合适的数字

给定一个数字N ，任务是检查N是否是一个可接受的数字。如果N是一个可接受的数字，则打印“是”，否则打印“否” 。

Amenable numbers are numbers if there exists a multiset of integers whose size, sum and product equal to the number. 
For example, 8 is an Amenable Number because there is a multiset of integers {-1, -1, 1, 1, 1, 1, 2, 4} whose size, sum and product is 8. 
 

例子：

Input: N = 8 
Output: Yes 
Explanation: 
8 is an Amenable Number because there is a multiset of integers {-1, -1, 1, 1, 1, 1, 2, 4} whose size, sum and product is 8.
Input: N = 30 
Output: No 
Explanation: 
30 is not an Amenable Number because there doesn’t exists a multiset of integers whose size, sum and product is 30. 
 

方法：
前几个可允许编号是1、5、8、9、12、13、16、17、20、21 …..可以表示为4K或4K +1形式。
因此，任何形式为4K或4K +1的数字N都是可接受的数字。
下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if N
// is Amenable number
bool isAmenableNum(int N)
{
 
    // Return true if N is of the form
    // 4K or 4K + 1
    return (N % 4 == 0
            || (N - 1) % 4 == 0);
}
 
// Driver Code
int main()
{
    // Given Number
    int n = 8;
 
    // Function Call
    if (isAmenableNum(n))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function to check if N
// is Amenable number
static boolean isAmenableNum(int N)
{
     
    // Return true if N is of the form
    // 4K or 4K + 1
    return (N % 4 == 0 || (N - 1) % 4 == 0);
}
 
// Driver code
public static void main(String[] args)
{
    int n = 8;
     
    if (isAmenableNum(n))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}
 
// This code is contributed by shubham

Python3

# Python3 program for the above approach
import math
 
# Function to check if N
# is Amenable number
def isAmenableNum(N):
 
    # Return true if N is of the
    # form 4K or 4K + 1
    return (N % 4 == 0 or
           (N - 1) % 4 == 0);
 
# Driver code
 
# Given number
N = 8;
 
# Function call
if (isAmenableNum(N)):
    print("Yes");
else:
    print("No");
 
# This code is contributed by rock_cool

C#

// C# program for the above approach
using System;
class GFG{
  
// Function to check if N
// is Amenable number
static bool isAmenableNum(int N)
{
      
    // Return true if N is of the form
    // 4K or 4K + 1
    return (N % 4 == 0 || (N - 1) % 4 == 0);
}
  
// Driver code
public static void Main(String[] args)
{
    int n = 8;
      
    if (isAmenableNum(n))
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
}
}
 
// This code is contributed by amal kumar choubey

Javascript

// Javascript program for the above approach
 
    // Function to check if N
    // is Amenable number
    function isAmenableNum( N)
    {
 
        // Return true if N is of the form
        // 4K or 4K + 1
        return (N % 4 == 0 || (N - 1) % 4 == 0);
    }
 
    // Driver code
    let n = 8;
 
    if (isAmenableNum(n)) {
        document.write("Yes");
    } else {
        document.write("No");
    }
 
 
// This code is contributed by Rajput-Ji.

输出：

Yes

时间复杂度： O(1)