小于或等于N的A，B或C的倍数计数

给定四个整数N ， A ， B和C。任务是从[1，N]范围内找到整数，该整数可以被A ， B或C整除。

例子：

Input: A = 2, B = 3, C = 5, N = 10 
Output: 8 
2, 3, 4, 5, 6, 8, 9 and 10 are the only number from the 
range [1, 10] which are divisible by wither 2, 3 or 5.

Input: A = 7, B = 3, C = 5, N = 100 
Output: 55 
 

方法：一种有效的方法是使用集合论的概念。因为我们必须找到可以被a或b或c整除的数字。

• 设n(a)：被a整除的数的计数。
• 令n(b)：被b整除的数的计数。
• 令n(c)：可被c整除的数字计数。
• n(a∩b)：被a和b整除的数字计数。
• n(a∩c)：被a和c整除的数字计数。
• n(b∩c)：被b和c整除的数字计数。
• n(a b b c)：被a和b与c整除的数字计数。

根据集合论

n(a b c)= n(a)+ n(b)+ n(c)– n(a b)– n(b c)– n(a c)+ n(a b ∩c)

所以。被A，B或C整除的数字计数为(num / A)+(num / B)+(num / C)–(num / lcm(A，B))–(num / lcm(A，B) ))–(num / lcm(A，C))+ –(num / lcm(A，B，C)))

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the
// gcd of a and b
long gcd(long a, long b)
{
    if (a == 0)
        return b;
 
    return gcd(b % a, a);
}
 
// Function to return the count of integers
// from the range [1, num] which are
// divisible by either a, b or c
long divTermCount(long a, long b, long c, long num)
{
    // Calculate the number of terms divisible by a, b
    // and c then remove the terms which are divisible
    // by both (a, b) or (b, c) or (c, a) and then
    // add the numbers which are divisible by a, b and c
    return ((num / a) + (num / b) + (num / c)
            - (num / ((a * b) / gcd(a, b)))
            - (num / ((c * b) / gcd(c, b)))
            - (num / ((a * c) / gcd(a, c)))
            + (num / ((a * b * c) / gcd(gcd(a, b), c))));
}
 
// Driver code
int main()
{
    long a = 7, b = 3, c = 5, n = 100;
 
    cout << divTermCount(a, b, c, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
     
class GFG
{
     
// Function to return the
// gcd of a and b
static long gcd(long a, long b)
{
    if (a == 0)
        return b;
 
    return gcd(b % a, a);
}
 
// Function to return the count of integers
// from the range [1, num] which are
// divisible by either a, b or c
static long divTermCount(long a, long b,
                         long c, long num)
{
    // Calculate the number of terms divisible by a, b
    // and c then remove the terms which are divisible
    // by both (a, b) or (b, c) or (c, a) and then
    // add the numbers which are divisible by a, b and c
    return ((num / a) + (num / b) + (num / c) -
                (num / ((a * b) / gcd(a, b))) -
                (num / ((c * b) / gcd(c, b))) -
                (num / ((a * c) / gcd(a, c))) +
                (num / ((a * b * c) / gcd(gcd(a, b), c))));
}
 
// Driver code
static public void main (String []arr)
{
    long a = 7, b = 3, c = 5, n = 100;
 
    System.out.println(divTermCount(a, b, c, n));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach
 
# Function to return the
# gcd of a and b
def gcd(a, b) :
 
    if (a == 0) :
        return b;
 
    return gcd(b % a, a);
 
def lcm (x, y):
    return (x * y) // gcd (x, y)
 
# Function to return the count of integers
# from the range [1, num] which are
# divisible by either a, b or c
def divTermCount(a, b, c, num) :
 
    # Calculate the number of terms divisible by a, b
    # and c then remove the terms which are divisible
    # by both (a, b) or (b, c) or (c, a) and then
    # add the numbers which are divisible by a, b and c
    return (num // a + num // b + num // c -
                 num // lcm(a, b) -
                 num // lcm(c, b) -
                 num // lcm(a, c) +
                 num // (lcm(lcm(a, b), c)))
 
# Driver code
if __name__ == "__main__" :
 
    a = 7; b = 3; c = 5; n = 100;
 
    print(divTermCount(a, b, c, n));
 
# This code is contributed by AnkitRai01

C#

// C# implementation for above approach
using System;
     
class GFG
{
     
// Function to return the
// gcd of a and b
static long gcd(long a, long b)
{
    if (a == 0)
        return b;
 
    return gcd(b % a, a);
}
 
// Function to return the count of integers
// from the range [1, num] which are
// divisible by either a, b or c
static long divTermCount(long a, long b,
                         long c, long num)
{
    // Calculate the number of terms divisible by a, b
    // and c then remove the terms which are divisible
    // by both (a, b) or (b, c) or (c, a) and then
    // add the numbers which are divisible by a, b and c
    return ((num / a) + (num / b) + (num / c) -
            (num / ((a * b) / gcd(a, b))) -
            (num / ((c * b) / gcd(c, b))) -
            (num / ((a * c) / gcd(a, c))) +
            (num / ((a * b * c) / gcd(gcd(a, b), c))));
}
 
// Driver code
static public void Main (String []arr)
{
    long a = 7, b = 3, c = 5, n = 100;
 
    Console.WriteLine(divTermCount(a, b, c, n));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript implementation of the approach
 
// Function to return the
// gcd of a and b
function gcd(a, b)
{
    if (a == 0)
        return b;
 
    return gcd(b % a, a);
}
 
// Function to return the count of integers
// from the range [1, num] which are
// divisible by either a, b or c
function divTermCount(a, b, c, num)
{
     
    // Calculate the number of terms divisible by a, b
    // and c then remove the terms which are divisible
    // by both (a, b) or (b, c) or (c, a) and then
    // add the numbers which are divisible by a, b and c
    return Math.ceil(((num / a) + (num / b) + (num / c) -
                      (num / ((a * b) / gcd(a, b))) -
                      (num / ((c * b) / gcd(c, b))) -
                      (num / ((a * c) / gcd(a, c))) +
                      (num / ((a * b * c) / gcd(gcd(a, b), c)))));
}
 
// Driver code
n = 13;
var a = 7, b = 3, c = 5, n = 100;
 
document.write(divTermCount(a, b, c, n));
 
// This code is contributed by SoumikMondal
 

输出：

55

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