【Codeforces 743 BChloe and the sequence 】

B. Chloe and the sequence
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard 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

Note

In the first sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1]. The number on the second position is 2.

In the second sample the obtained sequence is [1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1]. The number on the eighth position is 4.

不该打着4级的旗号~不做练习的~不做点题总感觉缺点什么~

思路~数列是对称的,N- 1次变换后,数列长度为M = 2 ^ N - 1,每次对 M取半即可～看代码更容易懂～

AC代码：

#include<cstdio>typedef long long LL;LL KSM(LL y){    LL cut = 1,sum = 2;    while(y){        if(y & 1)            cut *= sum;        sum *= sum;        y >>= (LL)1;    }    return cut;}int main(){    LL N,K;    scanf("%I64d %I64d",&N,&K);    LL M = KSM(N) - 1;    while(M > 1){        if(K > M / 2 + 1)            K -= M / 2 + 1;        if(K == M / 2 + 1)           break;        N--;        M /= 2;    }    printf("%I64d\n",N);    return 0;}