Intel Code Challenge Elimination Round (Div.1 + Div.2, combined) A. Broken Clock 水题-优快云博客

针对一个显示错误时间的损坏时钟，本篇介绍了一个通过最小化数字更改次数来修正时间的算法。该算法适用于12小时制和24小时制，并提供了一种简单有效的解决方法。

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;
}

转载于:https://www.cnblogs.com/qscqesze/p/5927530.html