从给定数组中找到最大nCr值的一对

给定一个nrr个正整数的数组arr [] 。任务是从数组中找到元素arr [i]和arr [j] ，以便最大可能地获得^{arr [i]} C _{arr [j]} 。如果有效对超过1对，请打印其中的任何对。

例子：

Input: arr[] = {3, 1, 2}
Output: 3 2
³C₁ = 3
³C₂ = 3
²C₁ = 2
(3, 1) and (3, 2) are the only pairs with maximum ⁿC_r.

Input: arr[] = {9, 2, 4, 11, 6}
Output: 11 6

方法： ⁿ C _r是单调递增函数，即^{n + 1} C _r > ⁿ C _r 。我们可以利用这个事实来接近答案，我们将在所有给定的整数中选择最大值n 。这样我们固定了n的值。
现在，我们必须寻找r 。由于我们知道ⁿ C _r = ⁿ C _{n – r} ，这意味着ⁿ C _r将首先达到最大值，然后减小。

如果n为奇数，则我们的最大值将出现在n / 2和n / 2 +1处。
对于n = 11 ，我们将获得¹¹ C ₅和¹¹ C ₆的最大值。

当n为偶数时，我们的最大值将出现在n / 2处。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to print the pair that gives maximum nCr
void printMaxValPair(vector& v, int n)
{
    sort(v.begin(), v.end());
  
    // This gives the value of N in nCr
    long long N = v[n - 1];
  
    // Case 1 : When N is odd
    if (N % 2 == 1) {
        long long first_maxima = N / 2;
        long long second_maxima = first_maxima + 1;
        long long ans1 = 3e18, ans2 = 3e18;
        long long from_left = -1, from_right = -1;
        long long from = -1;
        for (long long i = 0; i < n; ++i) {
            if (v[i] > first_maxima) {
                from = i;
                break;
            }
            else {
                long long diff = first_maxima - v[i];
                if (diff < ans1) {
                    ans1 = diff;
                    from_left = v[i];
                }
            }
        }
        from_right = v[from];
        long long diff1 = first_maxima - from_left;
        long long diff2 = from_right - second_maxima;
  
        if (diff1 < diff2)
            cout << N << " " << from_left;
        else
            cout << N << " " << from_right;
    }
  
    // Case 2 : When N is even
    else {
        long long maxima = N / 2;
        long long ans1 = 3e18;
        long long R = -1;
        for (long long i = 0; i < n - 1; ++i) {
            long long diff = abs(v[i] - maxima);
            if (diff < ans1) {
                ans1 = diff;
                R = v[i];
            }
        }
        cout << N << " " << R;
    }
}
  
// Driver code
int main()
{
    vector v = { 1, 1, 2, 3, 6, 1 };
    int n = v.size();
    printMaxValPair(v, n);
  
    return 0;
}

Java

// Java implementation of the approach 
import java.util.*;
  
class GFG
{
  
// Function to print the pair that gives maximum nCr 
static void printMaxValPair(Vector v, int n) 
{ 
    Collections.sort(v); 
  
    // This gives the value of N in nCr 
    long N = v.get((int)n - 1); 
  
    // Case 1 : When N is odd 
    if (N % 2 == 1) 
    { 
        long first_maxima = N / 2; 
        long second_maxima = first_maxima + 1; 
        long ans1 =(long) 3e18, ans2 = (long)3e18; 
        long from_left = -1, from_right = -1; 
        long from = -1; 
        for (long i = 0; i < n; ++i)
        { 
            if (v.get((int)i) > first_maxima) 
            { 
                from = i; 
                break; 
            } 
            else 
            { 
                long diff = first_maxima - v.get((int)i); 
                if (diff < ans1)
                { 
                    ans1 = diff; 
                    from_left = v.get((int)i); 
                } 
            } 
        } 
        from_right = v.get((int)from); 
        long diff1 = first_maxima - from_left; 
        long diff2 = from_right - second_maxima; 
  
        if (diff1 < diff2) 
            System.out.println( N + " " + from_left); 
        else
            System.out.println( N + " " + from_right); 
    } 
  
    // Case 2 : When N is even 
    else 
    { 
        long maxima = N / 2; 
        long ans1 =(int) 3e18; 
        long R = -1; 
        for (long i = 0; i < n - 1; ++i) 
        { 
            long diff = Math.abs(v.get((int)i) - maxima); 
              
            if (diff < ans1) 
            { 
                ans1 = diff; 
                R = v.get((int)i); 
            } 
        } 
        System.out.println( N + " " + R); 
    } 
} 
  
// Driver code 
public static void main(String args[]) 
{ 
    long arr[] = { 1, 1, 2, 3, 6, 1};
    Vector v = new Vector( ); 
      
    for(int i = 0; i < arr.length; i++)
    v.add(arr[i]);
      
    int n = v.size(); 
    printMaxValPair(v, n); 
}
}
  
// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach
  
# Function to print the pair that
# gives maximum nCr
def printMaxValPair(v, n):
  
    v.sort()
  
    # This gives the value of N in nCr
    N = v[n - 1]
  
    # Case 1 : When N is odd
    if N % 2 == 1: 
        first_maxima = N // 2
        second_maxima = first_maxima + 1
        ans1, ans2 = 3 * (10 ** 18), 3 * (10 ** 18)
        from_left, from_right = -1, -1
        _from = -1
        for i in range(0, n): 
            if v[i] > first_maxima: 
                _from = i
                break
              
            else:
                diff = first_maxima - v[i]
                if diff < ans1: 
                    ans1 = diff
                    from_left = v[i]
                  
        from_right = v[_from]
        diff1 = first_maxima - from_left
        diff2 = from_right - second_maxima
  
        if diff1 < diff2:
            print(N, from_left)
        else:
            print(N, from_right)
  
    # Case 2 : When N is even
    else:
        maxima = N // 2
        ans1 = 3 * (10 ** 18)
        R = -1
        for i in range(0, n - 1): 
            diff = abs(v[i] - maxima)
            if diff < ans1: 
                ans1 = diff
                R = v[i]
              
        print(N, R)
  
# Driver code
if __name__ == "__main__":
  
    v = [1, 1, 2, 3, 6, 1] 
    n = len(v)
    printMaxValPair(v, n)
  
# This code is contributed by 
# Rituraj Jain

C#

// C# implementation of the approach 
using System;
using System.Collections.Generic;
  
class GFG 
{ 
  
// Function to print the pair that gives maximum nCr 
static void printMaxValPair(List v, int n) 
{ 
    v.Sort(); 
  
    // This gives the value of N in nCr 
    long N = v[(int)n - 1]; 
  
    // Case 1 : When N is odd 
    if (N % 2 == 1) 
    { 
        long first_maxima = N / 2; 
        long second_maxima = first_maxima + 1; 
        long ans1 = (long) 3e18, ans2 = (long)3e18; 
        long from_left = -1, from_right = -1; 
        long from = -1; 
        for (long i = 0; i < n; ++i) 
        { 
            if (v[(int)i] > first_maxima) 
            { 
                from = i; 
                break; 
            } 
            else
            { 
                long diff = first_maxima - v[(int)i]; 
                if (diff < ans1) 
                { 
                    ans1 = diff; 
                    from_left = v[(int)i]; 
                } 
            } 
        } 
        from_right = v[(int)from]; 
        long diff1 = first_maxima - from_left; 
        long diff2 = from_right - second_maxima; 
  
        if (diff1 < diff2) 
            Console.WriteLine( N + " " + from_left); 
        else
            Console.WriteLine( N + " " + from_right); 
    } 
  
    // Case 2 : When N is even 
    else
    { 
        long maxima = N / 2; 
        long ans1 = (long)3e18; 
        long R = -1; 
        for (long i = 0; i < n - 1; ++i) 
        { 
            long diff = Math.Abs(v[(int)i] - maxima); 
              
            if (diff < ans1) 
            { 
                ans1 = diff; 
                R = v[(int)i]; 
            } 
        } 
        Console.WriteLine( N + " " + R); 
    } 
} 
  
// Driver code 
public static void Main(String []args) 
{ 
    long []arr = { 1, 1, 2, 3, 6, 1}; 
    List v = new List( ); 
      
    for(int i = 0; i < arr.Length; i++) 
    v.Add(arr[i]); 
      
    int n = v.Count; 
    printMaxValPair(v, n); 
} 
} 
  
// This code contributed by Rajput-Ji

PHP

 $first_maxima) 
            {
                $from = $i;
                break;
            }
            else 
            {
                $diff = $first_maxima - $v[$i];
                if ($diff < $ans1) 
                {
                    $ans1 = $diff;
                    $from_left = $v[$i];
                }
            }
        }
        $from_right = $v[$from];
        $diff1 = $first_maxima - $from_left;
        $diff2 = $from_right - $second_maxima;
  
        if ($diff1 < $diff2)
            echo $N . " " . $from_left;
        else
            echo $N . " " . $from_right;
    }
  
    // Case 2 : When N is even
    else 
    {
        $maxima = $N / 2;
        $ans1 = 3e18;
        $R = -1;
        for ($i = 0; $i < $n - 1; ++$i) 
        {
            $diff = abs($v[$i] - $maxima);
            if ($diff < $ans1) 
            {
                $ans1 = $diff;
                $R = $v[$i];
            }
        }
        echo $N . " " . $R;
    }
}
  
// Driver code
$v = array( 1, 1, 2, 3, 6, 1 );
$n = count($v);
printMaxValPair($v, $n);
  
// This code is contributed by mits
?>

输出：

6 3