宇宙第一小仙女\(^o^)/~~萌量爆表求带飞=≡Σ((( つ^o^)つ~dalao们点个关注呗~~
emmm都喊我铁头,哼!!!而我的床呢偏偏就是磁铁打的........but每天还是要起床.....比如更博,therefore....没事多来瞅瞅给我增点访客量咯~这个叫Jamie的小盆友还真是像我一样爱睡觉,这个闹铃延迟提醒当然也是很常见了,再来一分钟,一分钟之后就起.......嚯!还非得在带“7”的时候设闹钟。
不扯了不扯了,hint 一下咯~ 这题倒是不算难,但是很容易WA,特殊情况比较多,多试试特殊情况就差不多找到错了,比如23点之后没有24点,要变成0点,时分秒到60要进1。
Problem is coming ~~~~
Description
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.
Sample Input
3 11 23
2
5 01 07
0
Hint
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.
思路应该都有,就不发了,建议看代码之前还是自己先想想思路啦~
----------------------------------------------------------------我只是一条可爱的分界线---------------------------------------------------------------
#include<stdio.h>
int main()
{
int k=0,x,h,m,i,j;
scanf("%d",&x);
scanf("%d%d",&h,&m);
i=h;
for(j=m;;j-=x){
if(j<0){
i--;
j+=60;
}
if(i<0)
i+=24;
if(i%10==7||j%10==7){
break;
}
k++;
}
printf("%d\n",k);
return 0;
}
宇宙第一小仙女\(^o^)/~~萌量爆表求带飞=≡Σ((( つ^o^)つ~dalao们点个关注呗~~
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