SOLVE
LATER
Akash is interested in a new function F such that,
\(F(x) = \frac{1}{GCD(1, x)} + \frac{2}{GCD(2, x)} + ... + \frac{x}{GCD(x, x)}\)
where \(GCD\) is the Greatest Common Divisor.
Now, the problem is quite simple. Given an array A of size N, there are 2 types of queries:
Input:
First line of input contain integer N, size of the array.
Next line contain N space separated integers.
Next line contain integer Q, number of queries.
Next Q lines contain one of the two queries.
Output:
For each of the first type of query, output the required sum (mod \(10^9 + 7\)).
Constraints:
\(1 \le N \le 10^6\)
\(1 \le Q \le 10^5\)
\(1 \le A[i] \le 5 * 10^5\)
For \(1^{st}\) type of query,
\(1 \le X \le Y \le N\)
For \(2^{nd}\) type of query
\(1 \le X \le N\)
\(1 \le Y \le 5 * 10^5\)
\(A[1] = 3, A[2] = 4, A[3] = 3\)
\(F(3) = \frac{1}{GCD(1, 3)} + \frac{2}{GCD(2, 3)} + \frac{3}{GCD(3, 3)} = 1 + 2 + 1 = 4\)
\(F(4) = \frac{1}{GCD(1, 4)} + \frac{2}{GCD(2, 4)} + \frac{3}{GCD(3, 4)} + \frac{4}{GCD(4, 4)} = 1 + 1 + 3 + 1 = 6.\)
First query, the sum will be \(F(3) + F(4) = 4 + 6 = 10 (mod 10^9 + 7)\).
Second query, the sum will be \(F(3) + F(4) + F(3) = 4 + 6 + 4 = 14 (mod 10^9 + 7)\).
Third query, the sum will be \(F(3) = 4 (mod 10^9 + 7)\).
Fourth query will update \(A[1] = 4\).
Fifth query, the sum will be \(F(4) + F(4) + F(3) = 6 + 6 + 4 = 16 (mod 10^9 + 7)\).
Sixth query, the sum will be \(F(4) + F(4) = 6 + 6 = 12 (mod 10^9 + 7).\)
Challenge Name
HackerEarth Collegiate Cup - First Elimination
HackerEarth Collegiate Cup - First Elimination