Jamie and Alarm Snooze

Posted 2020-11-08 王陸

Description

Jamie loves sleeping. One day, he decides that he needs to wake up at exactly hh: mm. However, he hates waking up, so he wants to make waking up less painful by setting the alarm at a lucky time. He will then press the snooze button every x minutes until hh: mm is reached, and only then he will wake up. He wants to know what is the smallest number of times he needs to press the snooze button.

A time is considered lucky if it contains a digit ‘7‘. For example, 13: 07 and 17: 27 are lucky, while 00: 48 and 21: 34 are not lucky.

Note that it is not necessary that the time set for the alarm and the wake-up time are on the same day. It is guaranteed that there is a luckytime Jamie can set so that he can wake at hh: mm.

Formally, find the smallest possible non-negative integer y such that the time representation of the time x·y minutes before hh: mmcontains the digit ‘7‘.

Jamie uses 24-hours clock, so after 23: 59 comes 00: 00.

Input

The first line contains a single integer x (1 ≤ x ≤ 60).

The second line contains two two-digit integers, hh and mm (00 ≤ hh ≤ 23, 00 ≤ mm ≤ 59).

Output

Print the minimum number of times he needs to press the button.

Sample Input

Input

3
11 23

Output

2

Input

5
01 07

Output

0

Hint

In the first sample, Jamie needs to wake up at 11:23. So, he can set his alarm at 11:17. He would press the snooze button when the alarm rings at 11:17 and at 11:20.

In the second sample, Jamie can set his alarm at exactly at 01:07 which is lucky.

 

题目意思：一个人想要在hh：mm时刻起床，而他想从在离起床时刻最近的一个带有数字7的时刻起，每x分钟按一次按钮，问一共按了多少次

解题思路：模拟即可，注意小时与分钟间的切换。

 1 #include <iostream>
 2 #include <stdio.h>
 3 using namespace std;
 4 int main()
 5 {
 6     int n,h,m,count;
 7     while(scanf("%d%d%d",&n,&h,&m)!=EOF)
 8     {
 9         count=0;
10         while(1)
11         {
12             if(h==17||h==7||m%10==7)
13             {
14                 break;
15             }
16             else
17             {
18                 count++;
19                 m=m-n;
20                 if(m<0)
21                 {
22                     m=m+60;
23                     h--;
24                 }
25                 if(h<0)
26                 {
27                     h=h+24;
28                 }
29             }
30         }
31         printf("%d\n",count);
32     }
33     return 0;
34 }