hour + '0';

}

else

{

time1[i++] = hour/10 + '0';

time1[i++] = hour%10 + '0';

}

        time1[i++] = ':';

        if (minute < 10)

{

            time1[i++] = '0';

            time1[i++] = minute + '0';

}

        else

        {

time1[i++] = minute/10 + '0';

time1[i++] = minute%10 + '0';

}

        time1[i++] = ':';

        if (second < 10)

{

            time1[i++] = '0';

            time1[i++] = second + '0';

}

        else

        {

time1[i++] = second/10 + '0';

time1[i++] = second%10 + '0';

}

time1[i] = '\0';

        printf("%s\n", time1);

        /* time for 12 hour */

i = 0;

        hour1 = hour;

        if (hour1 > 12)

        {

            isAm = 0;

            hour1 -= 12;

        }

        if (hour1 < 10)

        {

            time2[i++] = '0';

            time2[i++] = hour1 + '0';

}

        else

            {

time2[i++] = hour1/10 + '0';

time2[i++] = hour1%10 + '0';

}

        time2[i++] = ':';

        if (minute < 10)

{

            time2[i++] = '0';

            time2[i++] = minute + '0';

}

        else

         {

time2[i++] = minute/10 + '0';

time2[i++] = minute%10 + '0';

}

        time2[i++] = ':';

        if (second < 10)

{

            time2[i++] = '0';

            time2[i++] = second + '0';

}

        else

        {

time2[i++] = second/10 + '0';

time2[i++] = second%10 + '0';

}

        if (isAm == 0)

            time2[i++] = 'p';

        else

            time2[i++] = 'a';

time2[i++] = 'm';

time2[i] = '\0';

        printf("%s\n", time2);

        /* difference to 22:00 today */

diff = (3600.0 * (22 - hour) - 60 * minute - second) / 60.0;

printf("there are %.2lf mins\n", diff);

    }

运行测试:

回答2:

#include

int main(void)
{
int h,m,s;
scanf("%2d%2d%2d",&h,&m,&s);
printf("%02d:%02d:%02d\n",h,m,s);
if(h<=12)
printf("%02d:%02d:%02dam\n",h,m,s);
else
printf("%02d:%02d:%02dpm\n",h-12,m,s);
double rest=24*60-(h*60+m+s/60.0);
printf("there are %.2lf mins\n",rest);
return 0;
}

回答3:

已经让人给做了,我就不发了吧。

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