Codeforces 743B Chloe and the sequence

竞赛编程挑战

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本文介绍了一个竞赛编程题目，涉及通过特定算法生成整数序列，并找出序列中指定位置的元素。提供了两种解决方案，一种为模拟法，另一种利用位运算优化。

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.

题意：

给定一个整数 n ，操作 n-1 次后，问第 k 位是多少。操作要求，必须把下一个置入序列的数字作为下一次操作结果后序列的中间。

示例：

1 初始化

1 2 1置入 2

1 2 1 3 1 2 1置入 3

1 2 1 3 1 2 1 4 1 2 1 3 1 2 1置入 4

1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1置入 5

1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 6 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1 5 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1置入 6

思路：模拟吧，二分应该也可以。

AC CODE：

#include<stdio.h>
#include<cstring>
#include<algorithm>
#define HardBoy main()
#define ForMyLove return 0;
using namespace std;
typedef __int64 LL;
const int MYDD = 1103;

int HardBoy {
	LL n, k;
	scanf("%I64d %I64d", &n, &k);
	LL ans = 1;
	while(k%2 == 0){
		k /= 2;
		ans++;
	} 
	printf("%I64d\n", ans);
	ForMyLove
}//From China

如果来扒拉扒拉 Code 的

看高手的吧，位运算技巧性真强

AC CODE：一高手的核心代码

	scanf("%I64d %I64d",&n,&k);
	k=k&-k;
	for(int i = 0; i <= 51; i++)
		if (k&(1ll<<i)) return printf("%d\n",i+1),0;