D. Small GCD
bitmasks; brute force; dp; math; number theory
Let \(a\), \(b\), and \(c\) be integers. We define function \(f(a, b, c)\) as follows:Order the numbers \(a\), \(b\), \(c\) in such a way that \(a \le b \le c\). Then return \(\gcd(a, b)\), where \(\gcd(a, b)\) denotes the greatest common divisor (GCD) of integers \(a\) and \(b\).So basically, we take the \(\gcd\) of the \(2\) smaller values and ignore the biggest one. You are given an array \(a\) of \(n\) elements. Compute the sum of \(f(a_i, a_j, a_k)\) for each \(i\), \(j\), \(k\), such that \(1 \le i < j < k \le n\). More formally, compute $$$\(\sum_{i = 1}^n \sum_{j = i+1}^n \sum_{k =j +1}^n f(a_i, a_j, a_k).\)$$$
Each test contains multiple test cases. The first line contains the number of test cases \(t\) (\(1 \le t \le 10\)). The description of the test cases follows.The first line of each test case contains a single integer \(n\) (\(3 \le n \le 8 \cdot 10^4\)) — length of the array \(a\).The second line of each test case contains \(n\) integers, \(a_1, a_2, \ldots, a_n\) (\(1 \le a_i \le 10^5\)) — elements of the array \(a\).It is guaranteed that the sum of \(n\) over all test cases does not exceed \(8 \cdot 10^4\).
For each test case, output a single number — the sum from the problem statement.
In the first test case, the values of \(f\) are as follows: \(i=1\), \(j=2\), \(k=3\), \(f(a_i,a_j,a_k)=f(2,3,6)=\gcd(2,3)=1\); \(i=1\), \(j=2\), \(k=4\), \(f(a_i,a_j,a_k)=f(2,3,12)=\gcd(2,3)=1\); \(i=1\), \(j=2\), \(k=5\), \(f(a_i,a_j,a_k)=f(2,3,17)=\gcd(2,3)=1\); \(i=1\), \(j=3\), \(k=4\), \(f(a_i,a_j,a_k)=f(2,6,12)=\gcd(2,6)=2\); \(i=1\), \(j=3\), \(k=5\), \(f(a_i,a_j,a_k)=f(2,6,17)=\gcd(2,6)=2\); \(i=1\), \(j=4\), \(k=5\), \(f(a_i,a_j,a_k)=f(2,12,17)=\gcd(2,12)=2\); \(i=2\), \(j=3\), \(k=4\), \(f(a_i,a_j,a_k)=f(3,6,12)=\gcd(3,6)=3\); \(i=2\), \(j=3\), \(k=5\), \(f(a_i,a_j,a_k)=f(3,6,17)=\gcd(3,6)=3\); \(i=2\), \(j=4\), \(k=5\), \(f(a_i,a_j,a_k)=f(3,12,17)=\gcd(3,12)=3\); \(i=3\), \(j=4\), \(k=5\), \(f(a_i,a_j,a_k)=f(6,12,17)=\gcd(6,12)=6\). The sum over all triples is \(1+1+1+2+2+2+3+3+3+6=24\).In the second test case, there are \(56\) ways to choose values of \(i\), \(j\), \(k\). The sum over all \(f(a_i,a_j,a_k)\) is \(203\).