范围内两个数字的公倍数的计数

给定从L到R的范围，并且在从L到R的范围内，每个Xth磁贴都被涂成黑色，并且每个Yth磁贴都被涂成白色。如果一个磁贴同时被涂成白色和黑色，则认为它被涂成灰色。任务是找到在L到R(包括两端)范围内呈灰色显示的图块数量。
例子：

Input: X = 2, Y = 3, L = 6, R = 18
Output: 3
The grey coloured tiles are numbered 6, 12, 18

Input: X = 1, Y = 4, L = 5, R = 10
Output: 1
The only grey coloured tile is 8.

方法：由于X的每个倍数都是黑色，Y的每个倍数都是白色。 X和Y的倍数的任何图块都是灰色的。可被X和Y整除的术语是可被X和Y的Icm整除的术语。
Lcm可以使用以下公式求出：

lcm = (x*y) / gcd(x, y)

可以使用Euclid算法在登录时间计算GCD。可以使用以下常见技巧找到L到R范围内的1cm的倍数：

count(L, R) = count(R) - count(L-1)

被K除以N的项数为：

floor(N/K)

以下是查找灰色图块数量的实现：

C++

// C++ implementation to find the number of
// grey tiles
#include 
using namespace std;
 
// Function to count the numbe ro fgrey tiles
int findTileCount(int x, int y, int l, int r)
{
    int lcm = (x * y) / __gcd(x, y);
 
    // Number multiple of lcm less than L
    int countl = (l - 1) / lcm;
 
    // Number of multiples of lcm less than R+1
    int countr = r / lcm;
    return countr - countl;
}
 
// Driver code
int main()
{
    int x = 2, y = 3, l = 6, r = 18;
    cout << findTileCount(x, y, l, r);
    return 0;
}

Java

// Java  implementation to find the
// number of grey tiles
 
import java.io.*;
 
class GFG {
    // Function to count the number
// of grey tiles
static int findTileCount(int x, int y,
                         int l, int r)
{
    int lcm = (x * y) / __gcd(x, y);
 
    // Number multiple of lcm less than L
    int countl = (l - 1) / lcm;
 
    // Number of multiples of
    // lcm less than R+1
    int countr = r / lcm;
    return countr - countl;
}
 
static int __gcd(int a, int b)
{
     
    // Everything divides 0
    if (a == 0)
    return b;
    if (b == 0)
    return a;
 
    // base case
    if (a == b)
        return a;
 
    // a is greater
    if (a > b)
        return __gcd(a - b, b);
         
    return __gcd(a, b - a);
}
 
// Driver code
     
     
    public static void main (String[] args) {
 
    int x = 2, y = 3, l = 6, r = 18;
        System.out.println(findTileCount(x, y, l, r));
}
}
 
// This code is contributed ajit

Python3

# Python3 implementation to find the number of
# grey tiles
 
# from math lib import gcd method
from math import gcd
 
# Function to count the numbe of grey tiles
def findTileCount(x, y, l, r) :
 
    lcm = (x * y) // gcd(x, y)
 
    # Number multiple of lcm less than L
    count1 = (l - 1) // lcm
 
    # Number of multiples of lcm less than R+1
    countr = r // lcm
 
    return countr - count1
 
 
 
# Driver code
if __name__ == "__main__" :
 
    x, y, l, r = 2, 3, 6, 18
    print(findTileCount(x, y, l, r))
 
# This code is contributed by
# ANKITRAI1

C#

// C# implementation to find the
// number of grey tiles
using System;
 
class GFG
{
     
// Function to count the number
// of grey tiles
static int findTileCount(int x, int y,
                         int l, int r)
{
    int lcm = (x * y) / __gcd(x, y);
 
    // Number multiple of lcm less than L
    int countl = (l - 1) / lcm;
 
    // Number of multiples of
    // lcm less than R+1
    int countr = r / lcm;
    return countr - countl;
}
 
static int __gcd(int a, int b)
{
     
    // Everything divides 0
    if (a == 0)
    return b;
    if (b == 0)
    return a;
 
    // base case
    if (a == b)
        return a;
 
    // a is greater
    if (a > b)
        return __gcd(a - b, b);
         
    return __gcd(a, b - a);
}
 
// Driver code
public static void Main()
{
    int x = 2, y = 3, l = 6, r = 18;
    Console.Write(findTileCount(x, y, l, r));
}
}
 
// This code is contributed
// by Kirti_Mangal

PHP

 $b)
        return __gcd($a - $b, $b);
         
    return __gcd($a, $b - $a);
}
 
// Driver code
$x = 2; $y = 3; $l = 6; $r = 18;
echo findTileCount($x, $y, $l, $r);
     
// This code is contributed
// by Akanksha Rai(Abby_akku)
?>

Javascript

 
// JavaScript implementation to find the
// number of grey tiles
 
// Function to count the number
// of grey tiles
function findTileCount(x,y,l,r)
{
    lcm = parseInt((x * y) / __gcd(x, y));
 
    // Number multiple of lcm less than L
    countl = parseInt((l - 1) / lcm);
 
    // Number of multiples of
    // lcm less than R+1
    countr = parseInt(r / lcm);
    return countr - countl;
}
 
function __gcd(a, b)
{
     
    // Everything divides 0
    if (a == 0)
    return b;
    if (b == 0)
    return a;
 
    // base case
    if (a == b)
        return a;
 
    // a is greater
    if (a > b)
        return __gcd(a - b, b);
         
    return __gcd(a, b - a);
}
 
// Driver code
let x = 2;
let y = 3;
let l = 6;
let r = 18;
document.write(findTileCount(x, y, l, r));
 
// This code is contributed by bobby
 

输出：

3

时间复杂度： O(log(x * y))