Codeforces816A Karen and Morning 2017-06-27 15:11 43人阅读 评论(0) 收藏

A. Karen and Morning

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard 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.

————————————————————————————————

题目的意思是给你一个时间，问过了多少时间才会让现在时间变成回文

模拟时间流逝 每一秒判一次是否回文

#include <iostream>

#include <cstdio>

#include <cstring>

#include <string>

#include <algorithm>

#include <vector>

#include <queue>

#include <stack>

#include <map>

#include <set>

#include <cmath>

using namespace std;

#define LL long long

const int inf=0x7fffffff;

bool check(int h,int m)

{

    if(h/10+(h%10)*10==m)

        return 1;

    return 0;

}

int main()

{

   int h,m;

   while(~scanf("%d:%d",&h,&m))

   {

       int ans=0;

       while(!check(h,m))

       {

           ans++;

           m++;

           if(m==60)

            h+=1,m=0;

           if(h==24)

            h=0;

       }

       printf("%d\n",ans);

   }

    return 0;

}

或者干脆蠢一点根据时间分类讨论

#include <iostream>

#include <cstdio>

#include <cstring>

#include <string>

#include <algorithm>

#include <vector>

#include <queue>

#include <stack>

#include <map>

#include <set>

#include <cmath>

using namespace std;

#define LL long long

const int inf=0x7fffffff;

int main()

{

    int h,m;

    while(~scanf("%d:%d",&h,&m))

    {

        int t=h*60+m;

        int ans;

        if(h<5||(h==5&&m<=50))

        {

            int fm=h*10;

            if(m<=fm)

            {

                ans=fm-m;

            }

            else

            {

                fm=(h+1)*10;

                ans=(h+1)*60+fm-t;

            }

        }

        else if(h<=9)

        {

            ans=10*60+1-t;

        }

        else if(h<15||(h==15&&m<=51))

        {

            int fm=(h%10)*10+(h/10);

            if(m<=fm)

            {

                ans=fm-m;

            }

            else

            {

                fm=((h+1)%10)*10+((h+1)/10);

                ans=(h+1)*60+fm-t;

            }

        }

        else if(h<=19)

        {

            ans=20*60+2-t;

        }

        else if(h<23||(h==23&&m<=32))

        {

            int fm=(h%10)*10+(h/10);

            if(m<=fm)

            {

                ans=fm-m;

            }

            else

            {

                fm=((h+1)%10)*10+((h+1)/10);

                ans=(h+1)*60+fm-t;

            }

        }

        else

        {

            ans=24*60-t;

        }

        printf("%d\n",ans);

    }

    return 0;

}

Codeforces816A Karen and Morning 2017-06-27 15:11 43人阅读 评论(0) 收藏的相关教程结束。