codefoeces 384 div2 b题（二分）

第一次打codeforces，好开心啊，二血。（虽然是水题。。而且因为longlong耽误了很多时间。）可是好吧。。这样说又不是很开心了。
B. Chloe and the sequence
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Chloe, the same as Vladik, is a competitive programmer. She didn’t have any problems to get to the olympiad like Vladik, but she was confused by the task proposed on the olympiad.

Let’s consider the following algorithm of generating a sequence of integers. Initially we have a sequence consisting of a single element equal to 1. Then we perform (n - 1) steps. On each step we take the sequence we’ve got on the previous step, append it to the end of itself and insert in the middle the minimum positive integer we haven’t used before. For example, we get the sequence [1, 2, 1] after the first step, the sequence [1, 2, 1, 3, 1, 2, 1] after the second step.

The task is to find the value of the element with index k (the elements are numbered from 1) in the obtained sequence, i. e. after (n - 1) steps.

Please help Chloe to solve the problem!

Input
The only line contains two integers n and k (1 ≤ n ≤ 50, 1 ≤ k ≤ 2n - 1).

Output
Print single integer — the integer at the k-th position in the obtained sequence.

Examples
input
3 2
output
2
input
4 8
output
4

很典型的二分，因为刚学了线段树，所以终于会做这种题了。

找规律，发现数字和递归的层数是一样的。

于是乎

#include <cstdio>#include <string>#include <cstring>#include <iostream>using namespace std;typedef long long LL;int main(){    LL n,m;    while(cin>>n>>m)    {        LL ans=1;        LL i=1;        while(ans<m)        {            ans=ans*2+1;            i++;        }        // printf("%d\n", i);        LL l=1,r=ans;        while(i)        {            LL mid=(r+l)>>1;            if(mid==m) break;            else if(mid<m) l=mid+1;            else  r=mid;            i--;        }        cout<<i<<endl;    }}