codeforces 743 B Chloe and the sequence(递归)

最新推荐文章于 2020-12-05 19:59:28 发布

HandsomeHow

最新推荐文章于 2020-12-05 19:59:28 发布

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本文介绍了一个竞赛编程题目，涉及通过递归方法找到特定序列中指定位置的整数值。序列由初始值1开始，通过一系列规则逐步扩展。文章提供了解题思路及C++实现代码。

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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.

题解：因为是对称的加上去，那递归就行了。

#include <bits/stdc++.h>
using namespace std;

#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define pb push_back
#define mp make_pair
#define cl(a) memset((a),0,sizeof(a))
#ifdef HandsomeHow
#define dbg(x) cerr << #x << " = " << x << endl
#else
#define dbg(x)
#endif
typedef long long ll;
typedef unsigned long long ull;
typedef pair <int, int> pii;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
const int mod=1000000007;
const double pi=acos(-1.0);
inline void gn(long long&x){
    int sg=1;char c;while(((c=getchar())<'0'||c>'9')&&c!='-');c=='-'?(sg=-1,x=0):(x=c-'0');
    while((c=getchar())>='0'&&c<='9')x=x*10+c-'0';x*=sg;}
inline void gn(int&x){long long t;gn(t);x=t;}
inline void gn(unsigned long long&x){long long t;gn(t);x=t;}
ll gcd(ll a,ll b){return a? gcd(b%a,a):b;}
ll powmod(ll a,ll x,ll mod){ll t=1ll;while(x){if(x&1)t=t*a%mod;a=a*a%mod;x>>=1;}return t;}
// (づ°ω°)づe★------------------------------------------------

ll solve(ll n, ll k){
	ll mid = (1ll)<<n;
	mid = (mid)/2;
	//printf("%lld %lld\n",n,mid);
	if(k == mid) return n;
	if(k>mid) return solve(n-1,k-mid);
	return solve(n-1,k);
}
int main(){
#ifdef HandsomeHow
	//freopen("data.in","r",stdin);
	//freopen("data.out","w",stdout);
#endif
	ll n,k;
	cin>>n>>k;
	cout<<solve(n,k)<<endl;
	return 0;
}