Codeforces Round #641 (Div. 1) 题解

Codeforces Round #641 (Div. 1)

Orac and LCM

每个质因子考虑一下即可。

namespace shai {
	const int N = 1e6 + 7;
	int p[N], v[N], phi[N], miu[N];

	inline void init(int n) {
		v[1] = phi[1] = miu[1] = 1;
		for (int i = 2; i <= n; i++) {
			if (!v[i]) p[++p[0]] = v[i] = i, phi[i] = i - 1, miu[i] = -1;
			for (int j = 1; j <= p[0] && i * p[j] <= n && p[j] <= v[i]; j++)
				v[i*p[j]] = p[j],
				phi[i*p[j]] = phi[i] * (p[j] - 1 + (p[j] == v[i])),
				miu[i*p[j]] = p[j] == v[i] ? 0 : -miu[i];
		}
	}
}
using shai::p;

const int N = 1e5 + 7, M = 2e5 + 7;
int n, a[N], c[M];
pq<int> q[M];

inline vector<pair<ll, int>> divide(ll n) {
	vector<pair<ll, int>> p;
	for (ll i = 2; i * i <= n; i++)
		if (!(n % i)) {
			p.pb(mp(i, 0));
			while (!(n % i)) ++p.back().se, n /= i;
		}
	if (n > 1) p.pb(mp(n, 1));
	return p;
}

int main() {
	rd(n), rda(a, n);
	shai::init(2e5);
	for (int i = 1; i <= n; i++) {
		auto p = divide(a[i]);
		for (auto o : p) {
			q[o.fi].push(o.se);
			if (q[o.fi].size() > 2u) q[o.fi].pop();
			++c[o.fi];
		}
	}
	ll ans = 1;
	for (int i = 1; i <= p[0]; i++) {
		int x = p[i];
		if (c[x] == n) {
			int y = q[x].top();
			while (y--) ans *= x;
		}
		if (c[x] == n - 1) {
			if (q[x].size() > 1u) q[x].pop();
			int y = q[x].top();
			while (y--) ans *= x;
		}
	}
	print(ans);
	return 0;
}

Orac and Medians

首先判一些特殊情况。

一般情况就是判是否存在连续三个数中有两个 $\ge k$。

const int N = 1e5 + 7;
int n, k, a[N];

inline void solve() {
	rd(n, k), rda(a, n);
	vi p;
	for (int i = 1; i <= n; i++)
		if (a[i] == k) p.pb(i);
	if (!p.size()) return prints("no");
	if (n == 1) return prints("yes");
	if (n == 2) return prints(a[1] >= k && a[2] >= k ? "yes" : "no");
	for (int i = 1; i <= n - 2; i++) {
		int t = (a[i] >= k) + (a[i+1] >= k) + (a[i+2] >= k);
		if (t >= 2) return prints("yes");
	}
	prints("no");
}

int main() {
	int T;
	rd(T);
	while (T--) solve();
	return 0;
}

Orac and Game of Life

随便 BFS 一下。

const int N = 1e3 + 7;
ll inf = 1e18 + 1;
const int dx[] = {0,0,1,-1};
const int dy[] = {1,-1,0,0};
int n, m, t;
ll d[N][N];
char s[N][N];

int main() {
	rd(n, m, t);
	for (int i = 1; i <= n; i++) rds(s[i], m);
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
			d[i][j] = inf;
	queue<pi> q;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++) {
			bool ok = 0;
			for (int o = 0; o < 4; o++) {
				int x = i + dx[o], y = j + dy[o];
				if (x < 1 || x > n || y < 1 || y > m) continue;
				if (s[x][y] == s[i][j]) ok = 1;
			}
			if (ok) d[i][j] = 0, q.push(mp(i, j));
		}
	while (q.size()) {
		int i = q.front().fi, j = q.front().se;
		q.pop();
		for (int o = 0; o < 4; o++) {
			int x = i + dx[o], y = j + dy[o];
			if (x < 1 || x > n || y < 1 || y > m) continue;
			if (d[x][y] > d[i][j] + 1) d[x][y] = d[i][j] + 1, q.push(mp(x, y));
		}
	}
	while (t--) {
		int x, y;
		ll z;
		rd(x, y);
		rd(z);
		if (z <= d[x][y]) print(s[x][y] - '0');
		else {
			int k = s[x][y] - '0';
			z -= d[x][y];
			k ^= z & 1;
			print(k);
		}
	}
	return 0;
}

Slime and Biscuits

题目太神仙了，先咕咕咕着。

Slime and Hats

题目太神仙了，先咕咕咕着。

Slime and Sequences

题目太神仙了，先咕咕咕着。