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Sum of all Perfect numbers lying in the range [L, R]
• Last Updated : 12 Apr, 2021

Given two numbers L, R which signifies the range [L, R], the task is to find the sum of all perfect numbers lying in the range [L, R].
Examples:

Input: L = 6, R = 10
Output:
Explanation:
From 6 to 10, the only perfect number is 6.
Input: L = 6, R = 28
Output: 34
Explanation:
There are two perfect numbers in the range [6, 28]. They are, {6, 28}
6 + 28 = 34.

Naive Approach: The naive approach for this problem is to check if a number is a perfect number or not by iterating through every number in the range [L, R]. If there are N numbers in the range, the time complexity for this approach is O(N * sqrt(K)) where K is the maximum number(R) in the range [L, R].
Efficient Approach: The idea is to use the concept of prefix sum array to perform precomputations and store the sum of all the numbers in an array.

• Initialize an array up to the max size such that every index ‘i’ of the array represents the sum of all the perfect numbers from [0, i].
• Iterate over the array and for every index, check if it is a perfect number or not.
• If it is a perfect number, then add the sum stored at the previous index (i – 1) and the current index ‘i’.
• If it is not a perfect number, then add the sum stored at the previous index (i – 1) and the value 0.
• Finally, for every query [L, R], return the value:

`sum = arr[R] - arr[L - 1]`

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the``// sum of all perfect numbers``// lying in the range [L, R]` `#include ``#define ll int``using` `namespace` `std;` `// Array to store the sum``long` `long` `pref;` `// Function to check if a number is``// a perfect number or not``int` `isPerfect(``int` `n)``{``    ``int` `sum = 1;` `    ``// Iterating till the square root``    ``// of the number and checking if``    ``// the sum of divisors is equal``    ``// to the number or not``    ``for` `(``int` `i = 2; i * i <= n; i++) {``        ``if` `(n % i == 0) {``            ``if` `(i * i != n)``                ``sum = sum + i + n / i;``            ``else``                ``sum = sum + i;``        ``}``    ``}` `    ``// If it is a perfect number, then``    ``// return the number``    ``if` `(sum == n && n != 1)``        ``return` `n;` `    ``// Else, return 0``    ``return` `0;``}` `// Function to precompute the sum``// of perfect squares and store``// then in an array``void` `precomputation()``{``    ``for` `(``int` `i = 1; i <= 100000; ++i) {``        ``pref[i] = pref[i - 1] + isPerfect(i);``    ``}``}` `int` `main()``{` `    ``int` `L = 6, R = 28;` `    ``precomputation();` `    ``cout << pref[R] - pref[L - 1];` `    ``return` `0;``}`

## Java

 `// Java implementation to find the``// sum of all perfect numbers``// lying in the range [L, R]``class` `GFG {``    ` `    ``// Array to store the sum``    ``static` `int` `pref [] = ``new` `int``[``10000``];``    ` `    ``// Function to check if a number is``    ``// a perfect number or not``    ``static` `int` `isPerfect(``int` `n)``    ``{``        ``int` `sum = ``1``;``    ` `        ``// Iterating till the square root``        ``// of the number and checking if``        ``// the sum of divisors is equal``        ``// to the number or not``        ``for` `(``int` `i = ``2``; i * i <= n; i++) {``            ``if` `(n % i == ``0``) {``                ``if` `(i * i != n)``                    ``sum = sum + i + n / i;``                ``else``                    ``sum = sum + i;``            ``}``        ``}``    ` `        ``// If it is a perfect number, then``        ``// return the number``        ``if` `(sum == n && n != ``1``)``            ``return` `n;``    ` `        ``// Else, return 0``        ``return` `0``;``    ``}``    ` `    ``// Function to precompute the sum``    ``// of perfect squares and store``    ``// then in an array``    ``static` `void` `precomputation()``    ``{``        ``for` `(``int` `i = ``1``; i < ``10000``; ++i) {``            ``pref[i] = pref[i - ``1``] + isPerfect(i);``        ``}``    ``}``    ` `    ``public` `static` `void` `main (String[] args)``    ``{``    ` `        ``int` `L = ``6``, R = ``28``;``    ` `        ``precomputation();``    ` `        ``System.out.println(pref[R] - pref[L - ``1``]);``    ` `    ``}` `}` `// This code is contributed by AnkitRai01`

## Python3

 `# Python3 implementation to find the``# sum of all perfect numbers``# lying in the range [L, R]` `from` `math ``import` `sqrt` `# Array to store the sum``pref ``=` `[``0``]``*``10000``;` `# Function to check if a number is``# a perfect number or not``def` `isPerfect(n)  :` `    ``sum` `=` `1``;` `    ``# Iterating till the square root``    ``# of the number and checking if``    ``# the sum of divisors is equal``    ``# to the number or not``    ``for` `i ``in` `range``(``2``, ``int``(sqrt(n)) ``+` `1``) :``        ``if` `(n ``%` `i ``=``=` `0``) :``            ``if` `(i ``*` `i !``=` `n) :``                ``sum` `=` `sum` `+` `i ``+` `n ``/``/` `i;``            ``else` `:``                ``sum` `=` `sum` `+` `i;``                ` `    ``# If it is a perfect number, then``    ``# return the number``    ``if` `(``sum` `=``=` `n ``and` `n !``=` `1``) :``        ``return` `n;` `    ``# Else, return 0``    ``return` `0``;` `# Function to precompute the sum``# of perfect squares and store``# then in an array``def` `precomputation() :` `    ``for` `i ``in` `range``(``1``, ``10000``) :``        ``pref[i] ``=` `pref[i ``-` `1``] ``+` `isPerfect(i);` `if` `__name__ ``=``=` `"__main__"` `:`  `    ``L ``=` `6``; R ``=` `28``;` `    ``precomputation();` `    ``print``(pref[R] ``-` `pref[L ``-` `1``]);` `# This code is contributed by AnkitRai01`

## C#

 `// C# implementation to find the``// sum of all perfect numbers``// lying in the range [L, R]``using` `System;` `public` `class` `GFG {``     ` `    ``// Array to store the sum``    ``static` `int` `[]pref = ``new` `int``;``     ` `    ``// Function to check if a number is``    ``// a perfect number or not``    ``static` `int` `isPerfect(``int` `n)``    ``{``        ``int` `sum = 1;``     ` `        ``// Iterating till the square root``        ``// of the number and checking if``        ``// the sum of divisors is equal``        ``// to the number or not``        ``for` `(``int` `i = 2; i * i <= n; i++) {``            ``if` `(n % i == 0) {``                ``if` `(i * i != n)``                    ``sum = sum + i + n / i;``                ``else``                    ``sum = sum + i;``            ``}``        ``}``     ` `        ``// If it is a perfect number, then``        ``// return the number``        ``if` `(sum == n && n != 1)``            ``return` `n;``     ` `        ``// Else, return 0``        ``return` `0;``    ``}``     ` `    ``// Function to precompute the sum``    ``// of perfect squares and store``    ``// then in an array``    ``static` `void` `precomputation()``    ``{``        ``for` `(``int` `i = 1; i < 10000; ++i) {``            ``pref[i] = pref[i - 1] + isPerfect(i);``        ``}``    ``}``     ` `    ``public` `static` `void` `Main(String[] args)``    ``{``     ` `        ``int` `L = 6, R = 28;``     ` `        ``precomputation();``     ` `        ``Console.WriteLine(pref[R] - pref[L - 1]);``     ` `    ``}`` ` `}``// This code contributed by Rajput-Ji`

## Javascript

 ``
Output:
`34`

Time Complexity:

• The time taken for the precomputation is O(K * sqrt(K)) where K is the number upto which we are performing the precomputation
• After precomputation, each query is answered in O(1).

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