Given a positive integer **N**, the task is to calculate the sum of all integers from **1 to N** but excluding the number which is a perfect power of 2.**Examples:**

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

**Naive Approach:**

The naive approach is to iterate every number from **1 to N** and compute the sum in the variable by excluding the number which is a perfect power of 2. But to compute the sum to the number **10^9**, the above approach will give Time Limit Error.**Time Complexity:** O(N)**Efficient Approach:**

To find desired sum, below are the steps:

- Find the sum of all the number till
**N**using the formula discussed in this article in**O(1)**time. - Since sum of all perfect power of 2 forms a Geometric Progression. Hence the sum of all powers of 2 less than N is calculated by the below formula:

The number of element with perfect power of 2 less than N is given by

log,_{2}N

Letr= log_{2}N

And the sum of all numbers which are perfect power of 2 is given by2.^{r}– 1

- Subtract the sum of all perfect powers of
**2**calculated above from the sum of first**N**numbers to get the result.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the` `// approach` `#include <bits/stdc++.h>` `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

`<script>` `// 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` `</script>` |

**Output:**

0

**Time Complexity:** O(1)

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