题目描述
Xiao Ming is a 2013-BUCTer.This time he has a Programming Contest for 2013th.So he has a lot of things to do recently.Attending classes,doing homework,programing,playing etc.In order to do things more efficient,he makes a time schedule,which has a start time and a end time,but no time of duration.Can you calculate it?
输入
There is more than one test case.Each test case has 2 strings,which means the start and the end time, like this x:y:z ,x is the number of days after the he made the plan,y is the hour and z is the minute.The end time is bigger than the start time.
输出
For each test case,output the time of duration like this Case k: x:y ,x is hour and y is the minute.
样例输入
1:8:00
1:9:30
1:22:20
2:6:40
3:8:00
3:9:00
1:0:00
3:0:05
样例输出
Case 1: 1:30
Case 2: 8:20
Case 3: 1:0
Case 4: 48:5
1 //此题是一道非常简单的题目，并不用多长的代码，只要利用C的特性，读入就可以了，有的同学可能会用读入字符串的方法，然后分离，这样非常的麻烦，不如直接来，下面是代码
2
3 #include <stdio.h>
4 int main () {
5 int day1,hour1,minute1;
6 int day2,hour2,minute2;
7 int pos=0;
8 char ch;
9 while (scanf ("%d%c%d%c%d",&day1,&ch,&hour1,&ch,&minute1)!=EOF) {
10 scanf ("%d%c%d%c%d",&day2,&ch,&hour2,&ch,&minute2);
11 pos++;
12 day1=day2-day1;
13 hour1=hour2-hour1;
14 minute1=minute2-minute1;
15 if (minute1<0) {minute1+=60;hour1-=1;}
16 if (hour1<0) {hour1+=24;day1-=1;}
17 hour1+=day1*24;
18 printf("Case %d: %d:%d\n",pos,hour1,minute1);
19 }
20 }
