题解：CF381B Sereja and Stairs

涉及知识点：构造。

解题思路

题目要求 \(a\) 的长度要最长，所以需要 \(a_1\sim a_i\) 和 \(a_i \sim a_{|a|}\) 最长，于是直接构造一个单调递增和单调递减的序列即可。

由于数据较小，可利用桶通过此题。

注意：

• \(a_1\sim a_{i-1}\) 与 \(a_{i + 1} \sim a_{|a|}\) 中可以有重复的元元素，但不能有两个 \(a_i\)。

代码

#include <bits/stdc++.h>
#define int long long
#define ll __int128
#define db double
#define ldb long double
#define vo void
#define endl '\n'
#define il inline
#define re register
#define ve vector
#define p_q priority_queue
#define PII pair<int, int>
#define u_m unordered_map
#define bt bitset

using namespace std;

//#define O2 1
#ifdef O2
#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3, "Ofast", "inline")
#endif

struct IO {
#define MAXSIZE (1 << 20)
#define isdigit(x) (x >= '0' && x <= '9')
	char buf[MAXSIZE], *p1, *p2;
	char pbuf[MAXSIZE], *pp;
	IO() : p1(buf), p2(buf), pp(pbuf) {}
	
	~IO() {
		fwrite(pbuf, 1, pp - pbuf, stdout);
	}
	char gc() {
		if (p1 == p2) p2 = (p1 = buf) + fread(buf, 1, MAXSIZE, stdin);
		return p1 == p2 ? ' ' : *p1++;
	}
	
	bool blank(char ch) {
		return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
	}
	
	template <class T>
	void read(T &x) {
		double tmp = 1;
		bool sign = 0;
		x = 0;
		char ch = gc();
		while (!isdigit(ch)) {
			if (ch == '-') sign = 1;
			ch = gc();
		}
		while (isdigit(ch)) {
			x = x * 10 + (ch - '0');
			ch = gc();
		}
		if (ch == '.') {
			ch = gc();
			while (isdigit(ch)) {
				tmp /= 10.0, x += tmp * (ch - '0');
				ch = gc();
			}
		}
		if (sign) x = -x;
	}
	
	void read(char *s) {
		char ch = gc();
		for (; blank(ch); ch = gc());
		for (; !blank(ch); ch = gc()) * s++ = ch;
		*s = 0;
	}
	
	void read(char &c) {
		for (c = gc(); blank(c); c = gc());
	}
	
	void push(const char &c) {
		if (pp - pbuf == MAXSIZE) fwrite(pbuf, 1, MAXSIZE, stdout), pp = pbuf;
		*pp++ = c;
	}
	
	template <class T>
	void write(T x) {
		if (x < 0) x = -x, push('-');
		static T sta[35];
		T top = 0;
		do {
			sta[top++] = x % 10, x /= 10;
		} while (x);
		while (top) push(sta[--top] + '0');
	}
	
	template <class T>
	void write(T x, char lastChar) {
		write(x), push(lastChar);
	}
} io;

namespace COMB {
	int fact[200000];
	int Triangle[1010][1010];
	void Fact(int n, int mod) {
		fact[0] = 1;
		for (int i = 1; i <= n; i ++ ) fact[i] = ((fact[i - 1]) % mod * (i % mod)) % mod;
	}
	void Pascal_s_triangle(int n, int mod) {
		for (int i = 0; i <= n; i ++ ) Triangle[i][0] = 1;
		for (int i = 1; i <= n; i ++ )
			for (int j = 1; j <= i; j ++ )
				Triangle[i][j] = (Triangle[i - 1][j] + Triangle[i - 1][j - 1]) % mod;
	}
	int pw(int x, int y, int mod) {
		int res = 1;
		while (y) {
			if (y & 1) res = ((res % mod) * (x % mod)) % mod;
			x = (x % mod) * (x % mod) % mod;
			y >>= 1;
		}
		return res;
	}
	int pw(int x, int y) {
		int res = 1;
		while (y) {
			if (y & 1) res *= x;
			x *= x;
			y >>= 1;
		}
		return res;
	}
	int GCD(int x, int y, int mod) {
		return __gcd(x, y) % mod;
	}
	int LCM(int x, int y, int mod) {
		return (((x % mod) * (y % mod)) % mod / (GCD(x, y, mod) % mod)) % mod;
	}
	int C(int n, int m, int mod) {
		if (m > n || m < 0) return 0;
		return fact[n] * pw(fact[m], mod - 2, mod) % mod * pw(fact[n - m], mod - 2, mod) % mod;
	}
	int Ask_triangle(int x, int y) {
		return Triangle[x][y];
	}
}
using namespace COMB;

//#define fre 1
#define IOS 1
//#define multitest 1

const int N = 4e6 + 10;
const int M = 4e5 + 10;
const int inf = 1e12;
const int Mod = 1e9 + 9;

namespace zla {
	int n, a[N], t[N], ans[N], len;
	
	il void Init() {
		cin >> n;
		for (int i = 1; i <= n; i ++ ) {
			cin >> a[i];
			t[a[i]] ++;
		}
	}
	
	il void Solve() {
		for (int i = 1; i <= 5000; i ++ ) {
			if (t[i]) {
				ans[++ len] = i;
				t[i] --;
			}
		}
		int T = ans[len - 1];
		for (int i = T; i >= 1; i -- )
			if (t[i]) {
				ans[++ len] = i;
				t[i] --;
			}
		cout << len << endl;
		for (int i = 1; i <= len; i ++ ) cout << ans[i] << " ";
	}
	
	il void main() {
		Init();
		Solve();
	}
}

signed main() {
	int T;
#ifdef IOS
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#endif
#ifdef fre
	freopen(".in", "r", stdin);
	freopen(".out", "w", stdout);
#endif
#ifdef multitest
	cin >> T;
#else
	T = 1;
#endif
	while (T--) zla::main();
	return 0;
}