C语言实现输入的时间,屏幕显示一秒后的时间。显示格式为HH:MM:SS。 #include <stdio.h>intmain(){intHH, MM, SS; scanf("%d:%d:%d",&HH,&MM,&SS); SS++;if(SS >=60) { SS =0; MM ++; }if(MM >=60) { MM =0; HH ++; }if(HH >=24) { HH =0; MM =0; SS =0; } printf(...
include<stdio.h>int main(void){ int hour; prntf("请输入小时数：") scanf("%d",&hour); switch(hour){ case 1: case 2: case 3: case 4: printf("深夜");break; case 5: case 6: printf("凌晨");break; case 7: case 8: print...
int main(){ SYSTEMTIME Stime,Etime;GetLocalTime(&Stime);//取得开始计时时间 Sleep(5000);//停5秒
include <stdio.h>int main (void){int a,b;printf ("请依次输入小时和分钟(24小时制输入)！\n");printf ("小时:");scanf ("%d",&a);printf ("分钟:");scanf ("%d",&b);if (a>0&&a<=6)printf ("现在是凌晨%d:%d\n",a,b);if (a>6&&a<=12)printf ("现在是上午%d:%d\...
int day = 0;int hour = 0;int minute = 0;int second = 0;int main(){ void inputDate(); /*输入年-月-日 时:分:秒*/ void nextSceond(); /*计算下一秒的时间*/ int leapYear(int year); /*判断是否为闰年*/ int dayMonth(int month); /*返回每个月份对应的天数...
include <stdio.h>int main(){int h,m,t;printf("请输入现在时间,格式如9:30\n");scanf("%d:%d",&h,&m);printf("请输入经过的分钟数");scanf("%d",&t);printf("经过%d分钟以后的时间是%d:%d",t,(h*60+m+t)/60%24,(m+t)%60);return 0;}//运行示例：...
计算输入时间是当年的第几天的问题可以使用C语言编写一个函数来解决。下面是一个示例实现: ```c #include <stdio.h> int isLeapYear(int year) if ((year % 4 == 0 && year % 100 != 0) , year % 400 == 0) return 1; } else return 0; } int getDayOfYear(int year, int month, int day...
int main(void){ int year, month, day;int m[] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};printf("Input year month day:");scanf("%d%d%d", &year, &month, &day);if (year % 4 == 0 && year % 100 != 0 || year % 400 == 0)m[1]++;(month...
/*跪求编写程序，输入一天24小时制的时间(0~23时)，输出对应的时段。规定[0~4]点深夜；(4,6]点为凌晨； (6,8]点为早晨；(8,12]点为上午； (12,18]点为下午； (18,24]点为晚上。*/# include <stdio.h>int main(){int n;scanf("%d",&n);if(n<=4)printf("对应的时段为深夜!"...