检查N阶乘是否可被X ^ Y整除

给定三个整数N，X和Y，任务是检查是否为N！可被X ^Y整除

例子：

Input: N = 10, X = 2, Y = 8
Output: YES
Explanation:
Factorial of 10 is – 3628800
and the value of X^Y = 2⁸ = 256
Since, 3628800 is divisible by 256, therefore answer is YES.

Input: N = 5, X = 2, Y = 4
Output: NO
Explanation:
The Factorial of 5 is – 120
and the value of X^Y = 2⁴ = 16
Since, 3628800 is not divisible by 16, therefore answer is NO.

方法：想法是分别找到N阶乘和X ^Y的值，然后检查N阶乘的值是否可整除X ^Y。

算法：

• 计算N阶乘的值
• 找到X ^Y的值。
• 检查N的阶乘是否可被X ^Y整除。

注意：此方法不适用于较大的N值。

下面是上述方法的实现：

C++

// CPP implementation to check if
// the value of the N! % X^Y == 0
#include
using namespace std;
  
      
    // Function to check if N! % X^Y == 0
    void check(int n,int x, int y){
        int fact = 1;
          
        // Loop to calculate N-factorial
        for (int i = 2; i <= n; i++) {
            fact *= i;
        }
  
        int divisor = pow(x, y);
          
        // Condition to check
        if (fact % divisor == 0)
            cout << "YES";
        else
            cout << "NO";
          
    }
      
    // Driver Code
        int main()
    {
        int n = 10;
        int x = 2;
        int y = 8;
          
        // Function Call
        check(n, x, y);
    }
  
// This code is contributed by Surendra_Gangwar

Java

// Java implementation to check if
// the value of the N! % X^Y == 0
import java.util.*;
import java.lang.*;
  
class divisible {
      
    // Function to check if N! % X^Y == 0
    public static void check(int n, 
                         int x, int y){
        long fact = 1;
          
        // Loop to calculate N-factorial
        for (int i = 2; i <= n; i++) {
            fact *= i;
        }
  
        long divisor = (long)Math.pow(x, y);
          
        // Condition to check
        if (fact % divisor == 0)
            System.out.println("YES");
        else
            System.out.println("NO");
          
    }
      
    // Driver Code
    public static void main(String args[])
    {
        int n = 10;
        int x = 2;
        int y = 8;
          
        // Function Call
        check(n, x, y);
    }
}

Python3

# Python3 implementation to check if 
# the value of the N! % X^Y == 0
      
# Function to check if N! % X^Y == 0 
def check(n, x, y) :
    fact = 1;
      
    # Loop to calculate N-factorial
    for i in range(2, n + 1) :
        fact *= i;
    divisor = x ** y;
          
    # Condition to check
    if (fact % divisor == 0) :
        print("YES");
    else :
        print("NO"); 
  
# Driver Code 
if __name__ == "__main__" : 
      
    n = 10; 
    x = 2; 
    y = 8; 
          
    # Function Call 
    check(n, x, y);
  
# This code is contributed by Yash_R

C#

// C# implementation to check if
// the value of the N! % X^Y == 0
using System;
  
class divisible {
       
    // Function to check if N! % X^Y == 0
    public static void check(int n, 
                         int x, int y){
        long fact = 1;
           
        // Loop to calculate N-factorial
        for (int i = 2; i <= n; i++) {
            fact *= i;
        }
   
        long divisor = (long)Math.Pow(x, y);
           
        // Condition to check
        if (fact % divisor == 0)
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
           
    }
       
    // Driver Code
    public static void Main(String []args)
    {
        int n = 10;
        int x = 2;
        int y = 8;
           
        // Function Call
        check(n, x, y);
    }
}
  
// This code is contributed by 29AjayKumar

输出：

YES

性能分析：

• 时间复杂度： O(N)
• 辅助空间： O(1)。