给定比赛场次，找出锦标赛中的队伍数量
给定一个整数 M ，这是一场锦标赛中的比赛次数，每个参赛队都与所有其他队进行了一场比赛。任务是找出锦标赛中有多少支队伍。 例:
输入: M = 3 输出: 3 如果有 A、B、C 三支队伍，那么 A 将与 B、C 进行一场比赛 B 将与 C 进行一场比赛(B 已经与 A 进行了一场比赛) C 已经与 A、B 队进行了比赛 总共进行了 3 场比赛 输入: M = 45 输出: 10 【T16
进场:由于每场比赛都是两队之间进行。所以这个问题类似于从给定的 N 个对象中选择 2 个对象。因此比赛总数将为 C(N，2)，其中 N 为参赛队伍数。因此，
M = C(N，2) M =(N (N–1))/2 N2–N–2 * M = 0 这是一个 ax 型二次方程 2 + bx + c = 0。这里 a = 1，b = -1，c = 2 * M .因此，应用公式 x =(-b+ sqrt(b2–4ac))/2a 和 x =(-b–sqrt(b2–4ac))/2a N =[(-1 -1)+sqrt((-1 -1)–(4 * 1 (2 * M)))]/2 N =(1+sqrt(M)
解完上面两个方程，我们会得到 n 的两个值，一个值为正，一个为负。忽略负值。因此，团队数量将是上述等式的正根。 以下是上述方法的实现:
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 语言(一种计算机语言，尤用于创建网站)
// 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..
Python 3
# 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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?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
?>
java 描述语言
<script>
// 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
</script>
Output:
10
版权属于：月萌API www.moonapi.com，转载请注明出处
