来自前 N 个自然数的总和为 N 的对的计数

给定一个整数N ，任务是计算前N 个自然数中对的数量，总和等于N 。

例子：

Input: N = 8
Output: 3
Explanation:
All possible pairs with sum 8 are { (1, 7), (2, 6), (3, 5)}

Input: N = 9
Output: 4

天真的方法：
解决问题的最简单方法是使用两个指针。请按照以下步骤解决问题：

• 最初设置i = 0和j = N – 1 。
• 迭代直到i >= j ，并且对于每对i, j ，检查它们的总和是否等于N 。如果是这样，请增加对的数量。
• 通过将i和j分别增加和减少1来移动到下一对。
• 最后，打印获得的对数。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
int numberOfPairs(int n)
{
 
    // Stores the count of
    // pairs
    int count = 0;
    // Set the two pointers
    int i = 1, j = n - 1;
 
    while (i < j) {
 
        // Check if the sum of
        // pairs is equal to N
        if (i + j == n) {
            // Increase the count
            // of pairs
            count++;
        }
 
        // Move to the next pair
        i++;
        j--;
    }
 
    return count;
}
 
// Driver Code
int main()
{
    int n = 8;
    cout << numberOfPairs(n);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
     
// Function to calculate the value of count
public static int numberOfPairs(int n)
{
 
    // Stores the count of pairs
    int count = 0;
 
    // Set the two pointers
    int i = 1, j = n - 1;
 
    while (i < j)
    {
         
        // Check if the sum of
        // pairs is equal to N
        if (i + j == n)
        {
             
            // Increase the count
            // of pairs
            count++;
        }
 
        // Move to the next pair
        i++;
        j--;
    }
    return count;
}
 
// Driver code
public static void main (String[] args)
{
    int n = 8;
     
    System.out.println(numberOfPairs(n));
}
}
 
// This code is contributed by piyush3010

Python3

# Python3 program for the
# above approach
def numberOfPairs(n):
   
  # Stores the count
  # of pairs
  count = 0
   
  # Set the two pointers
  i = 1
  j = n - 1
 
  while(i < j):
     
    # Check if the sum
    # of pirs is equal to n
    if (i + j) == n:
       
      # Increase the count of pairs
      count += 1
       
      # Move to the next pair
      i += 1
      j -= 1
       
  return count
 
# Driver code
if __name__=='__main__':
   
  n = 8
  print(numberOfPairs(n))
     
# This code is contributed by virusbuddah_

C#

// C# program for the above approach
using System;
class GFG{
      
// Function to calculate the value of count
public static int numberOfPairs(int n)
{
  
    // Stores the count of pairs
    int count = 0;
  
    // Set the two pointers
    int i = 1, j = n - 1;
  
    while (i < j)
    {
          
        // Check if the sum of
        // pairs is equal to N
        if (i + j == n)
        {
              
            // Increase the count
            // of pairs
            count++;
        }
  
        // Move to the next pair
        i++;
        j--;
    }
    return count;
}
  
// Driver code
public static void Main (string[] args)
{
    int n = 8;
      
    Console.Write(numberOfPairs(n));
}
}
  
// This code is contributed by rock_cool

Javascript

 
// Javascript program to implement
// the above approach
function numberOfPairs(n)
{
     
    // Stores the count of
    // pairs
    let count = 0;
     
    // Set the two pointers
    let i = 1, j = n - 1;
 
    while (i < j)
    {
         
        // Check if the sum of
        // pairs is equal to N
        if (i + j == n)
        {
             
            // Increase the count
            // of pairs
            count++;
        }
 
        // Move to the next pair
        i++;
        j--;
    }
    return count;
}
 
// Driver code
let n = 8;
 
document.write(numberOfPairs(n));
     
// This code is contributed by divyesh072019
     

输出

3

时间复杂度： O(N)
辅助空间： O(1)

有效的方法：
为了优化上述方法，我们只需要观察N是偶数还是奇数。如果N是偶数，则可能对的数量是N/2 – 1。否则，它是N/2。

Illustration:

N = 8
All possible pairs are (1, 7), (2, 6) and (3, 5)
Hence, count of possible pairs = 3 = 8/2 – 1

N = 9
All possible pairs are (1, 8), (2, 7), (3, 6) and (4, 5)
Hence, count of possible pairs = 4 = 9/2

下面是上述方法的实现：

C++

// C++ program to count the number
// of pairs among the first N
// natural numbers with sum N
#include 
using namespace std;
 
// Function to return the
// count of pairs
int numberOfPairs(int n)
{
    // If n is even
    if (n % 2 == 0)
 
        // Count of pairs
        return n / 2 - 1;
 
    // Otherwise
    else
        return n / 2;
}
 
// Driver Code
int main()
{
    int n = 8;
    cout << numberOfPairs(n);
 
    return 0;
}

Java

// Java program to count the number
// of pairs among the first N
// natural numbers with sum N
import java.io.*;
 
class GFG{
     
// Function to calculate the value of count
public static int numberOfPairs(int n)
{
 
    // If n is even
    if (n % 2 == 0)
     
        // Count of pairs
        return n / 2 - 1;
 
    // Otherwise
    else
        return n / 2;
}
 
// Driver code
public static void main (String[] args)
{
    int n = 8;
     
    System.out.println(numberOfPairs(n));
}
}
 
// This code is contributed by piyush3010

蟒蛇3

# Python3 program to count the number
# of pairs among the first N
# natural numbers with sum N
 
# Function to calculate the value of count
def numberOfPairs(n):
 
    # If n is even
    if (n % 2 == 0):
 
        # Count of pairs
        return n // 2 - 1;
 
    # Otherwise
    else:
        return n // 2;
 
# Driver code
n = 8;
 
print(numberOfPairs(n));
 
# This code is contributed by Rajput-Ji

C＃

// C# program to count the number
// of pairs among the first N
// natural numbers with sum N
using System;
class GFG{
      
// Function to calculate the value of count
public static int numberOfPairs(int n)
{
  
    // If n is even
    if (n % 2 == 0)
      
        // Count of pairs
        return n / 2 - 1;
  
    // Otherwise
    else
        return n / 2;
}
  
// Driver code
public static void Main (string[] args)
{
    int n = 8;
      
    Console.Write(numberOfPairs(n));
}
}
  
// This code is contributed by Ritik Bansal

Javascript

 
// Javascript program to count the number
// of pairs among the first N
// natural numbers with sum N
 
// Function to return the
// count of pairs
function numberOfPairs(n)
{
     
    // If n is even
    if (n % 2 == 0)
 
        // Count of pairs
        return (n / 2 - 1);
 
    // Otherwise
    else
        return (n / 2);
}
 
// Driver code
let n = 8;
 
document.write(numberOfPairs(n));
 
// This code is contributed by rameshtravel07
 

输出

3

时间复杂度： O(1)
辅助空间： O(1)