C - C CodeForces - 916A（模拟）

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.

Examples
Input
3
11 23
Output
2
Input
5
01 07
Output
0

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.

思路

• 在模拟时候，当h <0 的时候要让 h == 23 开始继续模拟（0:00 就是晚上的24:00）

代码

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

#define ll long long 
#define mod int(1e9 + 7)   //这里不要忘了加括号
#define rl rt << 1
#define rr rt << 1 | 1
const int mxn = 100005;
int n;
ll pre[mxn << 2];
int ar[mxn];

bool judge(int a, int b)
{
    if(a % 10 == 7 || b % 10 == 7 || a / 10 == 7 || b / 10 == 7)
        return true;
    return false;
}


int main()
{
    /* freopen("A.txt","r",stdin); */
    /* freopen("Ans.txt","w",stdout); */
    int x;
    scanf("%d", &x);
    int h, m;
    scanf("%d %d", &h, &m);
    for(int i = 0; ; i ++)
    {
        if(judge(h, m))
        {
            printf("%d\n", i);
            break;
        }
        m -= x;
        if(m < 0)
        {
            m += 60;
            h --;
            if(h < 0)
                h = 23;
        }
    }

    return 0;
}