前 N 个自然数和 2 的所有幂相加两次

给定一个整数N ，任务是计算前 N 个自然数的总和，将 2 的所有幂加到总和上。
例子：

Input: N = 4 
Output: 17 
Explanation: 
Sum = 2+4+3+8 = 17 
Since 1, 2 and 4 are 2 ⁰, 2 ¹ and 2 ² respectively, they are added twice to the sum.
Input: N = 5 
Output: 22 
Explanation: 
The sum is equal to 2+4+3+8+5 = 22, 
because 1, 2 and 4 are 2 ⁰, 2 ¹ and 2 ² respectively.

天真的方法：
解决这个问题最简单的方法是迭代到N ，并通过将每个数字相加一次来计算总和，除了2的幂需要相加两次。
时间复杂度： O(N)
辅助空间： O(1)
有效的方法：
按照以下步骤优化上述方法：

1. 通过公式(N * (N + 1)) / 2计算前 N 个自然数的总和。
2. 现在，需要再次添加 2 的所有幂。 2 到 N 的所有幂的总和可以计算为2 ^{log ₂ (N) + 1} – 1 。
3. 因此，所需的总和为：

(N * (N + 1)) / 2 + 2 ^{log₂(N) + 1} – 1

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to raise N to the
// power P and return the value
double power(int N, int P)
{
    return pow(N, P);
}
 
// Function to calculate the
// log base 2 of an integer
int Log2(int N)
{
 
    // Calculate log2(N) indirectly
    // using log() method
    int result = (int)(log(N) / log(2));
    return result;
}
 
// Function to calculate and
// return the required sum
double specialSum(int n)
{
 
    // Sum of first N natural
    // numbers
    double sum = n * (n + 1) / 2;
 
    // Sum of all powers of 2
    // up to N
    int a = Log2(n);
    sum = sum + power(2, a + 1) - 1;
 
    return sum;
}
 
// Driver code
int main()
{
    int n = 4;
 
    cout << (specialSum(n)) << endl;
 
    return 0;
}
 
// This code is contributed by divyeshrabadiya07

Java

// Java program to implement
// the above approach
import java.util.*;
import java.lang.Math;
 
class GFG {
 
    // Function to raise N to the
    // power P and return the value
    static double power(int N, int P)
    {
        return Math.pow(N, P);
    }
 
    // Function to calculate the
    // log base 2 of an integer
    public static int log2(int N)
    {
 
        // Calculate log2(N) indirectly
        // using log() method
        int result = (int)(Math.log(N)
                        / Math.log(2));
        return result;
    }
 
    // Function to calculate and
    // return the required sum
    static double specialSum(int n)
    {
 
        // Sum of first N natural
        // numbers
        double sum = n * (n + 1) / 2;
 
        // Sum of all powers of 2
        // up to N
        int a = log2(n);
        sum = sum + power(2, a + 1) - 1;
 
        return sum;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        int n = 4;
 
        System.out.println(specialSum(n));
    }
}

Python3

# Python3 program to implement
# the above approach
import math
 
# Function to raise N to the
# power P and return the value
def power(N, P):
     
    return math.pow(N, P)
 
# Function to calculate the
# log base 2 of an integer
def Log2(N):
     
    # Calculate log2(N) indirectly
    # using log() method
    result = (math.log(N) // math.log(2))
     
    return result
 
# Function to calculate and
# return the required sum
def specialSum(n):
     
    # Sum of first N natural
    # numbers
    sum = n * (n + 1) // 2
 
    # Sum of all powers of 2
    # up to N
    a = Log2(n)
    sum = sum + power(2, a + 1) - 1
     
    return sum
     
# Driver code   
if __name__=="__main__":
     
    n = 4
    print(specialSum(n))
 
# This code is contributed by rutvik_56

C#

// C# program to implement
// the above approach
using System;
class GFG
{
 
    // Function to raise N to the
    // power P and return the value
    static double power(int N, int P)
    {
        return Math.Pow(N, P);
    }
 
    // Function to calculate the
    // log base 2 of an integer
    public static int log2(int N)
    {
 
        // Calculate log2(N) indirectly
        // using log() method
        int result = (int)(Math.Log(N) /
                        Math.Log(2));
        return result;
    }
 
    // Function to calculate and
    // return the required sum
    static double specialSum(int n)
    {
 
        // Sum of first N natural
        // numbers
        double sum = (double)(n) * (n + 1) / 2;
 
        // Sum of all powers of 2
        // up to N
        int a = log2(n);
        sum = (sum) + power(2, a + 1) - 1;
 
        return sum;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int n = 4;
 
        Console.Write(specialSum(n));
    }
}
 
// This code is contributed by Ritik Bansal

Javascript

 
// Javascript program to implement
// the above approach
 
// Function to raise N to the
// power P and return the value
function power(N, P)
{
    return Math.pow(N, P);
}
 
// Function to calculate the
// log base 2 of an integer
function Log2(N)
{
 
    // Calculate log2(N) indirectly
    // using log() method
    let result = (Math.floor(Math.log(N) / Math.log(2)));
    return result;
}
 
// Function to calculate and
// return the required sum
function specialSum(n)
{
 
    // Sum of first N natural
    // numbers
    let sum = n * (n + 1) / 2;
 
    // Sum of all powers of 2
    // up to N
    let a = Log2(n);
    sum = sum + power(2, a + 1) - 1;
 
    return sum;
}
 
// Driver Code
 
    let n = 4;
 
    document.write(specialSum(n) + "<br>");
 
// This code is contributed by Mayank Tyagi
 

输出：

17.0

时间复杂度： O(log ₂ (N))
辅助空间： O(1)

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