本文主要是介绍CodeForces 387A George and Sleep，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
链接：http://codeforces.com/problemset/problem/387/A
George and Sleep
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)
05:50 05:44
00:06
00:00 01:00
23:00
00:01 00:00
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.
大意——给你当前的时间和持续睡眠的时间，问你上床睡觉的时间是何时？所有时间均采用24小时格式。
思路——直接模拟。用当前时间的小时和分钟直接减去持续睡眠时间的小时和分钟即可。最后分别判断一下小时和分钟的符号并做相应处理即可。
复杂度分析——时间复杂度:O(1),空间复杂度:O(1)
附上AC代码：
#include <iostream>
#include <cstdio>
#include <string>
#include <cmath>
#include <iomanip>
#include <ctime>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>
//#pragma comment(linker, "/STACK:102400000, 102400000")
using namespace std;
typedef long long ll;
const double pi = acos(-1.0);
const double e = exp(1.0);
const double eps = 1e-8;
short chour, cmin;
short lhour, lmin;int main()
{ios::sync_with_stdio(false);while (~scanf("%hd:%hd", &chour, &cmin)){scanf("%hd:%hd", &lhour, &lmin);int hour = chour-lhour;int min = cmin-lmin;if (min < 0){min += 60;hour -= 1;}if (hour < 0)hour += 24;printf("%02d:%02d\n", hour, min);}return 0;
}
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