George and Sleep（时间计算）

A. George and Sleep

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

George woke up and saw the current time s on the digital clock. Besides, George knows that he has slept for time t.

Help George! Write a program that will, given time s and t, determine the time p when George went to bed. Note that George could have gone to bed yesterday relatively to the current time (see the second test sample).

Input

The first line contains current time s as a string in the format "hh:mm". The second line contains time t in the format "hh:mm" — the duration of George's sleep. It is guaranteed that the input contains the correct time in the 24-hour format, that is, 00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59.

Output

In the single line print time p — the time George went to bed in the format similar to the format of the time in the input.

Sample test(s)

input

05:50
05:44

output

00:06

input

00:00
01:00

output

23:00

input

00:01
00:00

output

00:01

Note

In the first sample George went to bed at "00:06". Note that you should print the time only in the format "00:06". That's why answers "0:06", "00:6" and others will be considered incorrect.

In the second sample, George went to bed yesterday.

In the third sample, George didn't do to bed at all.

 

    题意：

    给出两个时间，求第二个时间到第一个时间总共过了多长。24小时制。

 

    思路：

    模拟。可以换成分钟来算。

 

    AC：

    1：

#include<stdio.h>
int main()
{
    int h1 = 0,h2 = 0,m1 = 0,m2 = 0;
    scanf("%d:%d",&h1,&m1);
    scanf("%d:%d",&h2,&m2);
    if(m1 < m2)
    {
        if(h1 == 0) h1 = 23;
        else        h1--;
        m1 += 60;
    }
    if(h1 < h2)      h1 += 24;
    if(h1 - h2 <= 9) printf("0%d:",h1 - h2);
    else             printf("%d:",h1 - h2);
    if(m1 - m2 <= 9) printf("0%d\n",m1 - m2);
    else             printf("%d\n",m1 - m2);
    return 0;
}

    2：

#include<stdio.h>
int main()
{
    int h1,h2,m1,m2,time;
    scanf("%d:%d",&h1,&m1);
    scanf("%d:%d",&h2,&m2);
    time = (h1 * 60 + m1 - h2 * 60 - m2 + 1440) % 1440;
    printf("%02d:%02d",time/60,time%60);
    return 0;
}