# Sum of all proper divisors of natural numbers in an array

Given an array of natural numbers count the sum of its proper divisors for every element in array.

Input : int arr[] = {8, 13, 24, 36, 59, 75, 87} Output : 7 1 36 55 1 49 21 Number 8 has 3 proper divisors 1, 2, 4 and their sum comes out to be 7.

A naive solution to this problem has been discussed in below post.

Sum of all proper divisors of a natural number

We can do this more efficiently by making use of sieve of Eratosthenes.

The idea is based on prime factorization of a number. By using sieve we can store all the prime factors of a number and their powers.

To find all divisors, we need to consider all powers of a prime factor and multiply it with all all powers of other prime factors. (For example, if the number is 36, its prime factors are 2 and 3 and all divisors are 1, 2, 3, 4, 6, 9, 12 and 18. Consider a number N can be written as P1^Q1 * P2^Q2 * P3^Q3 (here only 3 prime factors are considered but there can be more than that) then sum of its divisors will be written as: = P1^0 * P2^0 * P3^0 + P1^0 * P2^0 * P3^1 + P1^0 * P2^0 * P3^2 + ................ + P1^0 * P2^0 * P3^Q3 + P1^0 * P2^1 * P3^0 + P1^0 * P2^1 * P3^1 + P1^0 * P2^1 * P3^2 + ................ + P1^0 * P2^1 * P3^Q3 + . . . P1^Q1 * P2^Q2 * P3^0 + P1^Q1 * P2^Q2 * P3^1 + P1^Q1 * P2^Q2 * P3^2 + .......... + P1^Q1 * P2^Q2 * P3^Q3 Above can be written as, (((P1^(Q1+1)) - 1) / (P1 - 1)) * (((P2^(Q2+1)) - 1) / (P2 - 1)) * (((P3^(Q3 + 1)) - 1) / (P3 - 1))

Below is implementation based on above formula.

`// C++ program to find sum of proper divisors for ` `// every element in an array. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` `#define MAX 1000001 ` `#define pii pair<int, int> ` `#define F first ` `#define S second ` ` ` `// To store prime factors and their ` `// powers ` `vector<pii> factors[MAX]; ` ` ` `// Fills factors such that factors[i] is ` `// a vector of pairs containing prime factors ` `// (of i) and their powers. ` `// Also sets values in isPrime[] ` `void` `sieveOfEratothenese() ` `{ ` ` ` `// To check if a number is prime ` ` ` `bool` `isPrime[MAX]; ` ` ` `memset` `(isPrime, ` `true` `, ` `sizeof` `(isPrime)); ` ` ` `isPrime[0] = isPrime[1] = ` `false` `; ` ` ` ` ` `for` `(` `int` `i = 2; i < MAX; i++) ` ` ` `{ ` ` ` `// If i is prime, then update its ` ` ` `// powers in all multiples of it. ` ` ` `if` `(isPrime[i]) ` ` ` `{ ` ` ` `for` `(` `int` `j = i; j < MAX; j += i) ` ` ` `{ ` ` ` `int` `k, l; ` ` ` `isPrime[j] = ` `false` `; ` ` ` `for` `(k = j, l = 0; k % i == 0; l++, k /= i) ` ` ` `; ` ` ` `factors[j].push_back(make_pair(i, l)); ` ` ` `} ` ` ` `} ` ` ` `} ` `} ` ` ` `// Returns sum of proper divisors of num ` `// using factors[] ` `int` `sumOfProperDivisors(` `int` `num) ` `{ ` ` ` `// Applying above discussed formula for every ` ` ` `// array element ` ` ` `int` `mul = 1; ` ` ` `for` `(` `int` `i = 0; i < factors[num].size(); i++) ` ` ` `mul *= ((` `pow` `(factors[num][i].F, ` ` ` `factors[num][i].S + 1) - 1) / ` ` ` `(factors[num][i].F - 1)); ` ` ` `return` `mul - num; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `sieveOfEratothenese(); ` ` ` `int` `arr[] = { 8, 13, 24, 36, 59, 75, 91 }; ` ` ` `for` `(` `int` `i = 0; i < ` `sizeof` `(arr) / ` `sizeof` `(` `int` `); i++) ` ` ` `cout << sumOfProperDivisors(arr[i]) << ` `" "` `; ` ` ` `cout << endl; ` ` ` `return` `0; ` `} ` |

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Output:

7 1 36 55 1 49 21

This article is contributed by **Shubham Singh (singh_8)**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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