Codeforces Round #665 (Div. 2)

A

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
#define IO ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;
 
const int N = 1e5 + 5;
 
int n, m, _, k;
 
int main() {
    IO;
    for (cin >> _; _; --_) {
        cin >> n >> k;
		if (k >= n) cout << k - n << '\n';
		else if (k < n) cout << min(n - k, (n + k) & 1) << '\n';
    }
    return 0;
}

B

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
#define IO ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;
 
const int N = 1e5 + 5;
 
int n, m, _, k;
int x[3], y[3];
 
int main() {
    IO;
    for (cin >> _; _; --_) {
        rep (i, 0, 2) cin >> x[i];
		rep (i, 0, 2) cin >> y[i];
		ll ans = 0;
		if (x[2] >= y[1]) ans += (y[1] << 1), x[2] -= y[1], y[1] = 0;
		else ans += (x[2] << 1), y[1] -= x[2], x[2] = 0;
 
		if (x[2] >= y[2]) x[2] -= y[2], y[2] = 0;
		else y[2] -= x[2], x[2] = 0;
 
		if (x[2] >= y[0]) x[2] -= y[0], y[0] = 0;
		else y[0] -= x[2], x[2] = 0;
 
		if (x[1] >= y[1]) x[1] -= y[1], y[1] = 0;
		else y[1] -= x[1], x[1] = 0;
 
		if (x[1] >= y[0]) x[1] -= y[0], y[0] = 0;
		else y[0] -= x[1], x[1] = 0;
 
		ans -= (x[1] << 1);
		cout << ans << '\n'; 
    }
    return 0;
}

C

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
#define IO ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;
 
const int N = 1e5 + 5;
 
int n, m, _, k;
int a[N], b[N];
 
int main() {
    IO;
    for (cin >> _; _; --_) {
        cin >> n;
		rep (i, 1, n) cin >> a[i], b[i] = a[i];
		sort(b + 1, b + 1 + n);
 
		bool f = 1;
		rep (i, 1, n) {
			if (b[i] == a[i]) continue;
			f = (a[i] % b[1] == 0);
			if (!f) break;
		}
 
		if (f) cout << "YES\n";
		else cout << "NO\n";
    }
    return 0;
}

D

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define per(i,a,b) for(int i=(a);i>=(b);--i)
#define IO ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;
typedef vector<int> VI;
typedef double db;

const int N = 1e6 + 5, mod = 1e9 + 7;

int n, m, _, k;
int h[N], to[N << 1], ne[N << 1], tot;
ll b[N], siz[N];
ll tax[N], cnt;

void add(int u, int v) {
	ne[++tot] = h[u]; h[u] = tot; to[tot] = v;
}

void dfs(int u, int fa) {
	siz[u] = 1;
	for (int i = h[u]; i; i = ne[i]) {
		int y = to[i];
		if (y == fa) continue;
		dfs(y, u);
		siz[u] += siz[y];
	}

	if (fa) tax[++cnt] = siz[u] * (n - siz[u]);
}

int main() {
	IO;
	for (cin >> _; _; --_) {
		cin >> n; tot = cnt = 0;
		rep(i, 1, n) h[i] = 0;
		rep(i, 2, n) {
			int u, v; cin >> u >> v;
			add(u, v); add(v, u);
		}
		dfs(1, 0);
		sort(tax + 1, tax + n);

		cin >> m;
		rep(i, 1, m) cin >> b[i];
		sort(b + 1, b + 1 + m);

		ll ans = 0;
		while (m > n - 1) b[m - 1] = (b[m - 1] * b[m]) % mod, --m;
		per (i, n - 1, 1) {
			if (m > 0) ans = (ans + tax[i] % mod * b[m] % mod) % mod, --m;
			else ans = (ans + tax[i]) % mod;
		}
		cout << ans << '\n';
	}
	return 0;
}