Codeforces Round #457 (Div. 2) A. Jamie and Alarm Snooze简单模拟
2018-01-24 21:49
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A. Jamie and Alarm Snooze
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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
notlucky.
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 alucky 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: mmcontains 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
output
input
output
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.
题意:一个人想要在hh:mm时刻起床,而他想从在离起床时刻最近的一个带有数字7的时刻起,每x分钟按一次按钮,问一共按了多少次
模拟即可,注意小时与分钟间的切换,00:00的前时刻为23:59
#include <bits/stdc++.h>
using namespace std;
int main()
{
int x,hh,mm;
while (~scanf("%d %d %d",&x,&hh,&mm)){
int f=0,ans=0;
while (1){
if ((hh==17)||(hh==7)||(mm%10==7))
break;
ans++;
mm-=x;
if (mm<0){
mm+=60;
hh--;
}
if (hh<0)
hh=23;
}
printf("%d\n",ans);
}
return 0;
}
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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
notlucky.
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 alucky 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: mmcontains 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.
题意:一个人想要在hh:mm时刻起床,而他想从在离起床时刻最近的一个带有数字7的时刻起,每x分钟按一次按钮,问一共按了多少次
模拟即可,注意小时与分钟间的切换,00:00的前时刻为23:59
#include <bits/stdc++.h>
using namespace std;
int main()
{
int x,hh,mm;
while (~scanf("%d %d %d",&x,&hh,&mm)){
int f=0,ans=0;
while (1){
if ((hh==17)||(hh==7)||(mm%10==7))
break;
ans++;
mm-=x;
if (mm<0){
mm+=60;
hh--;
}
if (hh<0)
hh=23;
}
printf("%d\n",ans);
}
return 0;
}
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