Given three positive integers **a, b ** and **d**. You are currently at origin (0, 0) on infinite 2D coordinate plane. You are allowed to jump on any point in the 2D plane at euclidean distance either equal to **a** or **b** from your current position. The task is to find the minimum number of jump required to reach (d, 0) from (0, 0).

Examples:

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

First, observe we don’t need to find the intermediate points, we only want the minimum number of jumps required.

So, from (0, 0) we will move in direction of (d, 0) i.e vertically with the jump of length equal to **max(a, b)** until the Euclidean distance between the current position coordinate and (d, 0) either became less than **max(a, b)** or equal to 0. This will increase the minimum number of jumps required by 1 at each jump. So this value can be found by **floor(d/max(a, b))**.

Now, for the rest of the distance, if (d, 0) is **a** Euclidean distance away, we will make one jump of length **a** and reach the (d, 0). So, the minimum number of jumps required will be increased by 1 in this case.

Now, let’s solve the case if rest of the distance is not equal to **a**. Let the rest of distance left be **x**. Also, observe x will be greater than 0 and less than **max(a, b)**, therefore, 0 < x < 2*max(a, b). We can reach to (d, 0) in 2 steps.

**How ?**

Lets try to build a triangle ABC such that A is our current position, B is target position i.e (d, 0) and C will be the point such that AC = BC = max(a, b). And this is possible to triangle because sum of two side AC + BC is greater than third side AB. So, one jump is from A to C and another jump from C to B.

Below is the implementation of this approach:

## C++

`#include <bits/stdc++.h> ` `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; ` `} ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## PHP

`<?php ` `// PHP code 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` `, ` `$d` `) ` `{ ` ` ` ` ` `// Assigning maximum of ` ` ` `// a and b to b and ` ` ` `// assigning minimum of ` ` ` `// a and b to a. ` ` ` `$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` `== ` `$a` `) ` ` ` `return` `1; ` ` ` ` ` `// else make triangle, ` ` ` `// and only 2 ` ` ` `// steps required. ` ` ` `return` `2; ` `} ` ` ` ` ` `// Driver Code ` ` ` `$a` `= 3; ` ` ` `$b` `= 4; ` ` ` `$d` `= 11; ` ` ` `echo` `floor` `(minJumps(` `$a` `, ` `$b` `, ` `$d` `)); ` ` ` `// This code is contributed by anuj_67. ` `?> ` |

*chevron_right*

*filter_none*

**Output**

3

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.

## Recommended Posts:

- Check whether a given point lies inside a triangle or not
- How to check if a given point lies inside or outside a polygon?
- Maximum number of 2x2 squares that can be fit inside a right isosceles triangle
- Number of Integral Points between Two Points
- Triangle with no point inside
- Klee's Algorithm (Length Of Union Of Segments of a line)
- Optimum location of point to minimize total distance
- n'th Pentagonal Number
- Count of parallelograms in a plane
- Paper Cut into Minimum Number of Squares
- Check whether a point exists in circle sector or not.
- Find an Integer point on a line segment with given two ends
- Check if four segments form a rectangle
- Minimum block jumps to reach destination
- How to check if given four points form a square
- Number of Triangles that can be formed given a set of lines in Euclidean Plane
- Find the Missing Point of Parallelogram
- Program for Point of Intersection of Two Lines
- Centered cube number
- Find number of diagonals in n sided convex polygon

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.