The time interval for the brain

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Problem Description

从键盘输入两个时间点(24小时制），输出两个时间点之间的时间间隔，时间间隔用“小时:分钟:秒”表示。
如：3点5分25秒应表示为--03：05：25.假设两个时间在同一天内，时间先后顺序与输入无关。

Input

输入包括两行。
第一行为时间点1。
第二行为时间点2。

Output

以“小时:分钟:秒”的格式输出时间间隔。
格式参看输入输出（There is no blank line between the two lines of data）。

Example Input

12:01:12

13:09:43

Example Output

01:08:31

Although this problem is simple, but still stumped a lot of small partners, the key is to successfully convert the unit and then converted back.Many of the problems can use the idea of transformation to complete, or even a little conversion, you become the most familiar template.

//AC Code presented#include <stdio.h>#include <stdlib.h>int main(){    int h1,m1,s1,h2,m2,s2;    int t1,t2,t;    int ans1,ans2,ans3;    char ch;    scanf("%d%c%d%c%d\n",&h1,&ch,&m1,&ch,&s1);//Read    scanf("%d%c%d%c%d",&h2,&ch,&m2,&ch,&s2);    t1=3600*h1+60*m1+s1;//Calculate time 1    t2=3600*h2+60*m2+s2;//Calculate time 2    t=abs(t1-t2);//Does not rule out the time 2 than the time 1 early, so to calculate ABS    ans3=t%60;//Convert the time difference into the form of the subject requirement    ans2=((t-ans3)%3600)/60;    ans1=(t-ans3-ans2*60)/3600;    if (ans1<10) printf("0%d:",ans1); else printf("%d:",ans1);//The output format should be correct    if (ans2<10) printf("0%d:",ans2); else printf("%d:",ans2);    if (ans3<10) printf("0%d:",ans3); else printf("%d",ans3);    return 0;}

／／Look at the code of the little partners, please do not copy and paste directly, even if the copy is responsible for their own.

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