CF638div.2

A Phoenix and Balance

  由于\(\sum_{i=1}^{n-1}2^i=2^n-2\)，为了使重量差最小，将最重的与最轻的\(\frac{n}{2}-1\)个放在一起，因此答案为\((2^n+\sum_{i=1}^{\frac{n}{2}-1}2^i)-\sum_{i=\frac{n}{2}}^{n-1}2^i=2^{\frac{n}{2}+1}-2\)

#include <bits/stdc++.h>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 2e9;

int t, n;
int bit[20];

int main() 
{
	scanf("%d", &t);
	bit[0] = 1;
	for (int i = 1; i <= 16; ++i) bit[i] = bit[i-1]<<1;
	while (t--) {
		scanf("%d", &n);
		printf("%d\n", bit[n/2+1]-2);
	}
    return 0;
}

B Phoenix and Beauty

  漂亮数组以\(k\)为周期，那么如果原数组有超过\(k\)个不同的数，则无解，否则，将原数组每个数扩展至\(k\)个即可。

#include <bits/stdc++.h>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 2e9;

int t, n, k;
set<int> s;

int main() 
{
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d", &n, &k);
		s.clear();
		int x;
		for (int i = 1; i <= n; ++i) { scanf("%d", &x); s.insert(x); }
		if ((int)s.size() > k) { puts("-1"); continue; }
		printf("%d\n", n * k);
		for (int i = 1; i <= n; ++i) {
			for (int it : s) printf("%d ", it);
			for (int j = 1; j <= k - (int)s.size(); ++j) printf("%d ", 1);
		}
		puts("");

	}
    return 0;
}

C Phoenix and Distribution

  贪心，对字符串排序，要使分配的\(k\)个字符串中的字典序最大值最小，第一个字符必为第\(k\)小，如果最小的不与其相等，可将剩余字符全部分配至最小字符后。否则，如果最大字符不与第\(k\)小相等，则将第\(k\)小后面字符分配至其后，反正则平均分配后面字符。

#include <bits/stdc++.h>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 2e9;

int t, n, k;
string s;

int main() 
{
	cin >> t;
	while (t--) {
		cin >> n >> k;
		cin >> s;
		sort(s.begin(), s.end());
		if (s[0] != s[k-1]) { cout << s[k-1] << '\n'; continue; }
		cout << s[k-1];
		if (s[k] != s[n-1]) {
			for (int i = k; i < n; ++i) cout << s[i];
		}
		else {
			for (int i = 0; i < (n - 1) / k; ++i) cout << s[n-1];
		}
		puts("");
	}
    return 0;
}

D Phoenix and Science

  贪心，问题可转化为\(\{a_n\}\)满足\(a_0=1\)，使得\(a_i\leq a_{i+1}\leq2a_i\)且\(\sum_{i=1}^{len}a_i=n\)，要使\(len\)最小化，构造\(1,2,\cdots,2^k\)使得其和\(s\)小于或等于\(n\)，若小于\(sum\)，插入\(n-sum\)并排序即得一个满足题意的最短序列。

#include <bits/stdc++.h>
using namespace std;

#define debug(x) cout << #x << " is " << x << endl
#define inc(i, a, b) for (int i = a; i <= b; ++i)
typedef long long ll;
const int INF = 2e9;

int t, n;

int main()
{
	scanf("%d", &t);
	while (t--) {
		scanf("%d", &n);
		vector<int> v;
		for (int i = 1; i <= n; i <<= 1) {
			v.push_back(i);
			n -= i;
		}
		if (n) {
			v.push_back(n);
			sort(v.begin(), v.end());
		}
		printf("%d\n", (int)v.size() - 1);
		for (int i = 1; i < (int)v.size(); ++i) printf("%d ", v[i] - v[i-1]);
		puts("");
	}
    return 0;
}

E Phoenix and Berries

F