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;
}