1527A - And Then There Were K.cpp

/*
Given an integer n, find the maximum value of integer k such that the following condition holds:

n & (n-1) & (n-2) & (n-3) & ... (k) = 0
where & denotes the bitwise AND operation.
Input
The first line contains a single integer t (1=t=3·104). Then t test cases follow.

The first line of each test case contains a single integer n (1=n=109).

Output
For each test case, output a single integer — the required integer k.

Example
inputCopy
3
2
5
17
outputCopy
1
3
15
Note
In the first testcase, the maximum value for which the continuous & operation gives 0 value, is 1.

In the second testcase, the maximum value for which the continuous & operation gives 0 value, is 3. No value greater then 3, say for example 4, will give the & sum 0.

5&4?0,
5&4&3=0.
Hence, 3 is the answer.


*/



#include<bits/stdc++.h>
using namespace std;


int main(){
    ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	int t;
	cin >> t;
	
	while (t--) {
		long long n;
		cin >> n;
		
		long long p= log2(n);
		long long res = pow(2, p);
		
		cout << res - 1 << "\n";
	}
	

    return 0;
}




/*
#include<bits/stdc++.h>
using namespace std;
 
int main(){
    ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	int t;
	cin >> t;
	
	while (t--) {
		int n;
		cin >> n;
		
		int cnt = 0;
		
		while (n != 1) {
			cnt++;
			n /= 2;
		}
		
		long long res = (long long)pow(2, cnt) - 1;
		cout <<  res << "\n";
	}
 
    return 0;
}
*/