678A - Вася любит числа

The problem was suggested by Abdrakhman Ismail bash.

We should find minimal x, so x·k > n. Easy to see that . To learn more about floor/ceil functions I reccomend the book of authors Graham, Knuth, Patashnik "Concrete Mathematics". There is a chapter there about that functions and their properties.

li n, k;

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

void solve() {
	cout << (n / k + 1) * k << endl;
}

Complexity: O(1).

678B - Такой же календарь

The problem was suggested by Arthur Jaworski KingArthur.

Two calendars are same if and only if they have the same number of days and starts with the same day of a week. So we should simply iterate over years and maintain the day of a week of January, 1st (for example). Easy to see that the day of a week increases by one each year except of the leap years, when it increases by two.

int y;

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

bool leap(int y) {
	return y % 400 == 0 || (y % 4 == 0 && y % 100 != 0);
}

void solve() {
	bool is_leap = leap(y);
	int d = 0;
	do {
		d++;
		if (leap(y)) d++;
		y++;
		d %= 7;
	} while (d || leap(y) != is_leap);
	cout << y << endl;
}

Complexity: O(1) — easy to see that we will not iterate more than some small fixed constant times.

678C - Мила и шоколад

The problem was suggested by Sheikh Monir skmonir.

Easy to see that we can paint with both colours only tiles with the numbers multiple of lcm(a, b). Obviously that tiles should be painted with more expensive colour. So the answer equals to .

li n, a, b, p, q;

bool read() {
	return !!(cin >> n >> a >> b >> p >> q);
}

li gcd(li a, li b) { return !a ? b : gcd(b % a, a); }
li lcm(li a, li b) { return a / gcd(a, b) * b; }

void solve() {
	li ans = 0;
	ans += (n / a) * p;
	ans += (n / b) * q;
	ans -= (n / lcm(a, b)) * min(p, q);
	cout << ans << endl;
}

Complexity: O(log(max(a, b))).