A14343 | And Then There Were K
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题目描述
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.](https://en.wikipedia.org/wiki/Bitwise_operation#AND)
$n$ & ( $n-1$ ) & ( $n-2$ ) & ( $n-3$ ) & ... ( $k$ ) = $0$ where & denotes the [bitwise AND operation.](https://en.wikipedia.org/wiki/Bitwise_operation#AND)
输入格式
The first line contains a single integer $t$ ( $1 \le t \le 3 \cdot 10^4$ ). Then $t$ test cases follow.
The first line of each test case contains a single integer $n$ ( $1 \le n \le 10^9$ ).
The first line of each test case contains a single integer $n$ ( $1 \le n \le 10^9$ ).
输出格式
For each test case, output a single integer — the required integer $k$ .
输入输出样例
输入 #1
3 2 5 17
输出 #1
1 3 15
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.
- $5 \, \& \, 4 \neq 0$ ,
- $5 \, \& \, 4 \, \& \, 3 = 0$ .
Hence, 3 is the answer.
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.
- $5 \, \& \, 4 \neq 0$ ,
- $5 \, \& \, 4 \, \& \, 3 = 0$ .
Hence, 3 is the answer.
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评测结果：Accepted