当N除以[1，N]范围内的所有数字时，不同余数的计数

给定一个整数N ，任务是找到总的不同余数的计数，当N被范围[1，N]中的每个元素除时可以得到。

例子：

Input: N = 5
Output: 3
5 % 1 = 0
5 % 2 = 1
5 % 3 = 2
5 % 4 = 1
5 % 5 = 0
The distinct remainders are 0, 1 and 2.

Input: N = 44
Output: 22

方法：可以很容易地观察到，对于N的偶数，不同的余数为N / 2 ，对于N的奇数，其为1 +⌊N/2⌋ 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the count of distinct
// remainders that can be obtained when
// n is divided by every element
// from the range [1, n]
int distinctRemainders(int n)
{
  
    // If n is even
    if (n % 2 == 0)
        return (n / 2);
  
    // If n is odd
    return (1 + (n / 2));
}
  
// Driver code
int main()
{
    int n = 5;
  
    cout << distinctRemainders(n);
  
    return 0;
}

Java

// Java implementation of the above approach 
class GFG 
{ 
  
// Function to return the count of distinct
// remainders that can be obtained when
// n is divided by every element
// from the range [1, n]
static int distinctRemainders(int n)
{
  
    // If n is even
    if (n % 2 == 0)
        return (n / 2);
  
    // If n is odd
    return (1 + (n / 2));
}
  
// Driver code 
public static void main(String[] args) 
{ 
    int n = 5;
    System.out.println(distinctRemainders(n));
} 
} 
  
// This code is contributed by Mohit Kumar

Python3

# Python3 implementation of the approach
  
# Function to return the count of distinct 
# remainders that can be obtained when 
# n is divided by every element 
# from the range [1, n]
def distinctRemainders(n):
      
    # If n is even
    if n % 2 == 0:
        return n//2
      
    # If n is odd
    return ((n//2)+1)
  
# Driver code
if __name__=="__main__":
  
    n = 5
    print(distinctRemainders(n))

C#

// C# implementation of the above approach 
using System;
  
class GFG
{
      
    // Function to return the count of distinct
    // remainders that can be obtained when
    // n is divided by every element
    // from the range [1, n]
    static int distinctRemainders(int n)
    {
      
        // If n is even
        if (n % 2 == 0)
            return (n / 2);
      
        // If n is odd
        return (1 + (n / 2));
    }
      
    // Driver code 
    public static void Main() 
    { 
        int n = 5;
        Console.WriteLine(distinctRemainders(n));
    } 
}
  
// This code is contributed by AnkitRai01

输出：

3

时间复杂度： O(1)