第1、2、4、3、5、7、9、6、8、10、11、13 ..系列的和，直到第N个项

给定一系列数字1、2、4、3、5、7、9、6、8、10、11、13……的任务是找到直到第N个数字的所有数字之和。
例子：

Input: N = 4 
Output: 10 
1 + 2 + 4 + 3 = 10 
Input: N = 10 
Output: 55

方法：该系列基本上是2 ^0个奇数，2 ^1个偶数，2 ^2个偶数……。前N个奇数的总和为N * N ，前N个偶数的总和为(N *(N + 1)) 。计算总和的2个^我奇数或偶数，让他们增加的总和。
对2的每一个幂进行迭代，直到迭代次数超过N，并根据奇偶校验不断增加奇数或偶数的总和。对于每个段，该段的总和为(X个奇/偶数的当前总和-Y个奇/偶数的先前总和) ，其中X是直到该段的奇/偶数的总和，Y是总和直到奇数/偶数出现之前的奇数/偶数。
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 odd numbers
int sumodd(int n)
{
    return (n * n);
}
 
// Function to find the
// sum of first N even numbers
int sumeven(int n)
{
    return (n * (n + 1));
}
 
// Function to overall
// find the sum of series
int findSum(int num)
{
 
    // Initiall odd numbers
    int sumo = 0;
 
    // Initial even numbers
    int sume = 0;
 
    // First power of 2
    int x = 1;
 
    // Check for parity
    // for odd/even
    int cur = 0;
 
    // Counts the sum
    int ans = 0;
    while (num > 0) {
 
        // Get the minimum
        // out of remaining num
        // or power of 2
        int inc = min(x, num);
 
        // Decrease that much numbers
        // from num
        num -= inc;
 
        // If the segment has odd numbers
        if (cur == 0) {
 
            // Summate the odd numbers
            // By exclusion
            ans = ans + sumodd(sumo + inc) - sumodd(sumo);
 
            // Increase number of odd numbers
            sumo += inc;
        }
        // If the segment has evn numbers
        else {
 
            // Summate the even numbers
            // By exclusion
            ans = ans + sumeven(sume + inc) - sumeven(sume);
 
            // Increase number of even numbers
            sume += inc;
        }
 
        // Next set of numbers
        x *= 2;
 
        // Change parity for odd/even
        cur ^= 1;
    }
 
    return ans;
}
 
// Driver code
int main()
{
    int n = 4;
    cout << findSum(n);
 
    return 0;
}

Java

// Java program to implement
// the above approach
 
class GFG
{
 
    // Function to find the
    // sum of first N odd numbers
    static int sumodd(int n)
    {
        return (n * n);
    }
 
    // Function to find the
    // sum of first N even numbers
    static int sumeven(int n)
    {
        return (n * (n + 1));
    }
 
    // Function to overall
    // find the sum of series
    static int findSum(int num)
    {
 
        // Initiall odd numbers
        int sumo = 0;
 
        // Initial even numbers
        int sume = 0;
 
        // First power of 2
        int x = 1;
 
        // Check for parity
        // for odd/even
        int cur = 0;
 
        // Counts the sum
        int ans = 0;
        while (num > 0)
        {
 
            // Get the minimum
            // out of remaining num
            // or power of 2
            int inc = Math.min(x, num);
 
            // Decrease that much numbers
            // from num
            num -= inc;
 
            // If the segment has odd numbers
            if (cur == 0)
            {
 
                // Summate the odd numbers
                // By exclusion
                ans = ans + sumodd(sumo + inc) - sumodd(sumo);
 
                // Increase number of odd numbers
                sumo += inc;
            }
             
            // If the segment has evn numbers
            else
            {
 
                // Summate the even numbers
                // By exclusion
                ans = ans + sumeven(sume + inc) - sumeven(sume);
 
                // Increase number of even numbers
                sume += inc;
            }
 
            // Next set of numbers
            x *= 2;
 
            // Change parity for odd/even
            cur ^= 1;
        }
 
        return ans;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 4;
        System.out.println(findSum(n));
 
    }
}
 
// This code contributed by Rajput-Ji

Python

# Python3 program to implement
# the above approach
 
# Function to find the
# sum of first N odd numbers
def sumodd(n):
 
    return (n * n)
 
# Function to find the
# sum of first N even numbers
def sumeven(n):
 
    return (n * (n + 1))
 
 
# Function to overall
# find the sum of series
def findSum(num):
 
 
    # Initiall odd numbers
    sumo = 0
 
    # Initial even numbers
    sume = 0
 
    # First power of 2
    x = 1
 
    # Check for parity
    # for odd/even
    cur = 0
 
    # Counts the sum
    ans = 0
    while (num > 0):
 
        # Get the minimum
        # out of remaining num
        # or power of 2
        inc = min(x, num)
 
        # Decrease that much numbers
        # from num
        num -= inc
 
        # If the segment has odd numbers
        if (cur == 0):
 
            # Summate the odd numbers
            # By exclusion
            ans = ans + sumodd(sumo + inc) - sumodd(sumo)
 
            # Increase number of odd numbers
            sumo += inc
         
        # If the segment has evn numbers
        else:
 
            # Summate the even numbers
            # By exclusion
            ans = ans + sumeven(sume + inc) - sumeven(sume)
 
            # Increase number of even numbers
            sume += inc
         
 
        # Next set of numbers
        x *= 2
 
        # Change parity for odd/even
        cur ^= 1
     
    return ans
 
# Driver code
n = 4
print(findSum(n))
 
# This code is contributed by mohit kumar

C#

// C# program to implement
// the above approach
using System;
 
class GFG
{
 
    // Function to find the
    // sum of first N odd numbers
    static int sumodd(int n)
    {
        return (n * n);
    }
 
    // Function to find the
    // sum of first N even numbers
    static int sumeven(int n)
    {
        return (n * (n + 1));
    }
 
    // Function to overall
    // find the sum of series
    static int findSum(int num)
    {
 
        // Initiall odd numbers
        int sumo = 0;
 
        // Initial even numbers
        int sume = 0;
 
        // First power of 2
        int x = 1;
 
        // Check for parity
        // for odd/even
        int cur = 0;
 
        // Counts the sum
        int ans = 0;
        while (num > 0)
        {
 
            // Get the minimum
            // out of remaining num
            // or power of 2
            int inc = Math.Min(x, num);
 
            // Decrease that much numbers
            // from num
            num -= inc;
 
            // If the segment has odd numbers
            if (cur == 0)
            {
 
                // Summate the odd numbers
                // By exclusion
                ans = ans + sumodd(sumo + inc) - sumodd(sumo);
 
                // Increase number of odd numbers
                sumo += inc;
            }
             
            // If the segment has evn numbers
            else
            {
 
                // Summate the even numbers
                // By exclusion
                ans = ans + sumeven(sume + inc) - sumeven(sume);
 
                // Increase number of even numbers
                sume += inc;
            }
 
            // Next set of numbers
            x *= 2;
 
            // Change parity for odd/even
            cur ^= 1;
        }
 
        return ans;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int n = 4;
        Console.WriteLine(findSum(n));
 
    }
}
 
// This code has been contributed by 29AjayKumar

PHP

 0)
    {
 
        // Get the minimum
        // out of remaining num
        // or power of 2
        $inc = min($x, $num);
 
        // Decrease that much numbers
        // from num
        $num -= $inc;
 
        // If the segment has odd numbers
        if ($cur == 0)
        {
 
            // Summate the odd numbers
            // By exclusion
            $ans = $ans + sumodd($sumo + $inc) -
                          sumodd($sumo);
 
            // Increase number of odd numbers
            $sumo += $inc;
        }
         
        // If the segment has evn numbers
        else
        {
 
            // Summate the even numbers
            // By exclusion
            $ans = $ans + sumeven($sume + $inc) -
                          sumeven($sume);
 
            // Increase number of even numbers
            $sume += $inc;
        }
 
        // Next set of numbers
        $x *= 2;
 
        // Change parity for odd/even
        $cur ^= 1;
    }
 
    return $ans;
}
 
// Driver code
$n = 4;
echo findSum($n);
 
// This code contributed by princiraj1992
?>

Javascript

 
// javascript program to implement
// the above approach
 
// Function to find the
// sum of first N odd numbers
function sumodd( n)
{
    return (n * n);
}
 
// Function to find the
// sum of first N even numbers
function sumeven( n)
{
    return (n * (n + 1));
}
 
// Function to overall
// find the sum of series
function findSum( num)
{
 
    // Initiall odd numbers
    let sumo = 0;
 
    // Initial even numbers
    let sume = 0;
 
    // First power of 2
    let x = 1;
 
    // Check for parity
    // for odd/even
    let cur = 0;
 
    // Counts the sum
    let ans = 0;
    while (num > 0) {
 
        // Get the minimum
        // out of remaining num
        // or power of 2
        let inc = Math.min(x, num);
 
        // Decrease that much numbers
        // from num
        num -= inc;
 
        // If the segment has odd numbers
        if (cur == 0) {
 
            // Summate the odd numbers
            // By exclusion
            ans = ans + sumodd(sumo + inc) - sumodd(sumo);
 
            // Increase number of odd numbers
            sumo += inc;
        }
        // If the segment has evn numbers
        else {
 
            // Summate the even numbers
            // By exclusion
            ans = ans + sumeven(sume + inc) - sumeven(sume);
 
            // Increase number of even numbers
            sume += inc;
        }
 
        // Next set of numbers
        x *= 2;
 
        // Change parity for odd/even
        cur ^= 1;
    }
 
    return ans;
}
 
// Driver code
 
    let n = 4;
    document.write( findSum(n));
     
// This code is contributed by todaysgaurav
 
 

输出：

10