codeforces 916 A Jamie and Alarm Snooze

Description

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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

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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

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

Input	Output
3 11 23	2

Input	Output
5 01 07	0

Note

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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.

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题意：Jamie要在某一时刻起床，但希望在某一个含7的幸运时刻被闹钟吵醒，然后在恰好经过若干个x分钟的小睡后到达该时刻。
只需要模拟一个表从起床时刻开始，每次把时间向前调x分钟，最先遇到的含7的时刻就是答案。

#include 

class clock {
public:
    int x;
    int hh;
    int mm;
    bool luck() {
        if (hh % 10 == 7||mm%10==7)return true;//含7的只可能是个位
        else return false;
    }
    void dec() {//模拟时钟
        mm = mm - x;
        if (mm < 0) {
            mm += 60;
            hh--;
        }
        if (hh < 0)hh += 24;//前一天
    }
};

int main() {
    clock time;
    scanf("%d", &time.x);
    scanf("%d %d", &time.hh, &time.mm);
    int t;
    for (t = 0; !time.luck(); ++t)  time.dec();
    printf("%d", t);
}