Consider two cars A and B, running infinitely (either clockwise or anti-clockwise) on a circular road. Given the speed of both the cars a and b. If a or b is positive, indicate they are moving in clockwise, Else they are moving in the anti-clockwise direction. The task is to find the number of distinct points they will meet each other at.
Examples:
Input : a = 1, b = -1 Output : 2 Explanation Car A is moving clockwise while Car B is moving anti-clockwise but their speeds are same, so they will meet at two points i.e at the starting point and diametrically corresponding opposite point on the road. Input : a = 1, b = 2 Output : 1
Approach :
Let the circumference of the circular road be d.
Let the time taken by cars A and B be ta and tb respectively. Their relative speed is a – b.
A and B start from the starting point and after some time, they will meet at starting point again. This time can be calculated by the LCM of ta and tb. Within this time period, they may meet at certain points which needs to be found out. Observe that, after they meet at the starting point they keep on meeting at the same point.
Time taken to meet again at the starting point will be,
T1 = LCM(ta, tb) = LCM(d/a, d/b) = d/GCD(a, b)
Let them meet N times in the time period T1.
So, the time delay between their consecutive meets is, say T2 can be calculated as,
T2 = (T1 / N).
This time can be calculated by calculating the time taken to meet for the first time after they start.
So, the time taken by them to meet for the first time,
Therefore, T2 = (d / (a – b)).
Dividing T1 by T2, we get,
N = (T1 / T2) = ((a – b) / GCD(a, b))
Below is implementation for the above approach:
C++
// CPP Program to find number of distinct point of meet on a circular road#include <bits/stdc++.h>using namespace std;// Returns the GCD of two number.int gcd(int a, int b){ int c = a % b; while (c != 0) { a = b; b = c; c = a % b; } return b;}// Returns the number of distinct meeting points.int numberOfmeet(int a, int b){ int ans; // Find the relative speed. if (a > b) ans = a - b; else ans = b - a; // convert the negative value to positive. if (a < 0) a = a * (-1); if (b < 0) b = b * (-1); return ans / gcd(a, b);}// Driver Codeint main(){ int a = 1, b = -1; cout << numberOfmeet(a, b) << endl; return 0;} |
Java
// Java Program to find number // of distinct point of meet // on a circular roadimport java.io.*;class GFG { // Returns the GCD // of two number.static int gcd(int a, int b){ int c = a % b; while (c != 0) { a = b; b = c; c = a % b; } return b;}// Returns the number of// distinct meeting points.static int numberOfmeet(int a, int b){ int ans; // Find the relative speed. if (a > b) ans = a - b; else ans = b - a; // convert the negative // value to positive. if (a < 0) a = a * (-1); if (b < 0) b = b * (-1); return ans / gcd(a, b);}// Driver Codepublic static void main (String[] args){ int a = 1, b = -1; System.out.println(numberOfmeet(a, b));}}// This code is contributed by @ajit |
Python3
# Python3 Program to find # number of distinct point # of meet on a circular roadimport math# Returns the number of# distinct meeting points.def numberOfmeet(a, b): ans = 0; # Find the relative speed. if (a > b): ans = a - b; else: ans = b - a; # convert the negative # value to positive. if (a < 0): a = a * (-1); if (b < 0): b = b * (-1); return int(ans / math.gcd(a, b));# Driver Codea = 1;b = -1;print(numberOfmeet(a, b));# This code is contributed by mits |
C#
// C# Program to find number // of distinct point of meet // on a circular roadusing System;class GFG{// Returns the GCD // of two number.static int gcd(int a, int b){ int c = a % b; while (c != 0) { a = b; b = c; c = a % b; } return b;}// Returns the number of// distinct meeting points.static int numberOfmeet(int a, int b){ int ans; // Find the relative speed. if (a > b) ans = a - b; else ans = b - a; // convert the negative // value to positive. if (a < 0) a = a * (-1); if (b < 0) b = b * (-1); return ans / gcd(a, b);}// Driver Codestatic public void Main (){ int a = 1, b = -1; Console.WriteLine( numberOfmeet(a, b));}}// This code is contributed // by @ajit |
PHP
<?php// PHP Program to find number// of distinct point of meet// on a circular road// Returns the GCD of two number.function gcd($a, $b){ $c = $a % $b; while ($c != 0) { $a = $b; $b = $c; $c = $a % $b; } return $b;}// Returns the number of// distinct meeting points.function numberOfmeet($a, $b){ $ans; // Find the relative speed. if ($a > $b) $ans = $a - $b; else $ans = $b - $a; // convert the negative // value to positive. if ($a < 0) $a = $a * (-1); if ($b < 0) $b = $b * (-1); return $ans / gcd($a, $b);}// Driver Code$a = 1;$b = -1;echo numberOfmeet($a, $b)."\n"; // This code is contributed by mits?> |
Javascript
<script>// Javascript Program to find number of distinct// point of meet on a circular road// Returns the GCD of two number.function gcd(a, b){ var c = a % b; while (c != 0) { a = b; b = c; c = a % b; } return b;}// Returns the number of distinct meeting points.function numberOfmeet(a, b){ var ans; // Find the relative speed. if (a > b) ans = a - b; else ans = b - a; // convert the negative value to positive. if (a < 0) a = a * (-1); if (b < 0) b = b * (-1); return ans / gcd(a, b);}// Driver Codevar a = 1, b = -1;document.write( numberOfmeet(a, b));</script> |
Output:
2
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