Two people start races from two different points p1 and p2. They cover s1 and s2 meters in a jump. Find if they will ever meet at a point after the same number of jumps.

**Examples:**

Input : p1 = 6, s1 = 3, p2 = 8, s2 = 2 Output : Yes Explanation: 6->9->12 8->10->12 They meet after two jumps. Input : p1 = 4, s1 = 4, p2 = 8, s2 = 2 Output : Yes Explanation: 4->8->12 8->10->12 Input : p1 = 0, s1 = 2, p2 = 5, s2 = 3 Output : No Input : p1 = 42, s1 = 3, p2 = 94, s2 = 2 Output : Yes

A **simple solution** is to make them jump one by one. After every jump, see if they are same point or not.

An **efficient solution** is based on below facts:

Since starting points are always different, they will meet if following conditions are met.

(1) Speeds are not same

(2) Difference between speeds divide the total distance between initial points.

## C++

`// C++ program to find any one of them` `// can overtake the other` `#include<bits/stdc++.h>` `using` `namespace` `std;` ` ` `// function to find if any one of them can` `// overtake the other` `bool` `sackRace(` `int` `p1, ` `int` `s1, ` `int` `p2, ` `int` `s2){` ` ` ` ` `// Since starting points are always` ` ` `// different, they will meet if following ` ` ` `// conditions are met.` ` ` `// (1) Speeds are not same` ` ` `// (2) Difference between speeds divide the` ` ` `// total distance between initial points. ` ` ` `return` `( (s1 > s2 && (p2 - p1) % (s1 - s2) == 0) ||` ` ` `(s2 > s1 && (p1 - p2) % (s2 - s1) == 0));` `}` ` ` `// driver program` `int` `main()` `{` ` ` `int` `p1 = 4, s1 = 4, p2 = 8, s2 = 2;` ` ` `sackRace(p1, s1, p2, s2)? cout << ` `"Yes\n"` `:` ` ` `cout << ` `"No\n"` `;` ` ` `return` `0;` `}` |

## Java

`// java program to find any one of them` `// can overtake the other` `import` `java.util.Arrays;` ` ` `public` `class` `GFG {` ` ` ` ` `// function to find if any one of` ` ` `// them can overtake the other` ` ` `static` `boolean` `sackRace(` `int` `p1, ` `int` `s1,` ` ` `int` `p2, ` `int` `s2)` ` ` `{` ` ` ` ` `// Since starting points are ` ` ` `// always different, they will` ` ` `// meet if following conditions` ` ` `// are met.` ` ` `// (1) Speeds are not same` ` ` `// (2) Difference between speeds` ` ` `// divide the total distance ` ` ` `// between initial points. ` ` ` `return` `( (s1 > s2 && (p2 - p1) %` ` ` `(s1 - s2) == ` `0` `) ||` ` ` `(s2 > s1 && (p1 - p2)` ` ` `% (s2 - s1) == ` `0` `));` ` ` `}` ` ` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `int` `p1 = ` `4` `, s1 = ` `4` `, p2 = ` `8` `, s2 = ` `2` `;` ` ` ` ` `if` `(sackRace(p1, s1, p2, s2))` ` ` `System.out.println(` `"Yes"` `);` ` ` `else` ` ` `System.out.println(` `"No"` `);` ` ` ` ` `}` `}` ` ` `// This code is contributed by Sam007.` |

## Python3

`# python program to find any one of them` `# can overtake the other` ` ` `# function to find if any one of them can` `# overtake the other` `def` `sackRace(p1, s1, p2, s2):` ` ` ` ` `# Since starting points are always` ` ` `# different, they will meet if following ` ` ` `# conditions are met.` ` ` `# (1) Speeds are not same` ` ` `# (2) Difference between speeds divide the` ` ` `# total distance between initial points. ` ` ` `return` `( (s1 > s2 ` `and` `(p2 ` `-` `p1) ` `%` `(s1 ` `-` `s2) ` `=` `=` `0` `) ` ` ` `or` `(s2 > s1 ` `and` `(p1 ` `-` `p2) ` `%` `(s2 ` `-` `s1) ` `=` `=` `0` `))` ` ` ` ` `# driver program` `p1 ` `=` `4` `s1 ` `=` `4` `p2 ` `=` `8` `s2 ` `=` `2` `if` `(sackRace(p1, s1, p2, s2)):` ` ` `print` `(` `"Yes"` `)` `else` `:` ` ` `print` `(` `"No"` `)` ` ` `# This code is contributed by Sam007` |

## C#

`// C# program to find any one of them` `// can overtake the other` `using` `System;` ` ` `class` `GFG {` ` ` ` ` `// function to find if any one of` ` ` `// them can overtake the other` ` ` `static` `bool` `sackRace(` `int` `p1, ` `int` `s1,` ` ` `int` `p2, ` `int` `s2)` ` ` `{` ` ` ` ` `// Since starting points are ` ` ` `// always different, they will` ` ` `// meet if following conditions` ` ` `// are met.` ` ` `// (1) Speeds are not same` ` ` `// (2) Difference between speeds` ` ` `// divide the total distance ` ` ` `// between initial points. ` ` ` `return` `( (s1 > s2 && (p2 - p1) %` ` ` `(s1 - s2) == 0) ||` ` ` `(s2 > s1 && (p1 - p2)` ` ` `% (s2 - s1) == 0));` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `Main() ` ` ` `{` ` ` `int` `p1 = 4, s1 = 4, p2 = 8,` ` ` `s2 = 2;` ` ` ` ` `if` `(sackRace(p1, s1, p2, s2))` ` ` `Console.WriteLine(` `"Yes"` `);` ` ` `else` ` ` `Console.WriteLine(` `"No"` `);` ` ` ` ` `}` `}` ` ` `// This code is contributed by Sam007.` |

## PHP

`<?php` `// PHP program to find any one of them` `// can overtake the other` ` ` `// function to find if any one of ` `// them can overtake the other` `function` `sackRace(` `$p1` `, ` `$s1` `, ` `$p2` `, ` `$s2` `)` `{` ` ` ` ` `// Since starting points are always` ` ` `// different, they will meet if following ` ` ` `// conditions are met.` ` ` `// (1) Speeds are not same` ` ` `// (2) Difference between speeds divide the` ` ` `// total distance between initial points. ` ` ` `return` `((` `$s1` `> ` `$s2` `&& (` `$p2` `- ` `$p1` `) % (` `$s1` `- ` `$s2` `) == 0) ||` ` ` `(` `$s2` `> ` `$s1` `&& (` `$p1` `- ` `$p2` `) % (` `$s2` `- ` `$s1` `) == 0));` `}` ` ` ` ` `// Driver Code` ` ` `$p1` `= 4;` ` ` `$s1` `= 4;` ` ` `$p2` `= 8;` ` ` `$s2` `= 2;` ` ` `if` `(sackRace(` `$p1` `, ` `$s1` `, ` `$p2` `, ` `$s2` `))` ` ` `echo` `"Yes\n"` `;` ` ` `else` ` ` `echo` `"No\n"` `; ` ` ` `// This code is contributed by Sam007` `?>` |

**Output :**

Yes

This article is contributed by **Vishal Kumar Gupta**. 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.

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