Jamie and Alarm Snooze(暴力+思维)

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.
InputThe 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).
OutputPrint the minimum number of times he needs to press the button.
ExamplesInput

3
11 23

Output

2

Input

5
01 07

Output

0

NoteIn 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.

题意：一个人喜欢睡觉，他在设置某个时间起床，如果不是幸运数字的话，如果h或者m不带有7，就会每隔x分钟响一次。他不想被打扰，所i以不是幸运数的时间，就要按一次闹钟，才能继续睡。所以要求他定的闹钟时间和前面最近的的幸运数的时间，需要按的次数，就是最少需要按多少次闹钟。

#include<bits/stdc++.h>
using namespace std;
int main(){
int x,h,m;
while(cin>>x){
int ans=0;
cin>>h>>m;
if(m%10==7||h%10==7) cout<<"0"<<endl;
else{
while(h%10!=7&&m%10!=7){
m-=x;
ans++;
if(m<0){
m+=60;
h--;
if(h<0) h+=24;
}
}
cout<<ans<<endl;
}
}
return 0;
}