Given a number N, find sum of all GCDs that can be formed by selecting all the pairs from 1 to N.

Input : 4 Output : 7 Explanation: Numbers from 1 to 4 are: 1, 2, 3, 4 Result = gcd(1,2) + gcd(1,3) + gcd(1,4) + gcd(2,3) + gcd(2,4) + gcd(3,4) = 1 + 1 + 1 + 1 + 2 + 1 = 7 Input : 12 Output : 105 Input : 1 Output : 0 Input : 2 Output : 1

A **Naive approach** is to run two loops one inside the other. Select all pairs one by one, find GCD of every pair and then find sum of these GCDs. Time complexity of this approach is O(N^{2} * log(N))

**Efficient Approach** is based on following concepts:

**Euler’s Totient function Φ(n)**for an input n is count of numbers in {1, 2, 3, …, n} that are relatively prime to n, i.e., the numbers whose GCD (Greatest Common Divisor) with n is 1. For example, Φ(4) = 2, Φ(3) = 2 and Φ(5) = 4. There are 2 numbers smaller or equal to 4 that are relatively prime to 4, 2 numbers smaller or equal to 3 that are relatively prime to 3. And 4 numbers smaller than or equal to 5 that are relatively prime to 5.

The idea is to convert given problem into sum of Euler Totient Functions.

Sum of all GCDs where j is a part of pair is and j is greater element in pair:SumOur final result is_{j}= ∑(i=1 to j-1) gcd(i, j)Result = ∑(j=1 to N) SumThe above equation can be written as :_{j}SumFor every possible GCD 'g' of j. Here count(g) represents count of pairs having GCD equals to g. For every such pair(i, j), we can write : gcd(i/g, j/g) = 1 We can re-write our previous equation as_{j}= ∑ g * count(g)SumFor every divisor d of j and phi[] is Euler Totient number_{j}= ∑ d * phi(j/d)Example : j = 12 and d = 3is one of divisor of j so in order to calculate the sum of count of all pairs having 3 as gcd we can simple write it as => 3*phi[12/3] => 3*phi[4] => 3*2 => 6 Therefore sum of GCDs of all pairs where 12 is greater part of pair and 3 is GCD. GCD(3, 12) + GCD(9, 12) = 6.Complete Example :N = 4 Sum_{1}= 0 Sum_{2}= 1 [GCD(1, 2)] Sum_{3}= 2 [GCD(1, 3) + GCD(2, 3)] Sum_{4}= 4 [GCD(1, 4) + GCD(3, 4) + GCD(2, 4)] Result = Sum_{1}+ Sum_{2}+ Sum_{3}+ Sum_{4}= 0 + 1 + 2 + 4 = 7

Below is C++ implementation of above idea. We precompute Euler Totient Functions and result for all numbers till a maximum value. The idea used in implementation is based this post.

// C++ approach of finding sum of GCD of all pairs #include<bits/stdc++.h> using namespace std; #define MAX 100001 // phi[i] stores euler totient function for i // result[j] stores result for value j long long phi[MAX], result[MAX]; // Precomputation of phi[] numbers. Refer below link // for details : https://goo.gl/LUqdtY void computeTotient() { // Refer https://goo.gl/LUqdtY phi[1] = 1; for (int i=2; i<MAX; i++) { if (!phi[i]) { phi[i] = i-1; for (int j = (i<<1); j<MAX; j+=i) { if (!phi[j]) phi[j] = j; phi[j] = (phi[j]/i)*(i-1); } } } } // Precomputes result for all numbers till MAX void sumOfGcdPairs() { // Precompute all phi value computeTotient(); for (int i=1; i<MAX; ++i) { // Iterate throght all the divisors // of i. for (int j=2; i*j<MAX; ++j) result[i*j] += i*phi[j]; } // Add summation of previous calculated sum for (int i=2; i<MAX; i++) result[i] += result[i-1]; } // Driver code int main() { // Function to calculate sum of all the GCD // pairs sumOfGcdPairs(); int N = 4; cout << "Summation of " << N << " = " << result[N] << endl;; N = 12; cout << "Summation of " << N << " = " << result[N] << endl; N = 5000; cout << "Summation of " << N << " = " << result[N] ; return 0; }

Output:Summation of 4 = 7 Summation of 12 = 105 Summation of 5000 = 61567426

**Time complexity: **O(MAX*log(log MAX))

**Auxiliary space: **O(MAX)

**Reference:**

https://www.quora.com/How-can-I-solve-the-problem-GCD-Extreme-on-SPOJ-SPOJ-com-Problem-GCDEX

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