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

const int MAX = 1000000;
int N, L;
long long a[MAX];
long long c[MAX + 1];

int index_start, index_end;

void input(){
	cin >> N;
	for(int i = 0; i < N; i++){
		cin >> a[i];
	}
}

// check if positive elements ai make a connected component
bool connected(){
	int counter = 0;
	bool last_zero = true;
	for(int i = 0; i < N; i++){
		if(last_zero){
			if(a[i] > 0){
				last_zero = false;
				counter++;
			}
		} else {
			if(a[i] == 0){
				last_zero = true;
				counter++;
			}
		}
	}
	return counter < 3;
}

void calculate_c(){
	index_start = -1; // index pointing to first non-zero element.
	index_end = -1; // index pointing to last non-zero element.
	for(int i = 0; i < N; i++){
		if(a[i] > 0){
			index_start = i;
			break;
		}
	}
	for(int i = N - 1; i >= 0; i--){
		if(a[i] > 0){
			index_end = i;
			break;
		}
	}
	// so from now on we consider only the sequence a[index_start], ... , a[index_end]
	c[0] = 0;
	for(int i = index_start; i <= index_end; i++){
		c[i - index_start + 1] = a[i] - c[i - index_start];
	}
	L = index_end - index_start + 1;
}

void case0(){
	// in here, c[0] = 0 and c[L] = 0.
	// we also assume L >= 3.
	// now we have to essentially try out two parities,
	// see if any of them fit.
	bool WORKS = true;
	for(int i = 1; i <= L - 1; i++){
		if(c[i] < (i%2)){
			WORKS = false;
			break;
		}
	}
	if(L % 2 == 1){
		if(c[L - 1] <= 0){
			WORKS = false;
		}
	}
	if(WORKS){
		cout << "TAK\n";
		return;
	}

	WORKS = true;
	if(c[1] <= 0){
		WORKS = false;
	}
	for(int i = 2; i <= L - 1; i++){
		if(c[i] < ((i + 1)%2)){
			WORKS = false;
			break;
		}
	}
	if(L % 2 == 0){
		if(c[L - 1] <= 0){
			WORKS = false;
		}
	}
	if(WORKS){
		cout << "TAK\n";
		return;
	}

	cout << "NIE\n";
}

void case1(){
	// in here, c[0] = 0 and c[L] = 1.
	// this means that we need sequence:
	// 0   >0      >=0       >=0       >=0 >0        >0        >0   0
	// 0, ... , 0, -1, 0, -1, ... , -1, 0, -1, 1, -1, ... , -1, 1, -1
	// 0            x                       y                       L
	// in here, extending x as close to beginning as possible is optimal.
	// also, moving y as far to the end as possible is optimal as well.

	// first assume x = 1 (and y = L) but also 2 divides x - y.
	if(L % 2 == 1){
		bool WORKS = true;
		for(int i = 1; i <= L - 1; i++){
			if(c[i] < (i%2)){
				WORKS = false;
				break;
			}
		}
		if(WORKS){
			cout << "TAK\n";
			return;
		}
	}
	
	if(L % 2 == 0){
		bool WORKS = true;
		if(c[1] <= 0){
			WORKS = false;
		}
		for(int i = 2; i <= L - 1; i++){
			if(c[i] < ((i + 1)%2)){
				WORKS = false;
				break;
			}
		}
		if(WORKS){
			cout << "TAK\n";
			return;
		}
	}
	
	cout << "NIE\n";
}

void casem1(){
	// in here, c[0] = 0 and c[L] = -1.
	// this means that we need sequence:
	// 0   >0      >=0       >=0       >=0 >0        >0        >0      0
	// 0, ... , 0, -1, 0, -1, ... , -1, 0, -1, 1, -1, ... , -1, 1, -1, 1
	// 0            x                       y                          L
	// in here, extending x as close to beginning as possible is optimal.
	// also, moving y as far to the end as possible is optimal as well.

	// first assume x = 1 (and y = L - 1) but also 2 divides x - y.
	if(L % 2 == 0){
		bool WORKS = true;
		for(int i = 1; i <= L - 1; i++){
			if(c[i] < (i%2)){
				WORKS = false;
				break;
			}
		}
		if(c[L - 1] <= 1){
			WORKS = false;
		}
		if(WORKS){
			cout << "TAK\n";
			return;
		}
	}
	
	if(L % 2 == 1){
		bool WORKS = true;
		if(c[1] <= 0){
			WORKS = false;
		}
		for(int i = 2; i <= L - 1; i++){
			if(c[i] < ((i + 1)%2)){
				WORKS = false;
				break;
			}
		}
		if(c[L - 1] <= 1){
			WORKS = false;
		}
		if(WORKS){
			cout << "TAK\n";
			return;
		}
	}
	
	cout << "NIE\n";
}

int main(){
	cin.tie(NULL);
	ios_base::sync_with_stdio(0);
	int T;	
	cin >> T;
	for(int v = 0; v < T; v++){
		input();
		if(!connected()){
			cout << "NIE\n";
			continue;
		}
		calculate_c();
		if(L == 1){
			if(a[index_start] == 1){
				cout << "TAK\n";
			} else {
				cout << "NIE\n";
			}
			continue;
		} else if(L == 2){
			if(abs(a[index_start] - a[index_end]) <= 1){
				cout << "TAK\n";
			} else {
				cout << "NIE\n";
			}
			continue;
		}
		// and now we have 3 cases to consider...
		if(c[L] == 0){
			case0();
		} else if(c[L] == 1){
			case1();
		} else if(c[L] == -1){
			casem1();
		} else {
			cout << "NIE\n";
		}
	}
}