环形道路上不同的会合点数

考虑两辆汽车A和B ，它们在环形道路上无限地行驶(顺时针或逆时针)。给定汽车a和b的速度。如果a或b为正，则指示它们沿顺时针方向移动，否则指示它们沿逆时针方向移动。任务是找到彼此相遇的不同点的数量。

例子：

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

必备条件：GCD |液晶模组

方法：
令圆形道路的周长为d 。
假设汽车A和B花费的时间分别为t _a和t _b 。它们的相对速度为a – b 。

A和B从起点开始，一段时间后，它们将再次在起点相遇。可以通过t _a和t _b的LCM来计算该时间。在此时间段内，他们可能会在某些特定地点见面。观察到，他们在起点见面后便继续在同一地点见面。

在起点再次开会所需的时间是

T1 = LCM(ta, tb) = LCM(d/a, d/b) = d/GCD(a, b)

让他们在时间段T _1中相遇N次。
因此，他们两次连续见面之间的时间延迟为T _2，可以这样计算：

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))

以下是上述方法的实现：

C++

// CPP Program to find number of distinct point of meet on a circular road
#include 
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 Code
int 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 road
import 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 Code
public 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 road
import 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 Code
a = 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 road
using 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 Code
static public void Main ()
{
    int a = 1, b = -1;
    Console.WriteLine(
            numberOfmeet(a, b));
}
}
  
// This code is contributed 
// by @ajit

PHP

 $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
?>

输出：

2