Broken Clock Codeforces Intel Round-A

A. Broken Clock
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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


注意 如果你是通过改十位的话 注意12的时候 50:00 要变成 01:00

code:

<strong>import java.io.*;
import java.util.*;
public class Main {

	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		while(sc.hasNext()){
			int n=sc.nextInt();
			String cur=sc.next();
			int a=cur.charAt(1)-'0'+10*(cur.charAt(0)-'0');
			int b=cur.charAt(4)-'0'+10*(cur.charAt(3)-'0');
			if(n==12){
				if(a==0){
					++a;
				}
				while(a>12){
					a=a-10;
				}
				while(b>=60){
					b=b-10;
				}
			}
			else{
				while(a>23){
					a=a-10;
				}
				while(b>59){
					b=b-10;
				}
			}
			System.out.printf("%02d:%02d\n",a,b);
		}
	}

}
</strong>


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值