CodeForces 722A Broken Clock（破碎钟表）

最新推荐文章于 2020-05-18 08:50:45 发布

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本文介绍了一道编程题目，目标是最少修改破损时钟显示的时间，使其符合12或24小时制的标准格式。通过简单的逻辑判断和数值调整，实现合法时间的快速修正。

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You are given a broken clock. You know, that it is supposed to show time in 12- or 24-hours HH:MM format. In 12-hours format hours change from 1 to 12, while in 24-hours it changes from 0 to 23. In both formats minutes change from 0 to 59.

You are given a time in format HH:MM that is currently displayed on the broken clock. Your goal is to change minimum number of digits in order to make clocks display the correct time in the given format.

For example, if 00:99 is displayed, it is enough to replace the second 9 with 3 in order to get 00:39 that is a correct time in 24-hours format. However, to make 00:99 correct in 12-hours format, one has to change at least two digits. Additionally to the first change one can replace the second 0 with 1 and obtain 01:39.

Input

The first line of the input contains one integer 12 or 24, that denote 12-hours or 24-hours format respectively.

The second line contains the time in format HH:MM, that is currently displayed on the clock. First two characters stand for the hours, while next two show the minutes.

Output

The only line of the output should contain the time in format HH:MM that is a correct time in the given format. It should differ from the original in as few positions as possible. If there are many optimal solutions you can print any of them.

   Examples
 
     input
   
24
17:30

     output
   
17:30

     input
   
12
17:30

     output
   
07:30

     input
   
24
99:99

     output
   
09:09

题意：

给定一个时间，问最少需要修改多少位使其成为合法时间。

思路：

模拟即可啦，注意到10的整数倍。

AC code：

#include<stdio.h>
#include<cstring>
#include<algorithm>
#define AC main()
using namespace std;
const int MYDD = 1103;

int AC {
	int type, mm, hh, ans = 0;
	scanf("%d", &type);
	scanf("%d:%d", &hh, &mm);
	if(mm < 0) mm++;
	if(mm > 59) mm %= 10;

	if(type == 12) {
		if(hh > 12) {
			if(hh % 10)
				hh %= 10;
			else
				hh = 10;
		}
		if(hh <= 0)	hh = 1;
	} else {//32->22
		if(hh > 23)	{
			if(hh % 10)
				hh %= 10;
			else
				hh = 10;
		}
		if(hh < 0)	hh = 0;
	}
	
	printf("%02d:%02d\n", hh, mm);
	return 0;
}