665A - Buses Between Cities

The problem was suggested by Sergey Erlikh unprost.

Consider the time interval when Simion will be on the road strictly between cities (x₁, y₁) (x₁ = 60h + m, y₁ = x₁ + t_a). Let's iterate over the oncoming buses. Let (x₂, y₂) be the time interval when the oncoming bus will be strictly between two cities. If the intersection of that intervals (x = max(x₁, x₂), y = min(y₁, y₂)) is not empty than Simion will count that bus.

int a, ta;
int b, tb;
int h, m;

bool read() {
	if (!(cin >> a >> ta)) return false;
	assert(cin >> b >> tb);
	assert(scanf("%d:%d", &h, &m) == 2);
	return true;
}

void solve() {
	int x1 = h * 60 + m;
	int y1 = x1 + ta;

	int ans = 0;
	for (int x2 = 5 * 60 + 0; x2 < 24 * 60; x2 += b) {
		int y2 = x2 + tb;
		int x = max(x1, x2), y = min(y1, y2);
		if (x < y)
			ans++;
	}
	cout << ans << endl;
}

Complexity: O(1).

665B - Shopping

The problem was suggested by Ayush Anand JeanValjean01.

In this problem you should simply do what was written in the problem statement. There are no tricks.

const int N = 111;

int n, m, k;
int p[N];
int a[N][N];

bool read() {
	if (!(cin >> n >> m >> k)) return false;
	forn(i, k) {
		assert(scanf("%d", &p[i]) == 1);
		p[i]--;
	}
	forn(i, n)
		forn(j, m) {
			assert(scanf("%d", &a[i][j]) == 1);
			a[i][j]--;
		}
	return true;
}

void solve() {
	int ans = 0;
	forn(i, n)
		forn(j, m) {
			int pos = int(find(p, p + k, a[i][j]) - p);
			ans += pos + 1;
			nfor(l, pos) p[l + 1] = p[l];
			p[0] = a[i][j];
		}
	cout << ans << endl;
}

Complexity: O(nmk).

665C - Simple Strings

The problem was suggested by Zi Song Yeoh zscoder.

There are two ways to solve this problem: greedy approach and dynamic programming.

The first apprroach: Considerr some segment of consecutive equal characters. Let k be the length of that segment. Easy to see that we should change at least characters in the segment to remove all the pairs of equal consecutive letters. On the other hand we can simply change the second, the fourth etc. symbols to letter that is not equal to the letters before and after the segment.

const int N = 200200;

int n;
char s[N];

bool read() {
	if (!gets(s)) return false;
	n = int(strlen(s));
	return true;
}

void solve() {
	for (int i = 0, j = 0; i < n; i = j) {
		while (j < n && s[j] == s[i]) j++;
		char c = 'a';
		while (c == s[i] || (i > 0 && c == s[i - 1]) || (j < n && c == s[j]))
			c++;
		fore(k, i, j)
			if ((i + k) & 1)
				s[k] = c;
	}
	puts(s);
}

The second approach: Let z_ka be the minimal number of changes so that the prefix of length k has no equal consecutive letters and the symbol s'_k equals to a. Let's iterate over the letter on the position k + 1 and if it is not equal to a make transition. The cost of the transition is equal to 0 if we put the same letter as in the original string s on the position k + 1. Otherwise the cost is equal to 1.

const int N = 1200300, A = 27;

int n;
char s[N];

bool read() {
	if (!gets(s)) return false;
	n = int(strlen(s));
	return true;
}

int z[N][A];
int p[N][A];
char ans[N];

void solve() {
	memset(z, 63, sizeof(z));

	z[0][A - 1] = 0;
	forn(i, n) {
		int c = int(s[i] - 'a');
		forn(j, A) {
			int dv = z[i][j];
			if (dv > INF / 2) continue;
			forn(nj, A - 1) {
				if (nj == j) continue;
				int ct = nj != c;
				if (z[i + 1][nj] > dv + ct) {
					z[i + 1][nj] = dv + ct;
					p[i + 1][nj] = j;
				}
			}
		}
	}

	int idx = int(min_element(z[n], z[n] + A) - z[n]);

	for (int i = n, j = idx; i > 0; j = p[i][j], i--)
		ans[i - 1] = char('a' + j);
	ans[n] = 0;

	cerr << z[n][idx] << endl;
	puts(ans);
}

Complexity: O(n).

665D - Simple Subset

The problem was suggested by Zi Song Yeoh zscoder.

Consider the subset A that is the answer to the problem. Let a, b, c be the arbitrary three elements from A and let no more than one of them is equal to 1. By the pigeonhole principle two of three elements from a, b, c have the same parity. So we have two integers with even sum and only one of them is equal to 1, so their sum is also greater than 2. So the subset A is not simple. In this way A consists of only two numbers greater than one (with a prime sum) or consists of some number of ones and also maybe other value x, so that x + 1 is a prime.

We can simply process the first case in O(n²) time. The second case can be processed in linear time. Also we should choose the best answer from that two.

To check the value of order 2·10⁶ for primality in O(1) time we can use the simple or the linear Eratosthenes sieve.

const int N = 1010, X = 2100300;

int n, a[N];

bool read() {
	if (!(cin >> n)) return false;
	forn(i, n) assert(scanf("%d", &a[i]) == 1);
	return true;
}

int szp, p[X];
int mind[X];

void prepare() {
	fore(i, 2, X) {
		if (!mind[i]) {
			p[szp++] = i;
			mind[i] = i;
		}
		for (int j = 0; j < szp && p[j] <= mind[i] && i * p[j] < X; j++)
			mind[i * p[j]] = p[j];
	}
}

void printAns(int cnt1, int a = -1, int b = -1) {
	vector<int> ans;
	forn(i, cnt1) ans.pb(1);
	if (a != -1) ans.pb(a);
	if (b != -1) ans.pb(b);
	assert(!ans.empty());
	random_shuffle(all(ans));

	cout << sz(ans) << endl;
	forn(i, sz(ans)) {
		if (i) putchar(' ');
		printf("%d", ans[i]);
	}
	puts("");
}

void solve() {
	int cnt1 = (int) count(a, a + n, 1);

	function<bool(int)> isPrime = [](int p) { return mind[p] == p; };

	if (cnt1 > 0)
		forn(i, n)
			if (a[i] != 1 && isPrime(a[i] + 1)) {
				printAns(cnt1, a[i]);
				return;
			}

	if (cnt1 > 1) {
		printAns(cnt1);
		return;
	}

	forn(i, n)
		forn(j, i)
			if (isPrime(a[i] + a[j])) {
				printAns(0, a[i], a[j]);
				return;
			}

	printAns(0, a[0]);
}

Complexity: O(n² + X), where X is the maximal value in a.

665E - Beautiful Subarrays

The problem was suggested by Zi Song Yeoh zscoder.

The sign is used for the binary operation for bitwise exclusive or.

Let s_i be the xor of the first i elements on the prefix of a. Then the interval (i, j] is beautiful if . Let's iterate over j from 1 to n and consider the values s_j as the binary strings. On each iteration we should increase the answer by the value z_j — the number of numbers s_i (i < j) so . To do that we can use the trie data structure. Let's store in the trie all the values s_i for i < j. Besides the structure of the trie we should also store in each vertex the number of leaves in the subtree of that vertex (it can be easily done during adding of each binary string). To calculate the value z_j let's go down by the trie from the root. Let's accumulate the value cur equals to the xor of the prefix of the value s_j with the already passed in the trie path. Let the current bit in s_j be equal to b and i be the depth of the current vertex in the trie. If the number cur + 2ⁱ ≥ k then we can increase z_j by the number of leaves in vertex , because all the leaves in the subtree of tha vertex correspond to the values s_i that for sure gives . After that we should go down in the subtree b. Otherwise if cur + 2ⁱ < k then we should simply go down to the subtree and recalculate the value cur = cur + 2ⁱ.

const int N = 1200300, LOGN = 30, V = N * LOGN;

int n, k;
int a[N];

bool read() {
	if (!(cin >> n >> k)) return false;
	forn(i, n) assert(scanf("%d", &a[i]) == 1);
	return true;
}

int tsz;
int nt[V][2];
int cnt[V];

void clear() {
	forn(i, V) {
		nt[i][0] = nt[i][1] = -1;
		cnt[i] = 0;
	}
	tsz = 1;
}

void add(int x) {
	int v = 0;
	cnt[v]++;

	nfor(i, LOGN) {
		int b = (x >> i) & 1;
		if (nt[v][b] == -1) {
			assert(tsz < V);
			nt[v][b] = tsz++;
		}
		v = nt[v][b];
		cnt[v]++;
	}
}


int calc(int x) {
	int v = 0;
	int ans = 0;

	auto getCnt = [](int v) { return v == -1 ? 0 : cnt[v]; };

	int cur = 0;
	nfor(i, LOGN) {
		if (v == -1) break;
		int b = (x >> i) & 1;
		if ((cur | (1 << i)) >= k) {
			ans += getCnt(nt[v][b ^ 1]);
			v = nt[v][b];
		} else {
			v = nt[v][b ^ 1];
			cur |= (1 << i);
		}
	}
	if (cur >= k) ans += getCnt(v);
	return ans;
}

void solve() {
	clear();

	add(0);

	li ans = 0;
	int s = 0;
	forn(i, n) {
		s ^= a[i];
		li cur = calc(s);
		ans += cur;
		add(s);
	}
	cout << ans << endl;
}

Comlpexity by the time and the memory: O(nlogX), where X is the maximal xor on the prefixes.

665F - Four Divisors

The editorial for this problem is a little modification of the materials from the lecture of Mikhail Tikhomirov Endagorion of the autumn of 2015 in Moscow Institute of Physics and Technology. Thanks a lot to Endagorion for that materials.

Easy to see that only the numbers of the form p·q and p³ (for different prime p, q) have exactly four positive divisors.

We can easily count the numbers of the form p³ in , where n is the number from the problem statement.

Now let p < q and π(k) be the number of primes from 1 to k. Let's iterate over all the values p. Easy to see that . So for fixed p we should increase the answer by the value .

So the task is ot to find — the number of primes not exceeding , for all p.

Denote p_j the j-th prime number. Denote dp_n, j the number of k such that 1 ≤ k ≤ n, and all prime divisors of k are at least p_j (note that 1 is counted in all dp_n, j, since the set of its prime divisors is empty). dp_n, j satisfy a simple recurrence:

1. dp_n, 1 = n (since p₁ = 2)

2. dp_n, j = dp_{n, j + 1} + dp_{⌊ n / pj⌋, j}, hence dp_{n, j + 1} = dp_n, j - dp_{⌊ n / pj⌋, j}

Let p_k be the smallest prime greater than . Then π(n) = dp_n, k + k - 1 (by definition, the first summand accounts for all the primes not less than k).

If we evaluate the recurrence dp_n, k straightforwardly, all the reachable states will be of the form dp_{⌊ n / i⌋, j}. We can also note that if p_j and p_k are both greater than , then dp_n, j + j = dp_n, k + k. Thus, for each ⌊ n / i⌋ it makes sense to keep only values of dp_{⌊ n / i⌋, j}.

Instead of evaluating all DP states straightforwardly, we perform a two-step process:

1. Choose K.

2. Run recursive evaluation of dp_n, k. If we want to compute a state with n < K, memorize the query ``count the numbers not exceeding n with all prime divisors at least k''.

3. Answer all the queries off-line: compute the sieve for numbers up to K, then sort all numbers by the smallest prime divisor. Now all queries can be answered using RSQ structure. Store all the answers globally.

4. Run recurisive evaluation of dp_n, k yet again. If we want to compute a state with n < K, then we must have preprocessed a query for this state, so take it from the global set of answers.

The performance of this approach relies heavily on Q — the number of queries we have to preprocess.

Statement. .

Proof:

Each state we have to preprocess is obtained by following a dp_{⌊ n / pj⌋, j} transition from some greater state. It follows that Q doesn't exceed the total number of states for n > K.

The preprocessing of Q queries can be done in , and it is the heaviest part of the computation. Choosing optimal , we obtain the complexity .

li n;

bool read() {
	return !!(cin >> n);
}

const int K = 10 * 1000 * 1000;
const int N = K;
const int P = 700100;
const int Q = 25 * 1000 * 1000;

int szp, p[P];
int mind[N];

void prepare() {
	szp = 0;
	forn(i, N) mind[i] = -1;
	fore(i, 2, N) {
		if (mind[i] == -1) {
			assert(szp < P);
			mind[i] = szp;
			p[szp++] = i;
		}
		for (int j = 0; j < szp && j <= mind[i] && i * p[j] < N; j++)
			mind[i * p[j]] = j;
	}
}

inline int getk(li n) {
	int lf = 0, rg = szp - 1;
	while (lf != rg) {
		int md = (lf + rg) >> 1;
		if (p[md] * 1ll * p[md] > n) rg = md;
		else lf = md + 1;
	}
	assert(p[lf] * 1ll * p[lf] > n);
	return lf;
}

int t[K];
void inc(int i, int val) {
	for ( ; i < K; i |= i + 1)
		t[i] += val;
}
int sum(int i) {
	int ans = 0;
	for ( ; i >= 0; i = (i & (i + 1)) - 1)
		ans += t[i];
	return ans;
}

int szq;
pti q[Q];
int ans[Q];
vector<int> vs[P];

void process() {
	sort(q, q + szq, greater<pti> ());
	memset(t, 0, sizeof(t));

	forn(i, szp) vs[i].clear();
	fore(i, 2, K)
		vs[mind[i]].pb(i);

	inc(1, +1);

	int p = szp - 1;
	for (int i = 0, j = 0; i < szq; i = j) {
		while (p >= q[i].x) {
			for (auto v : vs[p])
				inc(v, +1);
			p--;
		}

		while (j < szq && q[j].x == q[i].x) {
			ans[j] = sum(q[j].y);
			j++;
		}
	}
}

map<pair<li, int>, li> z;

li solve(li n, int jj, bool fs) {
	if (!n) return 0;
	int j = min(jj, getk(n));
	if (!j) return n + j - jj;

	li ans = 0;
	if (n < K) {
		pti p(j, (int) n);
		if (fs) {
			assert(szq < Q);
			q[szq++] = p;
			ans = 0;
			return 0;
		} else {
			int idx = int(lower_bound(q, q + szq, p, greater<pti> ()) - q);
			assert(idx < szq && q[idx] == p);
			ans = ::ans[idx];
		}
	} else {
		if (!z.count(mp(n, j))) {
			ans = solve(n, j - 1, fs);
			ans -= solve(n / p[j - 1], j - 1, fs);
			z[mp(n, j)] = ans;
		} else {
			ans = z[mp(n, j)];
		}
	}

	ans += j - jj;

	return ans;
}

inline li pi(li n, bool fs) {
	int k = szp - 1;
	return solve(n, k, fs) + k - 1;
}

void solve() {
	szq = 0;
	z.clear();
	for (int j = 0; p[j] * 1ll * p[j] <= n; j++) {
		li nn = n / p[j];
		if (nn > p[j]) {
			pi(n / p[j], true);
		}
	}

	process();

	z.clear();
	li ans = 0;
	for (int j = 0; p[j] * 1ll * p[j] <= n; j++) {
		li nn = n / p[j];
		if (nn > p[j]) {
			ans += pi(n / p[j], false);
			ans -= j + 1;
		}
	}

	for (int i = 0; i < szp && p[i] * 1ll * p[i] * 1ll * p[i] <= n; i++)
		ans++;

	cout << ans << endl;
}

Complexity: .