Codeforces Round 905 (Div. 3) |
---|
Finished |
You are given an array of integers $$$a_1, a_2, \ldots, a_n$$$ and a number $$$k$$$ ($$$2 \leq k \leq 5$$$). In one operation, you can do the following:
Find the minimum number of operations needed to make the product of all the numbers in the array $$$a_1 \cdot a_2 \cdot \ldots \cdot a_n$$$ divisible by $$$k$$$.
Each test consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then follows the description of the test cases.
The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \leq n \leq 10^5$$$, $$$2 \leq k \leq 5$$$) — the size of the array $$$a$$$ and the number $$$k$$$.
The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 10$$$).
It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.
For each test case, output the minimum number of operations needed to make the product of all the numbers in the array divisible by $$$k$$$.
152 57 33 37 4 15 29 7 7 3 95 55 4 1 2 37 49 5 1 5 9 5 13 46 3 63 46 1 53 41 5 94 41 4 1 13 43 5 34 58 9 9 32 51 62 510 104 51 6 1 12 57 7
2 2 1 0 2 0 1 2 0 1 1 4 0 4 3
In the first test case, we need to choose the index $$$i = 2$$$ twice. After that, the array will be $$$a = [7, 5]$$$. The product of all the numbers in the array is $$$35$$$.
In the fourth test case, the product of the numbers in the array is $$$120$$$, which is already divisible by $$$5$$$, so no operations are needed.
In the eighth test case, we can perform two operations by choosing $$$i = 2$$$ and $$$i = 3$$$ in any order. After that, the array will be $$$a = [1, 6, 10]$$$. The product of the numbers in the array is $$$60$$$.
Name |
---|