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查找模拟时钟中时针和分针之间的角度

发布于 2024-08-31 03:35:58 字数 162 浏览 3 评论 0原文

最近我被问到这样一个面试问题：

给定一个 12 小时模拟时钟，计算时针和分针之间较小角度的度数。尽可能精确。

我想知道最简单、最易读、最精确的算法是什么。欢迎使用任何语言的解决方案（但如果您认为有必要，请稍微解释一下）。

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农村范ル 2024-09-07 03:35:59

该问题被称为“时钟角度问题”，我们需要找到时钟之间的角度模拟时钟在特定时间的指针（小时和分钟）。

C 编程语言的解决方案。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<inttypes.h>
#include<assert.h>

#define STRING_LENGTH 6

double angle_between_hour_min_hand(char[]);

int main(void) {
    uint8_t test;
    printf("Enter the number of test cases\n");
    scanf("%"SCNu8,&test);
    assert(test>0);
    while(test--) {
        char time_digital[STRING_LENGTH];
        printf("Enter the time\n");
        scanf("%s",time_digital);
        double angle_between_hands_deg = angle_between_hour_min_hand(time_digital);
        abs(angle_between_hands_deg) < angle_between_hands_deg ? printf("%0.1f\n",angle_between_hands_deg) : printf("%d\n",abs(angle_between_hands_deg));
    }
    return 0;
}

double angle_between_hour_min_hand(char time_digital[]) {
    uint8_t hr,min;
    double hr_angle_deg,min_angle_deg,angle_between_hands_deg;
    char*buffer = calloc(sizeof(char),STRING_LENGTH);
    if(buffer) {
        snprintf(buffer,STRING_LENGTH,"%s",time_digital);
        hr = atoi(__strtok_r(buffer,":",&buffer));
        min = atoi(__strtok_r(NULL,":",&buffer));
        buffer -= strlen(time_digital);
        free(buffer);
        hr_angle_deg = (double)(30*hr) + (double) (0.5*min);
        // printf("hr-angle: %f\n", hr_angle_deg);
        min_angle_deg = 6*min;
        // printf("min-angle: %f\n", min_angle_deg);
        angle_between_hands_deg = (hr_angle_deg > min_angle_deg) ? hr_angle_deg - min_angle_deg : min_angle_deg - hr_angle_deg;
        if(angle_between_hands_deg > 180) {
            angle_between_hands_deg = 360 - angle_between_hands_deg;
        }
    }
    else fprintf(stderr,"Memory not allocated to the buffer pointer!\n");
    return angle_between_hands_deg;
}

在你的系统中编译上面的程序，我使用的是Ubuntu 18.04 LTS Bionic Beaver，你可以使用任何安装了C编译器的系统。

gcc -Wall -g clock_angle_sol.c -o clock_angle_sol
./clock_angle_sol
Enter the time in 12-hour or 24 hour i.e (hr:min) format: 12:45
Angle: 112.00 degrees.

注释：
1. 等式 $\theta_{hr} = (30^\circ \times hour) + (0.5^\circ \times 分钟)$ 将为您提供 12 小时制时针所成的角度。
2. 如果要计算 24 小时制中时针所成的角度，请使用以下公式： $\theta_{hr} = (15^\circ \times 小时) + (0.25^\circ \times 分钟)$
3. 秒针对分针的旋转也有贡献，但我们忽略了它，因为贡献很小，即1/10 = 0.1。

The problem is known as a “Clock Angle Problem” where we need to find the angle between the hands (hour & minute) of an analog clock at a particular time.

Solution in C Programming Language.

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<inttypes.h>
#include<assert.h>

#define STRING_LENGTH 6

double angle_between_hour_min_hand(char[]);

int main(void) {
    uint8_t test;
    printf("Enter the number of test cases\n");
    scanf("%"SCNu8,&test);
    assert(test>0);
    while(test--) {
        char time_digital[STRING_LENGTH];
        printf("Enter the time\n");
        scanf("%s",time_digital);
        double angle_between_hands_deg = angle_between_hour_min_hand(time_digital);
        abs(angle_between_hands_deg) < angle_between_hands_deg ? printf("%0.1f\n",angle_between_hands_deg) : printf("%d\n",abs(angle_between_hands_deg));
    }
    return 0;
}

double angle_between_hour_min_hand(char time_digital[]) {
    uint8_t hr,min;
    double hr_angle_deg,min_angle_deg,angle_between_hands_deg;
    char*buffer = calloc(sizeof(char),STRING_LENGTH);
    if(buffer) {
        snprintf(buffer,STRING_LENGTH,"%s",time_digital);
        hr = atoi(__strtok_r(buffer,":",&buffer));
        min = atoi(__strtok_r(NULL,":",&buffer));
        buffer -= strlen(time_digital);
        free(buffer);
        hr_angle_deg = (double)(30*hr) + (double) (0.5*min);
        // printf("hr-angle: %f\n", hr_angle_deg);
        min_angle_deg = 6*min;
        // printf("min-angle: %f\n", min_angle_deg);
        angle_between_hands_deg = (hr_angle_deg > min_angle_deg) ? hr_angle_deg - min_angle_deg : min_angle_deg - hr_angle_deg;
        if(angle_between_hands_deg > 180) {
            angle_between_hands_deg = 360 - angle_between_hands_deg;
        }
    }
    else fprintf(stderr,"Memory not allocated to the buffer pointer!\n");
    return angle_between_hands_deg;
}

Compile the above program in your system, I used Ubuntu 18.04 LTS Bionic Beaver, you can use any system which has a C compiler installed.

gcc -Wall -g clock_angle_sol.c -o clock_angle_sol
./clock_angle_sol
Enter the time in 12-hour or 24 hour i.e (hr:min) format: 12:45
Angle: 112.00 degrees.

Notes:
1. The equation $\theta_{hr} = (30^\circ \times hour) + (0.5^\circ \times minute)$ will give you the angle made by hour hand in 12-hour clock.
2. If you want to calculate the angle made by hour hand in a 24-hour clock, then use the following equation: $\theta_{hr} = (15^\circ \times hour) + (0.25^\circ \times minute)$
3. Second hand also contribute to the rotation of the minute hand, but we ignored it because the contribution is insignificant i.e. 1/10 = 0.1.

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仅一夜美梦 2024-09-07 03:35:59

这是一种解决方案 (C#)。这是一个非常简单的解决方案，忽略了精度。希望解决方案是不言自明的。

public static double GetAngle(int hourHand, int minuteHand)
    {
        double oneMinuteAngle = (360 / 60);
        double oneHourAngle = (360 / 12);

        double hourAngle = oneHourAngle * hourHand;
        double minuteAngle = oneMinuteAngle * minuteHand;

        return (Math.Abs(hourAngle - minuteAngle));
    }

This is one solution (C#). This is a very simple solution and ignores precision. Hope the solution is self explanatory.

public static double GetAngle(int hourHand, int minuteHand)
    {
        double oneMinuteAngle = (360 / 60);
        double oneHourAngle = (360 / 12);

        double hourAngle = oneHourAngle * hourHand;
        double minuteAngle = oneMinuteAngle * minuteHand;

        return (Math.Abs(hourAngle - minuteAngle));
    }

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窝囊感情。 2024-09-07 03:35:59

不知道这样对不对。？

//m*360/60 - (h*360/24)+(m*360/(24*60)) ->
t = abs(25*m - 60*h)/4
t = min(t,360-t)

I do not know if it's right, .something like this?

//m*360/60 - (h*360/24)+(m*360/(24*60)) ->
t = abs(25*m - 60*h)/4
t = min(t,360-t)

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も让我眼熟你 2024-09-07 03:35:59

求时针和分针之间的角度是

angle=(hour*5-min)*6

for finding the angle between the hour hand and the minute hand is

angle=(hour*5-min)*6

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留一抹残留的笑 2024-09-07 03:35:58

事实证明，维基百科确实有最好的答案：

// h = 1..12, m = 0..59
static double angle(int h, int m) {
    double hAngle = 0.5D * (h * 60 + m);
    double mAngle = 6 * m;
    double angle = Math.abs(hAngle - mAngle);
    angle = Math.min(angle, 360 - angle);
    return angle;
}

基本上：

时针以每分钟0.5度的速度移动
分针以每分钟6度的速度移动每分钟

问题已解决。

并且精度不是问题，因为小数部分是 .0 或 .5，并且在 0..360 范围内，所有这些值都可以用double精确表示。

It turns out that Wikipedia does have the best answer:

// h = 1..12, m = 0..59
static double angle(int h, int m) {
    double hAngle = 0.5D * (h * 60 + m);
    double mAngle = 6 * m;
    double angle = Math.abs(hAngle - mAngle);
    angle = Math.min(angle, 360 - angle);
    return angle;
}

Basically:

The hour hand moves at the rate of 0.5 degrees per minute
The minute hand moves at the rate of of 6 degrees per minute

Problem solved.

And precision isn't a concern because the fractional part is either .0 or .5, and in the range of 0..360, all of these values are exactly representable in double.

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油饼 2024-09-07 03:35:58

为了找到时钟指针之间的角度，

30 * [HRS - (MIN/5)] + (MIN/2)

For finding the angle between the hands of a clock is ,

30 * [HRS - (MIN/5)] + (MIN/2)

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空城旧梦 2024-09-07 03:35:58

Polygenlubricants的java代码与我的类似。假设时钟是 12 小时制而不是 24

小时制。如果是 24 小时制，那就另当别论了。另外，还有另一个假设，假设在我们计算时时钟是否停止。

一个时钟周期是360度。

分针每分钟可以走多少度？ 360 / 60 = 每分钟 6 度。
时针每小时可以走多少度？ 360/12 = 每小时 30 度（因为时针走得比分钟慢）

由于以“分钟”为单位更容易计算，我们可以得到

“时针每分钟可以走多少度”？

30 / 60 = 每分钟 0.5 度。

因此，如果您知道如何获得这些数字，那么问题就可以通过这个部分数学解决方案得到解决。

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鱼窥荷 2024-09-07 03:35:58

试试这个代码：

import java.util.Scanner;

class Clock{

    public static void main(String args[]){
        int hours,mins;

    System.out.println("Enter the Time(hours) : ");
        Scanner dx = new Scanner(System.in);
        hours = dx.nextInt();

    System.out.println("Enter the time(mins) : ");
        Scanner fx = new Scanner(System.in);
        mins = fx.nextInt();

    if(hours>=0 && hours<=12){

        if(mins>=0 && mins<=59){
            double hDegrees = (hours * 30) + (mins * 0.5);
                    double mDegrees = mins * 6;
                    double diff  = Math.abs(hDegrees - mDegrees);

        System.out.println("The angle between sticks is (degrees) : "+diff);
                if (diff > 180){ 

                diff = 360 - diff;
        System.out.println("The angle between sticks is (degrees) : "+diff);
                }

        }

    }

    else{
        System.out.println("Wrong input ");
    }


}

}

Try this code :

import java.util.Scanner;

class Clock{

    public static void main(String args[]){
        int hours,mins;

    System.out.println("Enter the Time(hours) : ");
        Scanner dx = new Scanner(System.in);
        hours = dx.nextInt();

    System.out.println("Enter the time(mins) : ");
        Scanner fx = new Scanner(System.in);
        mins = fx.nextInt();

    if(hours>=0 && hours<=12){

        if(mins>=0 && mins<=59){
            double hDegrees = (hours * 30) + (mins * 0.5);
                    double mDegrees = mins * 6;
                    double diff  = Math.abs(hDegrees - mDegrees);

        System.out.println("The angle between sticks is (degrees) : "+diff);
                if (diff > 180){ 

                diff = 360 - diff;
        System.out.println("The angle between sticks is (degrees) : "+diff);
                }

        }

    }

    else{
        System.out.println("Wrong input ");
    }


}

}

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这个俗人 2024-09-07 03:35:58

分角（从 12 点钟开始）：360 * 分钟 / 60

时角（从 12 点钟开始）：360 * (小时 % 12) / 12 + 360 * (分钟 / 60) * (1 / 12)

小时和分钟之间的角度: (小时角度 - 分钟角度) % 360
通过简单的算术，这可以减少到 30 * 小时 - 5.5 * 分钟。

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妖妓 2024-09-07 03:35:58

    **php code for find angle via time (minutes and hour's)**

    echo calcAngle(3,70);

function calcAngle($h, $m)
{
    // validate the input
    if ($h <0 || $m < 0 || $h >12 || $m > 60)
      {
       return "Wrong input";
      }
      else {

    if ($h == 12) $h = 0;
    if ($m == 60) $m = 0;

    $hour_angle = 0.5 * ($h*60 + $m);
    $minute_angle = 6*$m;
    $angle = abs($hour_angle - $minute_angle);
    $angle = min(360-$angle, $angle);

    return $angle;
}
}

    **php code for find angle via time (minutes and hour's)**

    echo calcAngle(3,70);

function calcAngle($h, $m)
{
    // validate the input
    if ($h <0 || $m < 0 || $h >12 || $m > 60)
      {
       return "Wrong input";
      }
      else {

    if ($h == 12) $h = 0;
    if ($m == 60) $m = 0;

    $hour_angle = 0.5 * ($h*60 + $m);
    $minute_angle = 6*$m;
    $angle = abs($hour_angle - $minute_angle);
    $angle = min(360-$angle, $angle);

    return $angle;
}
}

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