Codeforces Round #638 (Div. 2) 题解

Codeforces Round #638 (Div. 2)

Phoenix and Balance

由于最大的要比其它所有加起来还要大，因此一定是最大的那个带 $\frac n2 -1$ 个最小的。

int n;

inline void solve() {
	rd(n);
	ll ans = 1 << n;
	for (int i = 1; i < (n >> 1); i++) ans += 1 << i;
	for (int i = n >> 1; i < n; i++) ans -= 1 << i;
	print(ans);
}

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

Phoenix and Beauty

直接构造长为 $nk$ 的序列即可。

const int N = 107;
int n, k, a[N];

inline void solve() {
	rd(n, k), rda(a, n);
	set<int> s;
	for (int i = 1; i <= n; i++) s.insert(a[i]);
	if ((int)s.size() > k) return print(-1);
	int t = *--s.end();
	while ((int)s.size() < k) {
		if (t == n) t = 0;
		s.insert(++t);
	}
	print(n * k);
	for (int i = 1; i <= n; i++)
		for (int x : s) print(x, ' ');
	prints("");
}

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

Phoenix and Distribution

分类贪心一下。

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

inline void solve() {
	rd(n, k), rds(s, n);
	sort(s + 1, s + n + 1);
	printc(s[k]);
	if (s[k] == s[1]) {
		if (s[k+1] == s[n])
			for (int i = k + 1; i <= n; i += k)
				printc(s[i]);
		else
			for (int i = k + 1; i <= n; i++)
				printc(s[i]);
	}
	prints("");
}

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

Phoenix and Science

按照 $2^i$ 随便构造一下即可。

int n;

inline void solve() {
	rd(n);
	int t = 1, o = 1;
	multiset<int> s;
	while (o + (1 << t) <= n) s.insert(1 << t), o += 1 << t, ++t;
	if (o < n) s.insert(n - o);
	o = 1;
	print(s.size());
	for (int x : s)
		print(x - o, ' '), o = x;
	prints("");
}

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

Phoenix and Berries

简单 DP。

const int N = 507;
const ll inf = 1e18;
int n, k, x, y;
ll f[N][N], s, ans;
bool v[N][N];

inline void upd(ll &x, ll y) {
	x = max(x, y);
}

int main() {
	rd(n, k);
	f[0][0] = 0, v[0][0] = 1;
	for (int i = 1; i <= n; i++) {
		rd(x, y);
		for (int j = 0; j < k; j++)
			if (v[i-1][j]) {
				upd(f[i][(j+x)%k], f[i-1][j] + (j + x) / k + (s - f[i-1][j] * k - j + y) / k);
				v[i][(j+x)%k] = 1;
				for (int u = 1; u < k; u++)
					if (u <= x && k - u <= y)
						upd(f[i][(j+x-u)%k], f[i-1][j] + (j + x - u) / k + (s - f[i-1][j] * k - j + y - (k - u)) / k + 1),
						v[i][(j+x-u)%k] = 1;
			}
		s += x + y;
	}
	for (int i = 0; i < k; i++)
		if (v[n][i]) upd(ans, f[n][i]);
	print(ans);
	return 0;
}

Phoenix and Memory

题目太神仙了，先咕咕咕着。