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
 //2016.10.1
#include <iostream>
#include <cstdio>
#include <cstring> using namespace std; int main()
{
int format, hh, mm;
while(scanf("%d", &format)!=EOF)
{
scanf("%d:%d", &hh, &mm);
if(format==)
{
if(hh == )hh++;
else if(hh< || hh > )
{
if(hh% == )hh = ;
else hh %= ;
}
if(mm< || mm > )mm%=;
printf("%02d:%02d\n", hh, mm);
}else
{
if(hh< || hh > )hh%=;
if(mm< || mm > )mm%=;
printf("%02d:%02d\n", hh, mm);
}
} return ;
}

CodeForces 722A的更多相关文章

  1. CodeForces 722A Broken Clock (水题)

    题意:给定一个时间,然后改最少的数字,使得它成为12进制或者24进制. 析:24进制主要判是不是大于23,如果是把第一位变成0,12进制判是不是大于12,如果是再看第二位是不是0,是0,第一位变成1, ...

  2. 【26.34%】【codeforces 722A】Broken Clock

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

  3. Codeforces水题集合[14/未完待续]

    Codeforces Round #371 (Div. 2) A. Meeting of Old Friends |B. Filya and Homework A. Meeting of Old Fr ...

  4. python爬虫学习(5) —— 扒一下codeforces题面

    上一次我们拿学校的URP做了个小小的demo.... 其实我们还可以把每个学生的证件照爬下来做成一个证件照校花校草评比 另外也可以写一个物理实验自动选课... 但是出于多种原因,,还是绕开这些敏感话题 ...

  5. 【Codeforces 738D】Sea Battle(贪心)

    http://codeforces.com/contest/738/problem/D Galya is playing one-dimensional Sea Battle on a 1 × n g ...

  6. 【Codeforces 738C】Road to Cinema

    http://codeforces.com/contest/738/problem/C Vasya is currently at a car rental service, and he wants ...

  7. 【Codeforces 738A】Interview with Oleg

    http://codeforces.com/contest/738/problem/A Polycarp has interviewed Oleg and has written the interv ...

  8. CodeForces - 662A Gambling Nim

    http://codeforces.com/problemset/problem/662/A 题目大意: 给定n(n <= 500000)张卡片,每张卡片的两个面都写有数字,每个面都有0.5的概 ...

  9. CodeForces - 274B Zero Tree

    http://codeforces.com/problemset/problem/274/B 题目大意: 给定你一颗树,每个点上有权值. 现在你每次取出这颗树的一颗子树(即点集和边集均是原图的子集的连 ...

随机推荐

  1. 如何删除要素类 IFeatureWorkspace 接口介绍(1)

    如何删除要素类 要想删除一个要素类,那么必须先得到这个,在得到这个要素类的时候,我们要学习一个新的接口IFeatureWorkspace. IFeatureWorkspace  接口介绍 这个接口主要 ...

  2. 左倾堆(C#)

    参考:http://www.cnblogs.com/skywang12345/p/3638384.html using System; using System.Collections.Generic ...

  3. 【supervisord】部署单进程服务的利器

    最近听了一场分享,里边同事介绍了一些python开发环境常用到的几种工具,其中之一就supervisord,分享后自己做了些功课,概括一下supervisord是一个什么东西呢 1. 它是一个独立的常 ...

  4. Python+Selenium定位不到元素常见原因及解决办法(报:NoSuchElementException)

         在做web应用的自动化测试时,定位元素是必不可少的,这个过程经常会碰到定位不到元素的情况(报selenium.common.exceptions.NoSuchElementException ...

  5. mysql 插入数据后返回当前的自增ID方法

    存储过程的写法: mysql>create procedure test( ->in username varchar(50), ->in password varchar(50), ...

  6. Android之事件分发

    网上总结的很详细了,有时间总结下做个笔记

  7. 日志文件 统计 网站PV IP

    1. 安装rrdtool yum install rrdtool 2. 创建 rrdtool 数据库 rrdtool create /opt/local/rrdtool/jicki.rrd -s 30 ...

  8. Linux下的强大工具之一sed(转),Shell必备

    sed命令基本用法sed是一个非交互式文本编辑器,它可以对文本文件和标准输入进行编辑,标准输入可以是来自键盘输入.文件重定向.字符串.变量.来自管道的文本等等.sed从文本的一个文本行或标准输入中读取 ...

  9. nginx location配置(URL)

    语法规则: location [=|~|~*|^~] /uri/ { … }= 表示精确匹配,这个优先级也是最高的^~ 表示uri以某个常规字符串开头,理解为匹配 url路径即可.nginx不对url ...

  10. window.location.href 和 document.location.href

    document表示的是一个文档对象,window表示的是一个窗口对象,一个窗口下可以有多个文档对象. 所以一个窗口下只有一个window.location.href,但是可能有多个document. ...