kth symbol in grammar.cpp

/*
///////////////////////////////////////////
//Question/Info

kth symbol in grammar

On the first row, we write a 0. Now in every subsequent row, we look at the previous row and replace each occurrence of 0 with 01, and each occurrence of 1 with 10.

Given row N and index K, return the K-th indexed symbol in row N. (The values of K are 1-indexed.) (1 indexed).

Examples:
Input: N = 1, K = 1
Output: 0

Input: N = 2, K = 1
Output: 0

Input: N = 2, K = 2
Output: 1

Input: N = 4, K = 5
Output: 1

Explanation:
row 1: 0
row 2: 01
row 3: 0110
row 4: 01101001
Note:

N will be an integer in the range [1, 30].
K will be an integer in the range [1, 2^(N-1)].

author: srj_v
///////////////////////////////////////////
*/

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

#define int 		   long long int
#define setbits(x) 	   __builtin_popcount(x)
#define pb(x) 		   push_back(x)
#define mp(x,y) 	   make_pair(x,y)
#define eb(x) 		   emplace_back(x)
#define ct(x) 		   cout << x << endl;
#define ct2(x,y) 	   cout << x << " " << y << endl;
#define forn(i,n) 	   for(int i = 0; i < (int)(n); ++i)
#define forx(i,x,n)    for(int i = x; i < (int)(n); ++i)
#define nfor(i,n)      for(int i = n-1; i >= 0; --i)
#define all(v) 		   v.begin(),v.end()
#define fsp(x,y)       fixed << setprecision(y) << x;
#define PI 			   3.1415926535897932384626433832795
#define MOD 		   1000000007 // (1e9+7)
#define pii 		   pair<int,int>
#define pis 		   pair<int,string>
#define vi  		   vector<int>
#define vii 		   vector<pii>
#define mii 		   map<int,int>
#define p_q            priority_queue // priority_queue<int> (&) priority_queue< int,vi,greater<int> >
#define _IOS      	   ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

typedef long double ld;

#pragma GCC optimize("Ofast")

void c_p_c()
{
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif
}

int gram(int n, int k) {

// base
	// if (n == 0 or k == 0) {
	// 	return -1;
	// }
	if (n == 1 and k == 1) {
		return 0;
	}

// since mid of n corressponds to th n-1 row...

	int mid = pow(2, n - 1) / 2;

	if (k <= mid) {
		// so equal to previous row
		return gram(n - 1, k);
	}
	else {

		return !gram(n - 1, k - mid);
	}

}


int32_t main() {
///////////
	c_p_c();
///////////
	_IOS
//////////

	// code

	/*
	int t ; cin >> t; while(t--){}

	*/

	int n = 3 ;
// nth row,

//  say we want kth index element
	int k = 2 ;

	ct(gram(n, k));

	// cerr << "time: " << clock() << " ms" << '\n';
	return 0;
}