CF1688B.Patchouli's Magical Talisman
普及/提高-
通过率:0%
AC君温馨提醒
该题目为【codeforces】题库的题目,您提交的代码将被提交至codeforces进行远程评测,并由ACGO抓取测评结果后进行展示。由于远程测评的测评机由其他平台提供,我们无法保证该服务的稳定性,若提交后无反应,请等待一段时间后再进行重试。
题目描述
She is skilled in all kinds of magics, and is keen on inventing new one.
—Perfect Memento in Strict Sense
Patchouli is making a magical talisman. She initially has n magical tokens. Their magical power can be represented with positive integers a1,a2,…,an .
Patchouli may perform the following two operations on the tokens.
- Fusion: Patchouli chooses two tokens, removes them, and creates a new token with magical power equal to the sum of the two chosen tokens.
- Reduction: Patchouli chooses a token with an even value of magical power x , removes it and creates a new token with magical power equal to 2x .
Tokens are more effective when their magical powers are odd values. Please help Patchouli to find the minimum number of operations she needs to make magical powers of all tokens odd values.
输入格式
Each test contains multiple test cases.
The first line contains a single integer t ( 1≤t≤103 ) — the number of test cases. The description of the test cases follows.
For each test case, the first line contains one integer n ( 1≤n≤2⋅105 ) — the initial number of tokens.
The second line contains n intergers a1,a2,…,an ( 1≤ai≤109 ) — the initial magical power of the n tokens.
It is guaranteed that the sum of n over all test cases does not exceed 2⋅105 .
输出格式
For each test case, print a single integer — the minimum number of operations Patchouli needs to make all tokens have an odd value of magical power.
It can be shown that under such restrictions the required sequence of operations exists.
输入输出样例
输入#1
4 2 1 9 3 1 1 2 3 2 4 8 3 1049600 33792 1280
输出#1
0 1 3 10
说明/提示
Test case 1:
a consists solely of odd numbers initially.
Test case 2:
Choose the tokens with magical power of 1 and 2 and perform Fusion. Now a=[1,3] , both are odd numbers.
Test case 3:
Choose the tokens with magical power of 2 and 8 and perform Fusion. Now a=[4,10] .
Choose the token with magical power of 10 and perform Reduction. Now a=[4,5] .
Choose the tokens with magical power of 4 and 5 and perform Fusion. Now a=[9] , and 9 is an odd number.
It can be shown that you can not make all the magical powers odd numbers in less than 3 moves, so the answer is 3 .