CF1271E.Common Number
普及/提高-
通过率:0%
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题目描述
At first, let's define function f(x) as follows: $$$$ \begin{matrix} f(x) & = & \left{ \begin{matrix} \frac{x}{2} & \mbox{if } x \text{ is even} \ x - 1 & \mbox{otherwise } \end{matrix} \right. \end{matrix} $$
We can see that if we choose some value v and will apply function f to it, then apply f to f(v) , and so on, we'll eventually get 1 . Let's write down all values we get in this process in a list and denote this list as path(v) . For example, path(1) = \[1\] , path(15) = \[15, 14, 7, 6, 3, 2, 1\] , path(32) = \[32, 16, 8, 4, 2, 1\] .
Let's write all lists path(x) for every x from 1 to n . The question is next: what is the maximum value y such that y is contained in at least k different lists path(x) ?
Formally speaking, you need to find maximum y such that left∣x ∣ 1lexlen,yinpath(x)right∣gek$$.
输入格式
The first line contains two integers n and k ( 1≤k≤n≤1018 ).
输出格式
Print the only integer — the maximum value that is contained in at least k paths.
输入输出样例
输入#1
11 3
输出#1
5
输入#2
11 6
输出#2
4
输入#3
20 20
输出#3
1
输入#4
14 5
输出#4
6
输入#5
1000000 100
输出#5
31248
说明/提示
In the first example, the answer is 5 , since 5 occurs in path(5) , path(10) and path(11) .
In the second example, the answer is 4 , since 4 occurs in path(4) , path(5) , path(8) , path(9) , path(10) and path(11) .
In the third example n=k , so the answer is 1 , since 1 is the only number occuring in all paths for integers from 1 to 20 .