task 14

def kek (a, b):
	ans = 0
	p = 1
	a = str (a)
	a = a[::-1]
	for i in range (len (a)):
		ans += int (a[i]) * p
		p *= b
	return ans

x = 3 * 16 ** kek (46, 8) + 5 * 4 ** kek (40, 16) - 8 ** (kek (47, 8) - 27) - 2 ** (kek (110101, 2) + kek (13, 8)) + 15
l = list ()
while (x > 0):
	l.append (x % 16)
	x //= 16
mx = 0
for t in l:
	mx = max (mx, t)
cnt = 0
# print (*l)
for t in l:
	if (t == mx):
		cnt += 1
print (cnt)

task 5

int kek (int n) {
	vector <int> a;
	while (n) {
		a.pb (n % 3);
		n /= 3;
	}
	reverse (all (a));
	int last = a.back ();
	fr (k, 3) {
		vector <int> cnt (3);
		for (auto& x: a) ++cnt[x];
		int f = 0;
		fr (i, 3) {
			fr (j, 3) {
				if (i == j) continue;
				f |= cnt[i] == cnt[j];
			}
		}
		if (f) a.pb (last);
		else {
			int mn = inf;
			for (auto& x: cnt) chkmin (mn, x);
			fr (i, 3) {
				if (cnt[i] == mn) a.pb (i);
			}
		}
	}
	reverse (all (a));
	int ans = 0;
	int p = 1;
	fr (i, sz (a)) {
		ans += a[i] * p;
		p *= 3;
	}
	return ans;
}

signed main () {
	ios_base::sync_with_stdio (false);
	cin.tie (nullptr);
	fl (i, 71, 10000) {
		int x = kek (i);
		if (x % 2 == 0 && x % 6) {
			dbg (i);
			break;
		}
	}
}

task 15

int imp (int a, int b) {
	if (a == 1 && b == 0) return 0;
	return 1;
}

int kek (int a, int x) {
	int aa = 110 % x != 0;
	int bb = a % x != 0;
	int c = imp (aa, bb);
	int d = 730 % x == 0;
	int e = imp (d, c);
	int f = a % 35 == 0;
	return f && e;
}

signed main () {
	ios_base::sync_with_stdio (false);
	cin.tie (nullptr);
	int cnt = 0;
	for (int a = 1; a <= 1000; ++a) {
		int f = 1;
		for (int x = 1; x <= 100000; ++x) {
			f &= kek (a, x);
		}
		cnt += f;
	}
	cout << cnt << '\n';
}

task 16

int dp[100000 + 100];

int f (int n) {
	if (dp[n] != -1) return dp[n];
	if (n < 2) return dp[n] = 1;
	if (n >= 2 && n % 2 == 0) return dp[n] = f (n / 2) + 1;
	if (n >= 2 && n % 2 == 1) return dp[n] = f (n - 3) + 3;
	assert (0);
}

signed main () {
	ios_base::sync_with_stdio (false);
	cin.tie (nullptr);
	for (auto& x: dp) x = -1;
	int cnt = 0;
	fl (i, 1, 100000 + 1) cnt += f (i) == 12;
	cout << cnt << '\n';
}

task 17

int kek (int x) {
	int f = x % 6 == 0;
	while (x) {
		f &= x % 5 == 2;
		x /= 5;
	}
	return f;
}

signed main () {
	ios_base::sync_with_stdio (false);
	cin.tie (nullptr);
	int cnt = 0, s = 0;
	fl (i, 10, 6000 + 1) {
		if (kek (i)) ++cnt, s += i;
	}
	cout << cnt << ' ' << s << '\n';
}

task 18

int n = 10;

int corr (int i, int j) {
	return i >= 0 && i < n && j >= 0 && j < n;
}

vector <pair <int, int>> d = {{-1, -2}, {-2, -1}, {1, -2}, {2, -1}};

int g[10][10], dp[10][10];

signed main () {
	ios_base::sync_with_stdio (false);
	cin.tie (nullptr);
	fr (i, n) fr (j, n) cin >> g[i][j];
	fr (i, n) fr (j, n) dp[i][j] = inf;
	fr (i, n) dp[i][0] = g[i][0];
	fr (j, n) {
		fr (i, n) {
			for (auto& x: d) {
				int ii = i + x.fi, jj = j + x.se;
				if (corr (ii, jj)) {
					chkmin (dp[i][j], dp[ii][jj] + g[i][j]);
				}
			}
		}
	}
	int mn = inf;
	fr (i, n) chkmin (mn, dp[i][n - 1]);
	cout << mn << '\n';
}