Jamie and Alarm Snooze

寻找幸运时间

最新推荐文章于 2018-03-27 20:20:55 发布

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本文介绍了一个有趣的算法问题，主角 Jamie 需要在特定时间醒来，为了减少起床痛苦，他选择在一个包含数字'7'的“幸运”时间设定闹钟，并通过按下小睡按钮每 x 分钟唤醒一次直到目标时间。文章探讨了如何找到最小次数按下小睡按钮的方法。

Jamie and Alarm Snooze

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 lucky time 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: mm contains 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.

Example

Input

3
11 23

Output

Input

5
01 07

Output

Note

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.

唯一需要注意的是如果小时变成零，再借为要变成23，用循环查找即可

code：

#include <iostream>
#include <cstdio>
using namespace std;
int judge(int hh,int mm){
    if(hh%10==7||hh/10==7||mm%10==7||mm/10==7)
        return 1;
    else
        return 0;
}
int main(){
    int x,y=0;
    int hh,mm;
    scanf("%d",&x);
    scanf("%d%d",&hh,&mm);
    while(1){
        if(judge(hh,mm)){
            printf("%d\n",y);
            break;
        }
        y++;
        mm -= x;
        if(mm<0){
            hh--;
            mm += 60;
        }
        if(hh<0){
            hh = 23;
        }
    }
    return 0;
}