在 2D 平面中从原点到达形状 (d, 0) 点所需的跳跃次数

给定三个正整数a, b和d 。您当前位于无限二维坐标平面上的原点 (0, 0)。您可以在 2D 平面上的任何点上跳跃，距离等于a或b 的欧几里得距离当前位置。任务是找到从 (0, 0) 到达 (d, 0) 所需的最小跳跃次数。
例子：

Input : a = 2, b = 3, d = 1 
Output : 2
First jump of length a = 2, (0, 0) -> (1/2, √15/2)
Second jump of length a = 2, (1/2, √15/2) -> (1, 0)
Thus, only two jump are required to reach 
(1, 0) from (0, 0).

Input : a = 3, b = 4, d = 11 
Output : 3
(0, 0) -> (4, 0) using length b = 4
(4, 0) -> (8, 0) using length b = 4
(8, 0) -> (11, 0) using length a = 3

首先，观察我们不需要找到中间点，我们只想要所需的最少跳跃次数。
所以，从 (0, 0) 我们将在 (d, 0) 的方向上移动，即垂直跳跃长度等于max(a, b)直到当前位置坐标和 (d, 0) 之间的欧几里得距离变得小于max(a, b)或等于 0。这将使每次跳跃所需的最小跳跃次数增加 1。所以这个值可以通过floor(d/max(a, b)) 找到。
现在，对于距离的休息，如果（d，0）是欧几里得距离，我们将长度的一个跳跃，到达（d，0）。因此，在这种情况下，所需的最小跳转次数将增加 1。
现在，让我们解决剩余距离不等于a 的情况。让剩下的距离为x 。此外，观察 x 将大于 0 且小于max(a, b) ，因此，0 < x < 2*max(a, b)。我们可以通过 2 步到达 (d, 0)。
如何 ？
让我们尝试构建一个三角形 ABC，其中 A 是我们当前的位置，B 是目标位置，即 (d, 0)，而 C 将是 AC = BC = max(a, b) 的点。这是可能的三角形，因为两条边 AC + BC 的总和大于第三边 AB。因此，一次跳转是从 A 到 C，另一次从 C 跳转到 B。
下面是这个方法的实现：

C++

#include 
using namespace std;
 
// Return the minimum jump of length either a or b
// required to reach (d, 0) from (0, 0).
int minJumps(int a, int b, int d)
{
    // Assigning maximum of a and b to b
    // and assigning minimum of a and b to a.
    int temp = a;
    a = min(a, b);
    b = max(temp, b);
 
    // if d is greater than or equal to b.
    if (d >= b)
        return (d + b - 1) / b;
 
    // if d is 0
    if (d == 0)
        return 0;
 
    // if d is equal to a.
    if (d == a)
        return 1;
 
    // else make triangle, and only 2
    // steps required.
    return 2;
}
 
int main()
{
    int a = 3, b = 4, d = 11;
    cout << minJumps(a, b, d) << endl;
    return 0;
}

Java

// Java code to find the minimum number
// of jump required to reach
// (d, 0) from (0, 0).
import java.io.*;
 
class GFG {
 
    // Return the minimum jump of length either a or b
    // required to reach (d, 0) from (0, 0).
    static int minJumps(int a, int b, int d)
    {
        // Assigning maximum of a and b to b
        // and assigning minimum of a and b to a.
        int temp = a;
        a = Math.min(a, b);
        b = Math.max(temp, b);
 
        // if d is greater than or equal to b.
        if (d >= b)
            return (d + b - 1) / b;
 
        // if d is 0
        if (d == 0)
            return 0;
 
        // if d is equal to a.
        if (d == a)
            return 1;
 
        // else make triangle, and only 2
        // steps required.
        return 2;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int a = 3, b = 4, d = 11;
        System.out.println(minJumps(a, b, d));
    }
}
 
// This code is contributed by vt_m

Python3

# Python3 code to find the minimum
# number of jump required to reach
# (d, 0) from (0, 0)
 
def minJumps(a, b, d):
     
    temp = a
    a = min(a, b)
    b = max(temp, b)
     
    if (d >= b):
        return (d + b - 1) / b
     
    # if d is 0
    if (d == 0):
        return 0
 
    # if d is equal to a.
    if (d == a):
        return 1
 
    # else make triangle, and
    # only 2  steps required.
    return 2
         
# Driver Code
a, b, d = 3, 4, 11
print (int(minJumps(a, b, d)))
 
# This code is contributed by _omg

C#

// C# code to find the minimum number
// of jump required to reach
// (d, 0) from (0, 0).
using System;
 
class GFG {
 
    // Return the minimum jump of length either a or b
    // required to reach (d, 0) from (0, 0).
    static int minJumps(int a, int b, int d)
    {
        // Assigning maximum of a and b to b
        // and assigning minimum of a and b to a.
        int temp = a;
        a = Math.Min(a, b);
        b = Math.Max(temp, b);
 
        // if d is greater than or equal to b.
        if (d >= b)
            return (d + b - 1) / b;
 
        // if d is 0
        if (d == 0)
            return 0;
 
        // if d is equal to a.
        if (d == a)
            return 1;
 
        // else make triangle, and only 2
        // steps required.
        return 2;
    }
     
    // Driver code
    public static void Main()
    {
        int a = 3, b = 4, d = 11;
        Console.WriteLine(minJumps(a, b, d));
    }
}
 
// This code is contributed by vt_m

PHP

= $b)
        return ($d + $b - 1) / $b;
 
    // if d is 0
    if ($d == 0)
        return 0;
 
    // if d is equal to a.
    if ($d == $a)
        return 1;
 
    // else make triangle,
    // and only 2
    // steps required.
    return 2;
}
 
    // Driver Code
    $a = 3;
    $b = 4;
    $d = 11;
    echo floor(minJumps($a, $b, $d));
     
// This code is contributed by anuj_67.
?>

Javascript

 
// JavaScript  program to find the minimum number
// of jump required to reach
// (d, 0) from (0, 0).
 
    // Return the minimum jump of length either a or b
    // required to reach (d, 0) from (0, 0).
    function minJumps(a, b, d)
    {
        // Assigning maximum of a and b to b
        // and assigning minimum of a and b to a.
        let temp = a;
        a = Math.min(a, b);
        b = Math.max(temp, b);
   
        // if d is greater than or equal to b.
        if (d >= b)
            return (d + b - 1) / b;
   
        // if d is 0
        if (d == 0)
            return 0;
   
        // if d is equal to a.
        if (d == a)
            return 1;
   
        // else make triangle, and only 2
        // steps required.
        return 2;
    }
 
// Driver Code
 
        let a = 3, b = 4, d = 11;
        document.write(Math.floor(minJumps(a, b, d)));
 

输出：

3

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。