算法设计：给定比赛次数，查找比赛中的球队数

本文概述

• C ++
• Java
• Python3
• C#
• 的PHP

给定一个整数中号这是一场比赛中进行比赛的次数, 每个参赛队都与其他所有队进行了比赛。任务是查找比赛中有多少支球队。

例子：

输入：M = 3
输出：3
如果有3支球队A, B和C, 则
A将与B和C进行一场比赛
B将与C进行比赛(B已经与A进行比赛)
C已经和A队、B队打过比赛了
比赛总数为3
输入：M = 45
输出：10

方法：由于每场比赛都是在两支球队之间进行。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, 比赛的总数将为C(N, 2), 其中N为参赛球队数。因此,

M = C(N, 2)
M =(N *(N – 1))/ 2
N^2 – N – 2 * M = 0
这是ax^2 + bx + c = 0的二次方程。这里a = 1 b = -1, c = 2 *M。因此, 应用公式
x =(-b + sqrt(b^2 – 4ac))/ 2a和x =(-b – sqrt(b^2 – 4ac))/ 2a
N = [( -1 * -1)+ sqrt((-1 * -1)–(4 * 1 *(-2 * M))))]/2
N =(1 + sqrt(1 +(8 * M))))/ 2和N =(1 – sqrt(1 +(8 * M)))/ 2

求解完上述两个方程序后, 我们将得到两个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序的正根。

下面是上述方法的实现：

C ++

//C++ implementation of the approach
#include <cmath>
#include <iostream>
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

的PHP

<?php
//PHP implementation of the approach
  
//Function to return the number of teams
function number_of_teams( $M )
{
     //To store both roots of the equation
  
     //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
$M = 45;
echo number_of_teams( $M );
  
//This code is contributed 
//by chandan_jnu
?>

输出如下：

10

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