Problem:
Score : 700 points

Problem Statement

Given is a sequence of N digits a_1a_2\ldots a_N, where each element is 1, 2, or 3.
Let x_{i,j} defined as follows:

x_{1,j} := a_j \quad (1 \leq j \leq N)

x_{i,j} := | x_{i-1,j} - x_{i-1,j+1} | \quad (2 \leq i \leq N and 1 \leq j \leq N+1-i)

Find x_{N,1}.

Constraints

2 \leq N \leq 10^6

a_i = 1,2,3 (1 \leq i \leq N)

Input

Input is given from Standard Input in the following format:

N
a_1a_2\ldotsa_N

Output

Print x_{N,1}.

Sample Input 1

4
1231

Sample Output 1

1

x_{1,1},x_{1,2},x_{1,3},x_{1,4} are respectively 1,2,3,1.

x_{2,1},x_{2,2},x_{2,3} are respectively |1-2| = 1,|2-3| = 1,|3-1| = 2.

x_{3,1},x_{3,2} are respectively |1-1| = 0,|1-2| = 1.

Finally, x_{4,1} = |0-1| = 1, so the answer is 1.

Sample Input 2

10
2311312312

Sample Output 2

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define sz(x) ((int)(x).size())
using namespace std;
typedef long long ll;
const int n = 1e6 + 5;

int N, M, K, T;
int A[n];
int F[n], TC[n];

bool bc(int n, int k){
	int r = F[n] - F[k] - F[n-k];
	//if(!(r >= 0)){
//		cerr << "n: " << n << " k:"
//	}
	return r == 0;
}

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cin >> N;
	F[0] = 0;
	for(int i = 0; i <= N; ++i){
		int z = i;
		while(z > 0 && (z % 2) == 0){
			TC[i]++;
			z = z >> 1;
		}
	}

	for(int i = 1; i <= N; ++i){
		F[i] = F[i-1] + TC[i];
	}
	/*
	for(int i = 0; i < 10; ++i){
		for(int j = 0; j <= i; ++j){
			cout << bc(i, j) << " ";
		}
		cout <<endl;
	}
*/
	string s; cin >> s;
	bool has1 = false;
	vector<int> Ones;

	for(int i = 0; i < N; ++i){
		A[i] = s[i] - '1';
		has1 |= A[i] == 1;
	}
	for(int i = 0; i < N; ++i){
		if(has1){
			if(A[i] == 2) A[i] = 0;
		}else{
			if(A[i] == 2) A[i] = 1;
		}
		if(A[i] == 1)
			Ones.push_back(i);
	}

	int o = 0;
	for(int i : Ones){
		o += bc(N-1, i);
	}
	o %= 2;
	int r = 0;
	if(has1 && o) r = 1;
	if(!has1 && o) r = 2;
	cout << r << endl;
	return 0;
}