计算从1到N的所有整数的和，不包括2的幂

给定正整数N ，任务是计算从1到N的所有整数之和，但不包括2的乘方幂。
例子：

Input: N = 2 
Output: 0
Input: N = 1000000000 
Output: 499999998352516354 
 

天真的方法：
天真的方法是迭代从1到N的每个数字，并通过排除作为2的乘方幂的数字来计算变量中的总和。但是要计算总和为10 ^ 9的数字，上述方法将给出时间限制错误。
时间复杂度： O(N)
高效方法：
要找到所需的总和，请执行以下步骤：

1. 使用本文讨论的公式在O(1)时间内找到直到N的所有数字的总和。
2. 由于2的所有完美幂的和形成几何级数。因此，通过以下公式计算所有小于N的2的幂的和：

The number of element with perfect power of 2 less than N is given by log₂N, 
Let r = log₂N 
And the sum of all numbers which are perfect power of 2 is given by 2^r – 1. 
 

2. 从前N个数字的总和中减去上面计算的2的所有完美幂的总和，即可得出结果。

下面是上述方法的实现：

C++

// C++ implementation of the
// approach
#include 
using namespace std;
 
// Function to find the required
// summation
void findSum(int N)
{
    // Find the sum of first N
    // integers using the formula
    int sum = (N) * (N + 1) / 2;
     
    int r = log2(N) + 1;
     
    // Find the sum of numbers
    // which are exact power of
    // 2 by using the formula
    int expSum = pow(2, r) - 1;   
 
    // Print the final Sum
    cout << sum - expSum << endl;
}
 
// Driver's Code
int main()
{
    int N = 2;
 
    // Function to find the
    // sum
    findSum(N);
    return 0;
}

Java

// Java implementation of the above approach
import java.lang.Math;
 
class GFG{
     
// Function to find the required
// summation
public static void findSum(int N)
{
 
    // Find the sum of first N
    // integers using the formula
    int sum = (N) * (N + 1) / 2;
         
    int r = (int)(Math.log(N) /
                  Math.log(2)) + 1;
         
    // Find the sum of numbers
    // which are exact power of
    // 2 by using the formula
    int expSum = (int)(Math.pow(2, r)) - 1;    
     
    // Print the final Sum
    System.out.println(sum - expSum);
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 2;
 
    // Function to find the sum
    findSum(N);
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python 3 implementation of the
# approach
from math import log2,pow
 
# Function to find the required
# summation
def findSum(N):
    # Find the sum of first N
    # integers using the formula
    sum = (N) * (N + 1) // 2
     
    r = log2(N) + 1
     
    # Find the sum of numbers
    # which are exact power of
    # 2 by using the formula
    expSum = pow(2, r) - 1
 
    # Print the final Sum
    print(int(sum - expSum))
 
# Driver's Code
if __name__ == '__main__':
    N = 2
 
    # Function to find the
    # sum
    findSum(N)
     
# This code is contributed by Surendra_Gangwar

C#

// C# implementation of the above approach
using System;
 
class GFG{
     
// Function to find the required
// summation
public static void findSum(int N)
{
 
    // Find the sum of first N
    // integers using the formula
    int sum = (N) * (N + 1) / 2;
         
    int r = (int)(Math.Log(N) /
                  Math.Log(2)) + 1;
         
    // Find the sum of numbers
    // which are exact power of
    // 2 by using the formula
    int expSum = (int)(Math.Pow(2, r)) - 1;
     
    // Print the final Sum
    Console.Write(sum - expSum);
}
 
// Driver Code
public static void Main(string[] args)
{
    int N = 2;
 
    // Function to find the sum
    findSum(N);
}
}
 
// This code is contributed by rutvik_56

Javascript

 
// Javascript implementation of the above approach
 
// Function to find the required
// summation
function findSum(N)
{
     
    // Find the sum of first N
    // integers using the formula
    var sum = (N) * (N + 1) / 2;
         
    var r = (Math.log(N) /
             Math.log(2)) + 1;
         
    // Find the sum of numbers
    // which are exact power of
    // 2 by using the formula
    var expSum = (Math.pow(2, r)) - 1;    
     
    // Print the final Sum
    document.write(sum - expSum);
}
     
// Driver code
var N = 2;
 
// Function to find the sum
findSum(N);
 
// This code is contributed by Kirti
 

输出：

0

时间复杂度： O(1)