给定比赛次数，查找比赛中的球队数

给定一个整数M ，它是锦标赛中进行比赛的次数，并且每个参与团队都与其他所有团队进行过比赛。任务是查找比赛中有多少支球队。
例子：

Input: M = 3 
Output: 3 
If there are 3 teams A, B and C then 
A will play a match with B and C 
B will play a match with C (B has already played a match with A) 
C has already played matches with team A and B 
Total matches played are 3
Input: M = 45 
Output: 10 
 

方法：由于每次比赛都是在两支球队之间进行的。因此，此问题类似于从给定的N个对象中选择2个对象。因此，比赛的总次数将为C(N，2)，其中N是参赛球队的数量。所以，

M = C(N, 2) 
M = (N * (N – 1)) / 2 
N² – N – 2 * M = 0 
This is a quadratic equation of type ax² + bx + c = 0. Here a = 1, b = -1, c = 2 * M. Therefore, applying formula 
x = (-b + sqrt(b² – 4ac)) / 2a and x = (-b – sqrt(b² – 4ac)) / 2a 
N = [(-1 * -1) + sqrt((-1 * -1) – (4 * 1 * (-2 * M)))] / 2 
N = (1 + sqrt(1 + (8 * M))) / 2 and N = (1 – sqrt(1 + (8 * M))) / 2 
 

解完上述两个方程式后，我们将获得两个N值。一个值将为正，而一个值为负。忽略负值。因此，团队数量将是上述方程式的正根。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
#include 
using namespace std;
 
// Function to return the number of teams
int number_of_teams(int M)
{
    // To store both roots of the equation
    int N1, N2, sqr;
 
    // sqrt(b^2 - 4ac)
    sqr = sqrt(1 + (8 * M));
 
    // First root (-b + sqrt(b^2 - 4ac)) / 2a
    N1 = (1 + sqr) / 2;
 
    // Second root (-b - sqrt(b^2 - 4ac)) / 2a
    N2 = (1 - sqr) / 2;
 
    // Return the positive root
    if (N1 > 0)
        return N1;
    return N2;
}
 
// Driver code
int main()
{
    int M = 45;
    cout << number_of_teams(M);
 
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
 
class GFG
{
 
// Function to return the number of teams
static int number_of_teams(int M)
{
    // To store both roots of the equation
    int N1, N2, sqr;
 
    // sqrt(b^2 - 4ac)
    sqr = (int)Math.sqrt(1 + (8 * M));
 
    // First root (-b + sqrt(b^2 - 4ac)) / 2a
    N1 = (1 + sqr) / 2;
 
    // Second root (-b - sqrt(b^2 - 4ac)) / 2a
    N2 = (1 - sqr) / 2;
 
    // Return the positive root
    if (N1 > 0)
        return N1;
    return N2;
}
 
    // Driver code
    public static void main (String[] args)
    {
        int M = 45;
        System.out.println( number_of_teams(M));
    }
}
 
// this code is contributed by vt_m..

Python3

# Python implementation of the approach
import math
 
# Function to return the number of teams
def number_of_teams(M):
     
    # To store both roots of the equation
    N1, N2, sqr = 0,0,0
 
    # sqrt(b^2 - 4ac)
    sqr = math.sqrt(1 + (8 * M))
 
    # First root (-b + sqrt(b^2 - 4ac)) / 2a
    N1 = (1 + sqr) / 2
 
    # Second root (-b - sqrt(b^2 - 4ac)) / 2a
    N2 = (1 - sqr) / 2
 
    # Return the positive root
    if (N1 > 0):
        return int(N1)
    return int(N2)
 
# Driver code
def main():
    M = 45
    print(number_of_teams(M))
if __name__ == '__main__':
    main()
     
# This code has been contributed by 29AjayKumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to return the number of teams
    static int number_of_teams(int M)
    {
        // To store both roots of the equation
        int N1, N2, sqr;
     
        // sqrt(b^2 - 4ac)
        sqr = (int)Math.Sqrt(1 + (8 * M));
     
        // First root (-b + sqrt(b^2 - 4ac)) / 2a
        N1 = (1 + sqr) / 2;
     
        // Second root (-b - sqrt(b^2 - 4ac)) / 2a
        N2 = (1 - sqr) / 2;
     
        // Return the positive root
        if (N1 > 0)
            return N1;
        return N2;
    }
     
    // Driver code
    public static void Main()
    {
        int M = 45;
        Console.WriteLine( number_of_teams(M));
    }
}
 
// This code is contributed by Ryuga

PHP

 0)
        return $N1;
    return $N2;
}
 
// Driver code
$M = 45;
echo number_of_teams($M);
 
// This code is contributed
// by chandan_jnu
?>

Javascript

// javascript implementation of the approach
 
    // Function to return the number of teams
    function number_of_teams(M) {
        // To store both roots of the equation
        var N1, N2, sqr;
 
        // sqrt(b^2 - 4ac)
        sqr = parseInt( Math.sqrt(1 + (8 * M)));
 
        // First root (-b + sqrt(b^2 - 4ac)) / 2a
        N1 = (1 + sqr) / 2;
 
        // Second root (-b - sqrt(b^2 - 4ac)) / 2a
        N2 = (1 - sqr) / 2;
 
        // Return the positive root
        if (N1 > 0)
            return N1;
        return N2;
    }
 
    // Driver code
     
        var M = 45;
        document.write(number_of_teams(M));
 
// This code contributed by gauravrajput1

输出：

10