Make Them Odd

time limit per test3 seconds
memory limit per test256 megabytes
input: standard input
output: standard output

There are n positive integers a1,a2,…,an. 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).

Input
The first line of the input contains one integer t(1≤t≤104) — the number of test cases in the input. Then t test cases follow.
The first line of a test case contains n(1≤n≤2⋅105) — the number of integers in the sequence a. The second line contains positive integers a1,a2,…,an(1≤ai≤109).The sum of n for all test cases in the input doesn’t exceed 2⋅105.
Output
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).

Example
Input
4
6
40 6 40 3 20 1
1
1024
4
2 4 8 16
3
3 1 7
Output
4
10
4
0

The meaning of problems:
the number t group, each number n, each array operation can be divided by 2 while the same until all of the array have become an odd number
as long as each even change process with an array of labeled down, if the loop current to a certain digit number has been marked on the end of the current cycle, and then in addition to recording a total of how many times just fine. A number from too large, do not define an int array, so use map like (but such a simple subject emmm should not somebody to blog it)
records for this problem is because of this approach reminds me of a brother had previously cc a question, we want to run five minutes to arrive at the answer locally, but because of the use of this approach would be quick ac. That question can not find it, use under this question instead, they have to make an impression -
or stick a pair of codes:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int t,n;
ll times;
 
int cmp(int a,int b){
	return a>b;
}
 
int main(){
	cin>>t;
	while(t--){
		cin>>n;
		times = 0;
		for(int i=1;i<=n;i++){
			cin>>a[i];
		}	
		sort(a+1,a+1+n,cmp);
		map<int,int> mp;
		for(int i=1;i<=n;i++){
			if(a[i]%2==1){
				continue;
			}else{
				while(a[i]%2==0 && mp[a[i]]==0){
					mp[a[i]]=1;
					a[i]/=2;
					times++;
				}
			}
		}
		cout<<times<<endl;
	}
	return 0;
	
}

　　

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CSDN link: https: //blog.csdn.net/weixin_43880627/article/details/103622672