[简单思维题] Jamie and Alarm Snooze

题目描述

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.

输入

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).

输出

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

样例输入

3
11 23

样例输出

2

提示

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.

题目大意：给定一个起床时间hhmm和每隔x分钟按一次闹钟，要你找到一个hhmm之前的幸运时刻使得起床前按闹钟的次数最少（从幸运时刻开始到起床时间结束按闹钟的次数最少）

这个题，因为涉及到时刻转化的问题，所以很容易错，需要考虑周到。

如下测试用例：

7

11 17（应输出0）

60

00 00 （应输出7）

7

20 50（应输出9）

基本思路还是很简单，就是将起床时间不断减小x分钟，并记录减了几次，直到一个幸运时刻为止，答案就是减的次数。（当然减的过程不能出现负数，需转化一下，容易出错）

AC代码：

#include<cstdio>
 
bool islucky(int m){
  if(m%10==7||m/10==7) return true;
  return false;
}
 
int main()
{
    int x;
    scanf("%d",&x);
    int h,m;
    scanf("%d%d",&h,&m);
    int ans=0;
    int sgn=0;
    if(islucky(h)||islucky(m)) {
        printf("0
");//如果hhmm就是幸运数，直接输出0就好了
        return 0;
    }
    while(!sgn){
      if(m<=0&&h>=1){
        m+=60;
        h-=1;
        if(islucky(h)||islucky(m)){
            if(m==60) ans++;//这个很重要，因为如果没有这句，如18：01减1后变到18：00也就是17：60这样的只算1次减就直接break掉了。
            sgn=1;
            break;
        }
      }
      else if(m<=0&&h<=0){
        m+=60;
        h=23;
        if(islucky(h)||islucky(m)){
            if(m==60) ans++;//同上
            sgn=1;
            break;
        }
      }
      else if(m>0){
        m-=x;
        ans++;
        if(islucky(h)||islucky(m)){
            sgn=1;
            break;
        }
      }
    }
    printf("%d
",ans);
    return 0;
}

总结：有很多细节要考虑，怎么说呢，多用极端用例测测吧，比如令x=60，令hhmm=00 00，同时令x=60 hhmm=00 00( = =！)