可用于表示N的不同正整数的最大数量

给定一个整数N ，任务是找到可用于表示N的最大不同正整数个数。

例子：

Input: N = 5 
Output: 2 
5 can be represented as 1 + 4, 2 + 3, 3 + 2, 4 + 1 and 5. 
So maximum integers that can be used in the representation are 2.

Input: N = 10 
Output: 4 

方法：我们总是可以贪婪地选择不同的整数，使其尽可能地小，以使可以使用的不同整数的数量最大化。如果我们使用的是前x个自然数，则将它们的和设为f(x)。
因此，我们需要找到一个最大值x，使得f(x)<= n。

1 + 2 + 3 + … n < = n 
x*(x+1)/2 < = n 
x^2+x-2n < = 0 
We can solve the above equation by using quadratic formula X = (-1 + sqrt(1+8*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 required count
int count(int n)
{
    return int((-1 + sqrt(1 + 8 * n)) / 2);
}
 
// Driver code
int main()
{
    int n = 10;
 
    cout << count(n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
    // Function to return the required count
    static int count(int n)
    {
        return (int)(-1 + Math.sqrt(1 + 8 * n)) / 2;
 
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 10;
     
        System.out.println(count(n));
    }
}
 
// This code is contributed by ihritik

Python3

# Python3 implementation of the approach
from math import sqrt
 
# Function to return the required count
def count(n) :
 
    return (-1 + sqrt(1 + 8 * n)) // 2;
 
# Driver code
if __name__ == "__main__" :
 
    n = 10;
 
    print(count(n));
 
# This code is contributed by AnkitRai01

C#

// C# implementation of approach
using System;
 
class GFG
{
     
    // Function to return the required count
    public static int count(int n)
    {
        return (-1 + (int)Math.Sqrt(1 + 8 * n)) / 2;
    }
 
    // Driver Code
    public static void Main()
    {
        int n = 10;
     
        Console.Write(count(n));
    }
}
 
// This code is contributed by Mohit Kumar

Javascript

 
// Javascript implementation of the approach
 
// Function to return the required count
function count(n)
{
    return parseInt((-1 + Math.sqrt(1 + 8 * n)) / 2);
}
 
// Driver code
var n = 10;
 
document.write(count(n));
 
// This code is contributed by rutvik_56
 

输出：

4

时间复杂度： O(1)