codeforces A Karen and Morning 傻瓜模拟

最新推荐文章于 2020-10-05 14:27:13 发布

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最新推荐文章于 2020-10-05 14:27:13 发布

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本文介绍了一个算法问题，即如何找到从当前时间开始最近的一个回文时间，并给出了具体的实现方法。通过遍历后续时间，判断是否为回文来确定最小等待时间。

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A. Karen and Morning
time limit per test2 seconds
memory limit per test512 megabytes
inputstandard input
outputstandard output
Karen is getting ready for a new school day!

It is currently hh:mm, given in a 24-hour format. As you know, Karen loves palindromes, and she believes that it is good luck to wake up when the time is a palindrome.

What is the minimum number of minutes she should sleep, such that, when she wakes up, the time is a palindrome?

Remember that a palindrome is a string that reads the same forwards and backwards. For instance, 05:39 is not a palindrome, because 05:39 backwards is 93:50. On the other hand, 05:50 is a palindrome, because 05:50 backwards is 05:50.

Input
The first and only line of input contains a single string in the format hh:mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59).

Output
Output a single integer on a line by itself, the minimum number of minutes she should sleep, such that, when she wakes up, the time is a palindrome.

Examples
input
05:39
output
11
input
13:31
output
0
input
23:59
output
1
Note
In the first test case, the minimum number of minutes Karen should sleep for is 11. She can wake up at 05:50, when the time is a palindrome.

In the second test case, Karen can wake up immediately, as the current time, 13:31, is already a palindrome.

In the third test case, the minimum number of minutes Karen should sleep for is 1 minute. She can wake up at 00:00, when the time is a palindrome.

让你找到后面最近的回文数。。。直接加就好

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int h,d;
    scanf("%d:%d",&h,&d);
    int sum=0;
    for(int j=d;j<=59;j++)
        {
            if((h/10==j%10)&&(h%10==j/10))
            {
                printf("%d\n",sum );
                return 0;
            }
            sum++;
        }
    if(h==23)
    {
        h=-1;
    }
    for(int i=h+1;i<=23;i++)
    {
        for(int j=0;j<=59;j++)
        {
            if((i/10==j%10)&&(i%10)==(j/10))
            {
                printf("%d\n",sum );
                return 0;
            }
            sum++;
        }
    }
}