Bead Challenge

You possess a vector A of size N, where A_i denotes the weight of the i-th bead among n beads on a table. Engaging in a competition against a machine, you have the liberty to select any two beads, keeping one for yourself (A_i) and surrendering the other to the machine (A_j). If you choose to retain A_i, you can partition its weight into k beads of equal weight (resulting in a new weight of A_i/k for the i-th bead). However, you must keep one of these new beads and discard the remaining k-1 beads. Subsequently, the machine will amplify the weight of the j-th bead (initially A_j) to k times its original weight (kA_j). The objective is to achieve equal weights for all beads through a series of operations. Return 1 if successful or 0 if unsuccessful in this challenge.

2 <= |A| <= 105

1 <= Ai <= 106

A = [2, 3, 1]

A = [100, 2, 50, 10, 1]

0

1

For Input 1:

We can never make all beads of equal size.

For Input 2:

Let's examine the vector A, which is [100, 2, 50, 10, 1], containing 5 elements. Two operations are performed on it:

Select a3 (which is 50) and a2 (which is 2), with k as 5. Replace a3 with a3 / k, resulting in 10, and replace a2 with a2 * k, resulting in 10. The modified array is now [100, 10, 10, 10, 1].

Choose a1 (which is 100) and a5 (which is 1), with k as 10. Replace a1 with a1 / k, resulting in 10, and replace a5 with a5 * k, resulting in 10. The final array is [10, 10, 10, 10, 10].