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

Javascript

 
// Javascript 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]
function distinctRemainders(n)
{
 
    // If n is even
    if (n % 2 == 0)
        return parseInt(n / 2);
 
    // If n is odd
    return (1 + parseInt(n / 2));
}
 
// Driver code
    let n = 5;
 
    document.write(distinctRemainders(n));
 

输出：

3

时间复杂度： O(1)