Sum of product of proper divisors of all Numbers lying in range [L, R]

Given an array arr[][] consisting of Q queries where every row consists of two numbers L and R which denotes the range [L, R]; the task is to find the sum of the product of proper divisors of all numbers lying in the range [L, R].

Note: Since the answer might be very big, perform % with 1000000007 for every query.

Examples:



Input: Q = 2, arr[] = { { 4, 6 }, { 8, 10 } }
Output: 9 21
Explanation:
Query 1: From 4 to 6
Product of proper divisors of 4 = 1 * 2 = 2
Product of proper divisors of 5 = 1
Product of proper divisors of 6 = 1 * 2 * 3 = 6
Sum of product of proper divisors from 4 to 6 = 2 + 1 + 6 = 9
Query 2: From 8 to 10
Product of proper divisors of 8 = 1 * 2 * 4 = 8
Product of proper divisors of 9 = 1 * 3 = 3
Product of proper divisors of 10 = 1 * 2 * 5 = 10
Sum of product of proper divisors from 8 to 10 = 8 + 3 + 10 = 21

Input: Q = 2, arr[] = { { 10, 20 }, { 12, 16 } }
Output: 975 238

Approach: Since there can be multiple queries and finding the divisors and product for every query is not feasible, the idea is to precompute and store every element along with its product of proper divisors in an array using a modification of Sieve of Eratosthenes.

Once the product of the proper divisors of all the numbers is stored in an array, the idea is to use the concept of prefix sum array. The sum of the product of proper divisors till that particular index is precomputed and stored in an array pref[] so that every query can be answered in constant time.

For each query, the sum of all product of proper divisors for the range [L, R] can be found as follows:

sum = pref[R] - pref[L - 1] 

Below is the implementation of the above approach:

CPP

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// C++ implementation to find the sum
// of the product of proper divisors of
// all the numbers lying in the range [L, R]
  
#include <bits/stdc++.h>
#define ll long long int
#define mod 1000000007
using namespace std;
  
// Vector to store the product
// of the proper divisors of a number
vector<ll> ans(100002, 1);
  
// Variable to store the prefix
// sum of the product array
long long pref[100002];
  
// Function to precompute the product
// of proper divisors of a number at
// it's corresponding index
void preCompute()
{
    // Modificatino of sieve to store the
    // product of the proper divisors
    for (int i = 2; i <= 100000 / 2; i++) {
        for (int j = 2 * i; j <= 100000; j += i) {
  
            // Multiplying the existing value
            // with i because i is the
            // proper divisor of ans[j]
            ans[j] = (ans[j] * i) % mod;
        }
    }
  
    // Loop to store the prefix sum of the
    // previously computed product array
    for (int i = 1; i < 100002; ++i) {
  
        // Computing the prefix sum
        pref[i] = pref[i - 1]
                  + ans[i];
        pref[i] %= mod;
    }
}
  
// Function to print the sum
// for each query
void printSum(int L, int R)
{
    cout << pref[R] - pref[L - 1]
         << " ";
}
  
// Function to print te sum of product
// of proper divisors of a number in
// [L, R]
void printSumProper(int arr[][2], int Q)
{
  
    // Calling the function that
    // pre computes
    // the sum of product
    // of proper divisors
    preCompute();
  
    // Iterate over all Queries
    // to print the sum
    for (int i = 0; i < Q; i++) {
        printSum(arr[i][0], arr[i][1]);
    }
}
  
// Driver code
int main()
{
    int Q = 2;
    int arr[][2] = { { 10, 20 },
                     { 12, 16 } };
  
    printSumProper(arr, Q);
    return 0;
}

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Java

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// Java implementation to find the sum
// of the product of proper divisors of
// all the numbers lying in the range [L, R]
import java.util.*;
  
class GFG{
  
static int mod = 1000000007;
  
// Vector to store the product
// of the proper divisors of a number
static int []ans = new int[100002];
  
// Variable to store the prefix
// sum of the product array
static int []pref = new int[100002];
  
// Function to precompute the product
// of proper divisors of a number at
// it's corresponding index
static void preCompute()
{
    // Modificatino of sieve to store the
    // product of the proper divisors
    Arrays.fill(ans, 1);
    for (int i = 2; i <= 100000 / 2; i++) {
        for (int j = 2 * i; j <= 100000; j += i) {
  
            // Multiplying the existing value
            // with i because i is the
            // proper divisor of ans[j]
            ans[j] = (ans[j] * i) % mod;
        }
    }
  
    // Loop to store the prefix sum of the
    // previously computed product array
    for (int i = 1; i < 100002; ++i) {
  
        // Computing the prefix sum
        pref[i] = pref[i - 1]
                + ans[i];
        pref[i] %= mod;
    }
}
  
// Function to print the sum
// for each query
static void printSum(int L, int R)
{
    System.out.print(pref[R] - pref[L - 1]+" ");
}
  
// Function to print te sum of product
// of proper divisors of a number in
// [L, R]
static void printSumProper(int [][]arr, int Q)
{
  
    // Calling the function that
    // pre computes
    // the sum of product
    // of proper divisors
    preCompute();
  
    // Iterate over all Queries
    // to print the sum
    for (int i = 0; i < Q; i++) {
        printSum(arr[i][0], arr[i][1]);
    }
}
  
// Driver code
public static void main(String args[])
{
    int Q = 2;
    int[][] arr = {{10, 20 },
                    { 12, 16 } };
  
    printSumProper(arr, Q);
}
}
  
// This code is contributed by Surendra_Gangwar

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Python3

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# Python3 implementation to find the sum
# of the product of proper divisors of
# all the numbers lying in the range [L, R]
  
mod = 1000000007
  
# Vector to store the product
# of the proper divisors of a number
ans = [1]*(100002)
  
# Variable to store the prefix
# sum of the product array
pref = [0]*100002
  
# Function to precompute the product
# of proper divisors of a number at
# it's corresponding index
def preCompute():
  
    # Modificatino of sieve to store the
    # product of the proper divisors
    for i in range(2,100000//2+1): 
        for j in range(2*i,100000+1,i): 
  
            # Multiplying the existing value
            # with i because i is the
            # proper divisor of ans[j]
            ans[j] = (ans[j] * i) % mod
          
    # Loop to store the prefix sum of the
    # previously computed product array
    for i in range(1,100002): 
  
        # Computing the prefix sum
        pref[i] = pref[i - 1]+ ans[i]
        pref[i] %= mod
      
# Function to prthe sum
# for each query
def printSum(L, R):
  
    print(pref[R] - pref[L - 1],end=" ")
  
# Function to prte sum of product
# of proper divisors of a number in
# [L, R]
def printSumProper(arr, Q):
  
    # Calling the function that
    # pre computes
    # the sum of product
    # of proper divisors
    preCompute()
  
    # Iterate over all Queries
    # to prthe sum
    for i in range(Q): 
        printSum(arr[i][0], arr[i][1])
      
# Driver code
if __name__ == '__main__':
    Q = 2
    arr= [ [ 10, 20 ],
            [ 12, 16 ] ]
  
    printSumProper(arr, Q)
  
# This code is contributed by mohit kumar 29

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Output:

975 238



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