Make Them Odd

题意翻译

### 题意简述 有一个长度为 $n$ 的数列 $a_1,a_2,...,a_n$。 您每次可以选择任意一个偶数 $c$，将数列中等于 $c$ 的数都除以 $2$，求至少需要多少次才能把数列中的所有数都变成奇数。 ### 输入格式 第一行一个正整数 $t(1\leq t \leq 10^4)$，表示数据的组数。 对于每组数据，第一行一个正整数 $n(1\leq n \leq 2·10^5)$。 接下来一行 $n$ 个正整数 $a_1,a_2,...,a_n(1\leq a_i \leq 10^9)$。 保证输入的 $n$ 之和不超过 $2·10^5$。 ### 输出格式 对于每组数据，输出一行，表示最少的操作次数。 翻译贡献者 U108949

题目描述

There are $ n $ positive integers $ a_1, a_2, \dots, a_n $ . For the one move you can choose any even value $ c $ and divide by two all elements that equal $ c $ . For example, if $ a=[6,8,12,6,3,12] $ and you choose $ c=6 $ , and $ a $ is transformed into $ a=[3,8,12,3,3,12] $ after the move. You need to find the minimal number of moves for transforming $ a $ to an array of only odd integers (each element shouldn't be divisible by $ 2 $ ).

输入输出格式

输入格式

The first line of the input contains one integer $ t $ ( $ 1 \le t \le 10^4 $ ) — the number of test cases in the input. Then $ t $ test cases follow. The first line of a test case contains $ n $ ( $ 1 \le n \le 2\cdot10^5 $ ) — the number of integers in the sequence $ a $ . The second line contains positive integers $ a_1, a_2, \dots, a_n $ ( $ 1 \le a_i \le 10^9 $ ). The sum of $ n $ for all test cases in the input doesn't exceed $ 2\cdot10^5 $ .

输出格式

For $ t $ test cases print the answers in the order of test cases in the input. The answer for the test case is the minimal number of moves needed to make all numbers in the test case odd (i.e. not divisible by $ 2 $ ).

输入输出样例

输入样例 #1

4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7

输出样例 #1

4
10
4
0

说明

In the first test case of the example, the optimal sequence of moves can be as follows: - before making moves $ a=[40, 6, 40, 3, 20, 1] $ ; - choose $ c=6 $ ; - now $ a=[40, 3, 40, 3, 20, 1] $ ; - choose $ c=40 $ ; - now $ a=[20, 3, 20, 3, 20, 1] $ ; - choose $ c=20 $ ; - now $ a=[10, 3, 10, 3, 10, 1] $ ; - choose $ c=10 $ ; - now $ a=[5, 3, 5, 3, 5, 1] $ — all numbers are odd. Thus, all numbers became odd after $ 4 $ moves. In $ 3 $ or fewer moves, you cannot make them all odd.