# Sum of all even factors of numbers in the range [l, r]

Given a range [l, r], the task is to find the sum of all the even factors of the numbers from the given range.

Examples:

Input: l = 6, r = 8
Output: 22
factors(6) = 1, 2, 3, 6, evenfactors(6) = 2, 6 sumEvenFactors(6) = 2 + 6 = 8
factors(7) = 1, 7, No even factors
factors(8) = 1, 2, 4, 8, evenfactors(8) = 2, 4, 8 sumEvenFactors(8) = 2 + 4 + 8 = 14
Therefore sum of all even factors = 8 + 14 = 22

Input: l = 1, r = 10
Output: 42

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: We can modify Sieve Of Eratosthenes to store the sum of all even factors of a number at it’s corresponding index. Then we will make a prefix array to store sum upto that index. And now each query can be answered in O(1) using prefix[r] – prefix[l – 1].

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` `#define ll long long int ` ` `  `const` `int` `MAX = 100000; ` ` `  `ll prefix[MAX]; ` ` `  `// Function to calculate the prefix sum ` `// of all the even factors ` `void` `sieve_modified() ` `{ ` `    ``for` `(``int` `i = 2; i < MAX; i += 2) { ` ` `  `        ``// Add i to all the multiples of i ` `        ``for` `(``int` `j = i; j < MAX; j += i) ` `            ``prefix[j] += i; ` `    ``} ` ` `  `    ``// Update the prefix sum ` `    ``for` `(``int` `i = 1; i < MAX; i++) ` `        ``prefix[i] += prefix[i - 1]; ` `} ` ` `  `// Function to return the sum of ` `// all the even factors of the ` `// numbers in the given range ` `ll sumEvenFactors(``int` `L, ``int` `R) ` `{ ` `    ``return` `(prefix[R] - prefix[L - 1]); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``sieve_modified(); ` `    ``int` `l = 6, r = 10; ` `    ``cout << sumEvenFactors(l, r); ` ` `  `    ``return` `0; ` `} `

 `// Java implementation of the approach ` `class` `GFG { ` ` `  `    ``static` `final` `int` `MAX = ``100000``; ` `    ``static` `long` `prefix[] = ``new` `long``[MAX]; ` ` `  `    ``// Function to calculate the prefix sum ` `    ``// of all the even factors ` `    ``static` `void` `sieve_modified() ` `    ``{ ` `        ``for` `(``int` `i = ``2``; i < MAX; i += ``2``) { ` ` `  `            ``// Add i to all the multiples of i ` `            ``for` `(``int` `j = i; j < MAX; j += i) ` `                ``prefix[j] += i; ` `        ``} ` ` `  `        ``// Update the prefix sum ` `        ``for` `(``int` `i = ``1``; i < MAX; i++) ` `            ``prefix[i] += prefix[i - ``1``]; ` `    ``} ` ` `  `    ``// Function to return the sum of ` `    ``// all the even factors of the ` `    ``// numbers in the given range ` `    ``static` `long` `sumEvenFactors(``int` `L, ``int` `R) ` `    ``{ ` `        ``return` `(prefix[R] - prefix[L - ``1``]); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``sieve_modified(); ` `        ``int` `l = ``6``, r = ``10``; ` `        ``System.out.print(sumEvenFactors(l, r)); ` `    ``} ` `} `

 `# Python3 implementation of the approach.  ` ` `  `# Function to calculate the prefix sum  ` `# of all the even factors  ` `def` `sieve_modified():  ` ` `  `    ``for` `i ``in` `range``(``2``, ``MAX``, ``2``):  ` ` `  `        ``# Add i to all the multiples of i  ` `        ``for` `j ``in` `range``(i, ``MAX``, i):  ` `            ``prefix[j] ``+``=` `i  ` ` `  `    ``# Update the prefix sum  ` `    ``for` `i ``in` `range``(``1``, ``MAX``):  ` `        ``prefix[i] ``+``=` `prefix[i ``-` `1``]  ` ` `  `# Function to return the sum of  ` `# all the even factors of the  ` `# numbers in the given range  ` `def` `sumEvenFactors(L, R):  ` ` `  `    ``return` `(prefix[R] ``-` `prefix[L ``-` `1``])  ` ` `  `# Driver code  ` `if` `__name__ ``=``=` `"__main__"``: ` `     `  `    ``MAX` `=` `100000` `    ``prefix ``=` `[``0``] ``*` `MAX` `    ``sieve_modified()  ` `    ``l, r ``=` `6``, ``10` `    ``print``(sumEvenFactors(l, r))  ` ` `  `# This code is contributed by Rituraj Jain `

 `using` `System; ` `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `    ``public` `const` `int` `MAX = 100000; ` `    ``public` `static` `long``[] prefix = ``new` `long``[MAX]; ` ` `  `    ``// Function to calculate the prefix sum ` `    ``// of all the even factors ` `    ``public` `static` `void` `sieve_modified() ` `    ``{ ` `        ``for` `(``int` `i = 2; i < MAX; i += 2) ` `        ``{ ` ` `  `            ``// Add i to all the multiples of i ` `            ``for` `(``int` `j = i; j < MAX; j += i) ` `            ``{ ` `                ``prefix[j] += i; ` `            ``} ` `        ``} ` ` `  `        ``// Update the prefix sum ` `        ``for` `(``int` `i = 1; i < MAX; i++) ` `        ``{ ` `            ``prefix[i] += prefix[i - 1]; ` `        ``} ` `    ``} ` ` `  `    ``// Function to return the sum of ` `    ``// all the even factors of the ` `    ``// numbers in the given range ` `    ``public` `static` `long` `sumEvenFactors(``int` `L, ``int` `R) ` `    ``{ ` `        ``return` `(prefix[R] - prefix[L - 1]); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main(``string``[] args) ` `    ``{ ` `        ``sieve_modified(); ` `        ``int` `l = 6, r = 10; ` `        ``Console.Write(sumEvenFactors(l, r)); ` `    ``} ` `} ` ` `  `// This code is contributed by Shrikant13 `

 ` `

Output:
```34
```

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