A. Broken Clock

题目连接:

http://codeforces.com/contest/722/problem/A

Description

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.

Sample Input

24

17:30

Sample Output

17:30

Hint

题意

12小时制的话,时钟是从1开始到12的。

24小时制的话,时钟是从0开始到23的

然后给你一个小时制度,然后再给你一个时间,问你这个时间是否合法,如果不合法的话,你需要修改最少的数字,使得这个时间合法 。

题解:

智力低下的人,就像我一样,直接暴力枚举就好了,不要去想那么多……

代码

#include<bits/stdc++.h>
using namespace std; int main()
{
int h;scanf("%d",&h);
string s;cin>>s;
int ans1 = 1e9;
string ans2;
if(h==24){
for(int i=0;i<h;i++)
{
for(int j=0;j<60;j++)
{
string s2;
s2+=(i/10)%10+'0';
s2+=i%10+'0';
s2+=':';
s2+=(j/10)%10+'0';
s2+=j%10+'0';
int tmp = 0;
for(int k=0;k<s2.size();k++)
if(s2[k]!=s[k])tmp++;
if(ans1>tmp)ans1=tmp,ans2=s2;
}
}
}
else
{
for(int i=1;i<=h;i++)
{
for(int j=0;j<60;j++)
{
string s2;
s2+=(i/10)%10+'0';
s2+=i%10+'0';
s2+=':';
s2+=(j/10)%10+'0';
s2+=j%10+'0';
int tmp = 0;
for(int k=0;k<s2.size();k++)
if(s2[k]!=s[k])tmp++;
if(ans1>tmp)ans1=tmp,ans2=s2;
}
}
}
cout<<ans2<<endl;
}

Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) A. Broken Clock 水题的更多相关文章

  1. Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) A B C D 水 模拟 并查集 优先队列

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

  2. Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) B. Verse Pattern 水题

    B. Verse Pattern 题目连接: http://codeforces.com/contest/722/problem/B Description You are given a text ...

  3. Intel Code Challenge Elimination Round (Div.1 + Div.2, combined)

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

  4. Intel Code Challenge Elimination Round (Div.1 + Div.2, combined)(set容器里count函数以及加强for循环)

    题目链接:http://codeforces.com/contest/722/problem/D 1 #include <bits/stdc++.h> #include <iostr ...

  5. 二分 Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) D

    http://codeforces.com/contest/722/problem/D 题目大意:给你一个没有重复元素的Y集合,再给你一个没有重复元素X集合,X集合有如下操作 ①挑选某个元素*2 ②某 ...

  6. 线段树 或者 并查集 Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) C

    http://codeforces.com/contest/722/problem/C 题目大意:给你一个串,每次删除串中的一个pos,问剩下的串中,连续的最大和是多少. 思路一:正方向考虑问题,那么 ...

  7. Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) D. Generating Sets 贪心

    D. Generating Sets 题目连接: http://codeforces.com/contest/722/problem/D Description You are given a set ...

  8. Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) C. Destroying Array 带权并查集

    C. Destroying Array 题目连接: http://codeforces.com/contest/722/problem/C Description You are given an a ...

  9. Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) C. Destroying Array

    C. Destroying Array time limit per test 1 second memory limit per test 256 megabytes input standard ...

随机推荐

  1. HTML的文档类型:<!DOCTYPE >

    <!DOCTYPE> 声明:它不是 HTML 标签而且对大小写不敏感,而是指示 web 浏览器关于页面使用哪个 HTML 版本进行编写的指令.而且 声明必须是 HTML 文档的第一行,位于 ...

  2. php银行卡校验

    前言银行金卡,维萨和万事达.银联品牌,如果是贷记卡或准贷记卡,一定为16位卡号.而借记卡可以16-19位不等.美国运通卡则不论金卡或是白金卡.普通卡,都是15位卡号.16-19 位卡号校验位采用 Lu ...

  3. caoha

  4. 阿里云OSS 中文名称地址不对

    oss中将该中文名称重命名.再输入一次

  5. Longest Words

    Given a dictionary, find all of the longest words in the dictionary. Example Given { "dog" ...

  6. [uart]1.Linux中tty框架与uart框架之间的调用关系剖析

    转自:http://developer.51cto.com/art/201209/357501_all.htm 目录 1.tty框架 2.uart框架 3.自底向上 4.自顶向下 5.关系图 在这期间 ...

  7. Batch Normalization 与 Caffe中的 相关layer

    在机器学习领域,通常假设训练数据与测试数据是同分布的,BatchNorm的作用就是深度神经网络训练过程中, 使得每层神经网络的输入保持同分布. 原因:随着深度神经网络层数的增加,训练越来越困难,收敛越 ...

  8. Dede更新提示DedeTag Engine Create File False的解决办法

    第一种情况:列表.频道.文章等命名规则未填写或填写错误 此种情况较为少见,因为初级用户一般不会去修改这些东西,情况可以大致分为: 命名规则未填写(即为空)解决方法:只需填好相应的规则即可,重新选择栏目 ...

  9. 数组中累加和小于等于k的最长子数组

    问题描述: 给定一个无序数组arr,其中元素可正.可负.可0,给定一个整数 k.求arr所有的子数组中累加和小于或等于k的最长子数组长度.例如:arr=[3,-2,-4,0,6],k=-2,相加和小于 ...

  10. unit测试出现异常:Exception in thread "main" java.lang.NoSuchMethodError: org.junit.platform.commons.util

    在进行单元测试时,测试出现异常 Exception in thread "main" java.lang.NoSuchMethodError: org.junit.platform ...