前N个自然数与交替符号的总和

给定一个整数N ，任务是找到具有备用符号(即1 – 2 + 3 – 4 + 5 – 6 +…)的前N个自然数的总和。

例子：

Input: N = 6 
Output: -3 
Explanation: 
1 – 2 + 3 – 4 + 5 – 6 = -3 
Therefore, the required output is -3.

Input: N = 5 
Output: 3 
Explanation: 
1 – 2 + 3 – 4 + 5 = 3 
Therefore, the required output = 3

天真的方法：请按照以下步骤解决问题：

• 初始化一个变量，例如alternateSum，以存储前N个自然数的替代符号之和。
• 使用变量i遍历[1，N]范围，并检查i是否为偶数。如果发现是真的，则更新alterSum + = -i 。
• 否则，更新alterSum + = i 。
• 最后，打印alternateSum的价值。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to find the sum of first
// N natural numbers with alternate signs
int alternatingSumOfFirst_N(int N)
{
    // Stores sum of alternate sign
    // of first N natural numbers
    int alternateSum = 0;
     
    for (int i = 1; i <= N; i++) {
 
        // If is an even number
        if (i % 2 == 0) {
 
            // Update alternateSum
            alternateSum += -i;
        }
 
        // If i is an odd number
        else {
 
            // Update alternateSum
            alternateSum += i;
        }
    }
    return alternateSum;
     
}
 
// Driver Code
int main()
{
 
    int N = 6;
    cout<

---

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to find the sum of first
// N natural numbers with alternate signs
static int alternatingSumOfFirst_N(int N)
{
     
    // Stores sum of alternate sign
    // of first N natural numbers
    int alternateSum = 0;
     
    for(int i = 1; i <= N; i++)
    {
         
        // If is an even number
        if (i % 2 == 0)
        {
             
            // Update alternateSum
            alternateSum += -i;
        }
 
        // If i is an odd number
        else
        {
             
            // Update alternateSum
            alternateSum += i;
        }
    }
    return alternateSum;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 6;
     
    System.out.print(alternatingSumOfFirst_N(N));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program to implement
# the above approach
 
# Function to find the sum of
# First N natural numbers with
# alternate signs
def alternatingSumOfFirst_N(N):
     
    # Stores sum of alternate sign
    # of First N natural numbers
    alternateSum = 0
     
    for i in range(1, N + 1):
         
        # If is an even number
        if (i % 2 == 0):
             
            # Update alternateSum
            alternateSum += -i
             
        # If i is an odd number
        else:
            alternateSum += i
 
    return alternateSum
 
# Driver Code 
if __name__ == "__main__" :
     
    N = 6
     
    print(alternatingSumOfFirst_N(N))
        
# This code is contributed by Virusbuddah_

C#

// C# program to implement
// the above approach
using System;
class GFG{
 
// Function to find the sum of first
// N natural numbers with alternate signs
static int alternatingSumOfFirst_N(int N)
{
     
    // Stores sum of alternate sign
    // of first N natural numbers
    int alternateSum = 0;
     
    for(int i = 1; i <= N; i++)
    {
         
        // If is an even number
        if (i % 2 == 0)
        {
             
            // Update alternateSum
            alternateSum += -i;
        }
 
        // If i is an odd number
        else
        {
             
            // Update alternateSum
            alternateSum += i;
        }
    }
    return alternateSum;
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 6;
     
    Console.Write(alternatingSumOfFirst_N(N));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript program to implement
// the above approach
 
    // Function to find the sum of first
    // N natural numbers with alternate signs
    function alternatingSumOfFirst_N(N)
    {
 
        // Stores sum of alternate sign
        // of first N natural numbers
        var alternateSum = 0;
 
        for (i = 1; i <= N; i++) {
 
            // If is an even number
            if (i % 2 == 0) {
 
                // Update alternateSum
                alternateSum += -i;
            }
 
            // If i is an odd number
            else {
 
                // Update alternateSum
                alternateSum += i;
            }
        }
        return alternateSum;
    }
 
    // Driver Code
     
        var N = 6;
 
        document.write(alternatingSumOfFirst_N(N));
 
// This code is contributed by Rajput-Ji
 

---

输出：

-3

时间复杂度： O(N)
辅助空间： O(1)

高效方法：为了优化上述方法，该思想基于以下观察结果：

If N is an even number then the sum of alternate sign of first N natural numbers are = (-N) / 2.

If N is an odd number then the sum of alternate sign of first N natural numbers are = (N + 1) / 2.

请按照以下步骤解决问题：

• 初始化一个变量，例如alternateSum，以存储前N个自然数的替代符号之和。
• 检查N是否为偶数。如果发现为真，则更新alterSum =(-N)/ 2 。
• 否则，更新alterSum =(N + 1)/ 2 。
• 最后，打印alternateSum的价值。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to find the sum of first
// N natural numbers with alternate signs
int alternatingSumOfFirst_N(int N)
{
    // Stores sum of alternate sign
    // of first N natural numbers
    int alternateSum = 0;
 
    // If N is an even number
    if (N % 2 == 0) {
 
        // Update alternateSum
        alternateSum = (-N) / 2;
    }
 
    // If N is an odd number
    else {
        // Update alternateSum
        alternateSum = (N + 1) / 2;
    }
    return alternateSum;
}
 
// Driver Code
int main()
{
 
    int N = 6;
    cout<

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG
{
 
// Function to find the sum of first
// N natural numbers with alternate signs
static int alternatingSumOfFirst_N(int N)
{
    // Stores sum of alternate sign
    // of first N natural numbers
    int alternateSum = 0;
 
    // If N is an even number
    if (N % 2 == 0)
    {
 
        // Update alternateSum
        alternateSum = (-N) / 2;
    }
 
    // If N is an odd number
    else
    {
       
        // Update alternateSum
        alternateSum = (N + 1) / 2;
    }
    return alternateSum;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 6;
    System.out.print(alternatingSumOfFirst_N(N));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python program to implement
# the above approach
 
# Function to find the sum of first
# N natural numbers with alternate signs
def alternatingSumOfFirst_N(N):
   
    # Stores sum of alternate sign
    # of first N natural numbers
    alternateSum = 0;
 
    # If N is an even number
    if (N % 2 == 0):
 
        # Update alternateSum
        alternateSum = (-N) // 2;
 
 
    # If N is an odd number
    else:
 
        # Update alternateSum
        alternateSum = (N + 1) // 2;
 
    return alternateSum;
 
# Driver Code
if __name__ == '__main__':
    N = 6;
    print(alternatingSumOfFirst_N(N));
 
# This code contributed by shikhasingrajput

C＃

// C# program to implement
// the above approach
using System;
class GFG
{
 
// Function to find the sum of first
// N natural numbers with alternate signs
static int alternatingSumOfFirst_N(int N)
{
    // Stores sum of alternate sign
    // of first N natural numbers
    int alternateSum = 0;
 
    // If N is an even number
    if (N % 2 == 0)
    {
 
        // Update alternateSum
        alternateSum = (-N) / 2;
    }
 
    // If N is an odd number
    else
    {
       
        // Update alternateSum
        alternateSum = (N + 1) / 2;
    }
    return alternateSum;
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 6;
    Console.Write(alternatingSumOfFirst_N(N));
}
}
 
// This code is contributed by 29AjayKumar

Java脚本

 
// Javascript program to implement
// the above approach
 
    // Function to find the sum of first
    // N natural numbers with alternate signs
    function alternatingSumOfFirst_N(N)
    {
        // Stores sum of alternate sign
        // of first N natural numbers
        var alternateSum = 0;
 
        // If N is an even number
        if (N % 2 == 0) {
 
            // Update alternateSum
            alternateSum = (-N) / 2;
        }
 
        // If N is an odd number
        else {
 
            // Update alternateSum
            alternateSum = (N + 1) / 2;
        }
        return alternateSum;
    }
 
    // Driver Code
     
        var N = 6;
        document.write(alternatingSumOfFirst_N(N));
 
// This code contributed by Rajput-Ji
 

输出：

-3

时间复杂度： O(1)
辅助空间： O(1)