Given a number N, the task is to find the sum of all the perfect square divisors of numbers from 1 to N.
Examples:
Input: N = 5
Output: 9
Explanation: N = 5
Perfect square divisors of 1 = 1.
Similarly, perfect square divisors of 2, 3 = 1.
Perfect square divisors of 4 = 1, 4.
Perfect square divisors of 5 = 1 (of course for any prime only 1 will be the perfect square divisor)
So, total sum = 1+1+1+(1+4)+1 = 9.Input: N = 30
Output: 126Input: N = 100
Output: 910
Naive Approach: This approach is based on the approach implemented in this article
The above problem can be solved in O(N1/k) for any Kth power divisors, where N is the number up to which we have to find the sum. This is because, in this sum, every number will contribute floor(N/p) or int(N/p) times. Thus, while iterating through these perfect powers, we just need to add [p * int(N/p)] to the sum.
Time Complexity: O(?N)
Efficient Approach:
- Let us start from start = 2, find the largest range (start to end) for which floor(N/(start2)) = floor(N/(end2))
- The contribution of all perfect squares in the interval [start, end] will contribute floor(N/(start2)) times, hence we can do update for this range at once.
- Contribution for range [start, end] can be given as:
floor(N/(start2))*(sumUpto(end) – sumUpto(start-1))
- How to find range?
For a given value of start, end can be found by
sqrt(N/K), where K = floor(N/(start^2))
- Now the next range can be found by substituting start = end+1.
Time complexity: O(N1/3) as N/(x2) cannot take more than N1/3 different values for a fixed value of N.
Below is the implementation of the above approach:
C++
// C++ Program to find the// sum of all perfect square// divisors of numbers from 1 to N#include <bits/stdc++.h>using namespace std;#define MOD 1000000007#define int unsigned long long// Function for finding inverse// of a number iteratively// Here we will find the inverse// of 6, since it appears as// denominator in the formula of// sum of squares from 1 to Nint inv(int a){ int o = 1; for (int p = MOD - 2; p > 0; p >>= 1) { if ((p & 1) == 1) o = (o * a) % MOD; a = (a * a) % MOD; } return o;}// Store the value of the inverse// of 6 once as we don't need to call// the function again and againint inv6 = inv(6);// Formula for finding the sum// of first n squaresint sumOfSquares(int n){ n %= MOD; return (((n * (n + 1)) % MOD * (2 * n + 1)) % MOD * inv6) % MOD;}int sums(int n){ // No perfect square // exists which is // less than 4 if (n < 4) return 0; // Starting from 2, present value // of start is denoted here as // curStart int curStart = 2, ans = 0; int sqrtN = sqrt(n); while (curStart <= n / curStart) { int V = n / (curStart * curStart); // Finding end of the segment // for which the contribution // will be same int end = sqrt(n / V); // Using the above mentioned // formula to find ans % MOD ans += (n / (curStart * curStart) % MOD * (sumOfSquares(end) + MOD - sumOfSquares(curStart - 1))) % MOD; if (ans >= MOD) ans -= MOD; // Now for the next iteration // start will become end+1 curStart = end + 1; } // Finally returning the answer return ans;}// Driver Codeint32_t main(){ int input[] = { 5 }; for (auto x : input) { cout << "sum of all perfect" << " square divisors from" << " 1 to " << x << " is: "; // Here we are adding x // because we have not // counted 1 as perfect // squares so if u want to // add it you can just add // that number to the ans cout << x + sums(x) << endl; } return 0;} |
Java
// Java program to find the// sum of all perfect square// divisors of numbers from 1 to Nimport java.util.*;class GFG{static final int MOD = 7;// Function for finding inverse// of a number iteratively// Here we will find the inverse// of 6, since it appears as// denominator in the formula of// sum of squares from 1 to Nstatic int inv(int a){ int o = 1; for(int p = MOD - 2; p > 0; p >>= 1) { if ((p & 1) == 1) o = (o * a) % MOD; a = (a * a) % MOD; } return o;}// Store the value of the inverse// of 6 once as we don't need to call// the function again and againstatic int inv6 = inv(6);// Formula for finding the sum// of first n squaresstatic int sumOfSquares(int n){ n %= MOD; return (((n * (n + 1)) % MOD * (2 * n + 1)) % MOD * inv6) % MOD;}static int sums(int n){ // No perfect square // exists which is // less than 4 if (n < 4) return 0; // Starting from 2, present value // of start is denoted here as // curStart int curStart = 2, ans = 0; int sqrtN = (int)Math.sqrt(n); while (curStart <= n / curStart) { int V = n / (curStart * curStart); // Finding end of the segment // for which the contribution // will be same int end = (int)Math.sqrt(n / V); // Using the above mentioned // formula to find ans % MOD ans += (n / (curStart * curStart) % MOD * (sumOfSquares(end) + MOD - sumOfSquares(curStart - 1))) % MOD; if (ans >= MOD) ans -= MOD; // Now for the next iteration // start will become end+1 curStart = end + 1; } // Finally returning the answer return ans;}// Driver Codepublic static void main(String[] args){ int input[] = {5}; for(int x : input) { System.out.print("sum of all perfect " + "square divisors from " + "1 to " + x + " is: "); // Here we are adding x // because we have not // counted 1 as perfect // squares so if u want to // add it you can just add // that number to the ans System.out.print(x + sums(x) + "\n"); }}}// This code is contributed by Amit Katiyar |
Python3
# Python3 program to find the # sum of all perfect square # divisors of numbers from 1 to N from math import *MOD = 1000000007# Function for finding inverse # of a number iteratively # Here we will find the inverse # of 6, since it appears as # denominator in the formula of # sum of squares from 1 to N def inv (a): o = 1 p = MOD - 2 while (p > 0): if (p % 2 == 1): o = (o * a) % MOD a = (a * a) % MOD p >>= 1 return o # Store the value of the inverse # of 6 once as we don't need to call # the function again and again inv6 = inv(6)# Formula for finding the sum # of first n squaresdef sumOfSquares (n): n %= MOD return (((n * (n + 1)) % MOD * (2 * n + 1)) % MOD * inv6) % MODdef sums (n): # No perfect square exists which # is less than 4 if (n < 4): return 0 # Starting from 2, present value # of start is denoted here as curStart curStart = 2 ans = 0 sqrtN = int(sqrt(n)) while (curStart <= n // curStart): V = n // (curStart * curStart) # Finding end of the segment for # which the contribution will be same end = int(sqrt(n // V)) # Using the above mentioned # formula to find ans % MOD ans += ((n // (curStart * curStart) % MOD * (sumOfSquares(end) + MOD - sumOfSquares(curStart - 1))) % MOD) if (ans >= MOD): ans -= MOD # Now for the next iteration # start will become end+1 curStart = end + 1 # Finally return the answer return ans # Driver Codeif __name__ == '__main__': Input = [5] for x in Input: print("sum of all perfect "\ "square " , end = '') print("divisors from 1 to", x, "is: ", end = '') # Here we are adding x because we have # not counted 1 as perfect squares so if u # want to add it you can just add that # number to the ans print(x + sums(x))# This code is contributed by himanshu77 |
C#
// C# program to find the// sum of all perfect square// divisors of numbers from 1 to Nusing System;class GFG{static readonly int MOD = 7;// Function for finding inverse// of a number iteratively// Here we will find the inverse// of 6, since it appears as// denominator in the formula of// sum of squares from 1 to Nstatic int inv(int a){ int o = 1; for(int p = MOD - 2; p > 0; p >>= 1) { if ((p & 1) == 1) o = (o * a) % MOD; a = (a * a) % MOD; } return o;}// Store the value of the inverse// of 6 once as we don't need to call// the function again and againstatic int inv6 = inv(6);// Formula for finding the sum// of first n squaresstatic int sumOfSquares(int n){ n %= MOD; return (((n * (n + 1)) % MOD * (2 * n + 1)) % MOD * inv6) % MOD;}static int sums(int n){ // No perfect square // exists which is // less than 4 if (n < 4) return 0; // Starting from 2, present // value of start is denoted // here as curStart int curStart = 2, ans = 0; int sqrtN = (int)Math.Sqrt(n); while (curStart <= n / curStart) { int V = n / (curStart * curStart); // Finding end of the segment // for which the contribution // will be same int end = (int)Math.Sqrt(n / V); // Using the above mentioned // formula to find ans % MOD ans += (n / (curStart * curStart) % MOD * (sumOfSquares(end) + MOD - sumOfSquares(curStart - 1))) % MOD; if (ans >= MOD) ans -= MOD; // Now for the next iteration // start will become end+1 curStart = end + 1; } // Finally returning // the answer return ans;}// Driver Codepublic static void Main(String[] args){ int []input = {5}; foreach(int x in input) { Console.Write("sum of all perfect " + "square divisors from " + "1 to " + x + " is: "); // Here we are adding x // because we have not // counted 1 as perfect // squares so if u want to // add it you can just add // that number to the ans Console.Write(x + sums(x) + "\n"); }}}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript program to find the// sum of all perfect square// divisors of numbers from 1 to Nlet MOD = 7;// Function for finding inverse// of a number iteratively// Here we will find the inverse// of 6, since it appears as// denominator in the formula of// sum of squares from 1 to Nfunction inv(a) { let o = 1; for (let p = MOD - 2; p > 0; p >>= 1) { if ((p & 1) == 1) o = (o * a) % MOD; a = (a * a) % MOD; } return o;}// Store the value of the inverse// of 6 once as we don't need to call// the function again and againlet inv6 = inv(6);// Formula for finding the sum// of first n squaresfunction sumOfSquares(n) { n %= MOD; return (((n * (n + 1)) % MOD * (2 * n + 1)) % MOD * inv6) % MOD;}function sums(n) { // No perfect square // exists which is // less than 4 if (n < 4) return 0; // Starting from 2, present value // of start is denoted here as // curStart let curStart = 2, ans = 0; let sqrtN = Math.floor(Math.sqrt(n)); while (curStart <= Math.floor(n / curStart)) { let V = Math.floor(n / (curStart * curStart)); // Finding end of the segment // for which the contribution // will be same let end = Math.floor(Math.sqrt(n / V)); // Using the above mentioned // formula to find ans % MOD ans += (Math.floor(n / (curStart * curStart)) % MOD * (sumOfSquares(end) + MOD - sumOfSquares(curStart - 1))) % MOD; if (ans >= MOD) ans -= MOD; // Now for the next iteration // start will become end+1 curStart = end + 1; } // Finally returning the answer return ans;}// Driver Codelet input = [5];for (let x of input) { document.write("sum of all perfect " + "square divisors from " + "1 to " + x + " is: "); // Here we are adding x // because we have not // counted 1 as perfect // squares so if u want to // add it you can just add // that number to the ans document.write(x + sums(x) + "<br>");}// This code is contributed by Saurabh Jaiswal</script> |
Output:
sum of all perfect square divisors from 1 to 5 is: 9
Time Complexity: O(N1/3)
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