Educational Codeforces Round 97 (Rated for Div. 2) F. Emotional Fishermen

题目链接
nnn fishermen have just returned from a fishing vacation. The iii-th fisherman has caught a fish of weight aia_iai.

Fishermen are going to show off the fish they caught to each other. To do so, they firstly choose an order in which they show their fish (each fisherman shows his fish exactly once, so, formally, the order of showing fish is a permutation of integers from 111 to nnn). Then they show the fish they caught according to the chosen order. When a fisherman shows his fish, he might either become happy, become sad, or stay content.

Suppose a fisherman shows a fish of weight xxx, and the maximum weight of a previously shown fish is yyy (y=0y = 0y=0 if that fisherman is the first to show his fish). Then:

if x≥2yx \ge 2yx≥2y, the fisherman becomes happy;
if 2x≤y2x \le y2x≤y, the fisherman becomes sad;
if none of these two conditions is met, the fisherman stays content.

Let’s call an order in which the fishermen show their fish emotional if, after all fishermen show their fish according to this order, each fisherman becomes either happy or sad. Calculate the number of emotional orders modulo 998244353998244353998244353.

Input

The first line contains one integer nnn (2≤n≤50002 \le n \le 50002≤n≤5000).

The second line contains nnn integers a1a_1a1, a2a_2a2, …, ana_nan (1≤ai≤1091 \le a_i \le 10^91≤ai≤109).

Output

Print one integer — the number of emotional orders, taken modulo 998244353998244353998244353.

Examples

input

4
1 1 4 9

output

input

4
4 3 2 1

output

input

3
4 2 1

output

input

8
42 1337 13 37 420 666 616 97

output

最后序列一定是如下形式：
b1⋯⋯⋯⏟≤b12b2⋯⋯⋯⏟≤b22b3⋯⋯⋯⏟≤b32\boxed{b_1} \begin{matrix}\underbrace{\cdots\cdots\cdots}\\\leq \frac{b_1}{2}\end{matrix} \boxed{b_2} \begin{matrix}\underbrace{\cdots\cdots\cdots}\\\leq \frac{b_2}{2}\end{matrix} \boxed{b_3} \begin{matrix}\underbrace{\cdots\cdots\cdots}\\\leq \frac{b_3}{2}\end{matrix} b1⋯⋯⋯≤2b1b2⋯⋯⋯≤2b2b3⋯⋯⋯≤2b3
其中 bi≤bi+12b_i\le \frac{b_{i+1}}{2}bi≤2bi+1 。
首先将序列 aaa 按升序排序。
考虑逐渐将序列 aaa 中的元素填入长度为 nnn 的空序列。
设 f[i]f[i]f[i] 为 aia_iai 作为当前序列中最后一个 bbb 时的方案数，lim[i]lim[i]lim[i] 为满足 aj≤ai2a_j\le \frac{a_i}{2}aj≤2ai 的最大的 jjj 。
则前一个 bbb 是 aj(j≤lim[i])a_j \ (j\le lim[i])aj (j≤lim[i]) 。
对于固定的 jjj ，此时序列中已经填好了 aja_jaj ， aia_iai 和 a1∼lim[j]a_{1\sim lim[j]}a1∼lim[j] ，其中 aia_iai 一定放在序列中第一个空余位置，此时序列中还有 n−lim[j]−2n-lim[j]-2n−lim[j]−2 个空余位置。
在将元素 aia_iai 加入序列的同时要将 alim[j]+1∼lim[i]a_{lim[j]+1\sim lim[i]}alim[j]+1∼lim[i] 加入序列，这些元素可以再空余位置中任意选择位置安放，因此共有f[j]⋅A(n−lim[j]−2,lim[i]−lim[j]−1)f[j]\cdot A(n-lim[j]-2,lim[i]-lim[j]-1)f[j]⋅A(n−lim[j]−2,lim[i]−lim[j]−1) 种情况。
因此转移方程为：
f[i]=∑j=0lim[i]f[j]⋅A(n−lim[j]−2,lim[i]−lim[j]−1)f[i]=\sum_{j=0}^{lim[i]}f[j]\cdot A(n-lim[j]-2,lim[i]-lim[j]-1) f[i]=j=0∑lim[i]f[j]⋅A(n−lim[j]−2,lim[i]−lim[j]−1)
其中 f[0]=1,lim[0]=−1f[0]=1,lim[0]=-1f[0]=1,lim[0]=−1 。
因为最后 ana_nan 一定是最后一个 bbb ，如果 lim[n]≠n−1lim[n]\not =n-1lim[n]=n−1 ，那么一定无解，否则答案为 f[n]f[n]f[n] 。
时间复杂度为 O(n2)O(n^2)O(n2) 。

#include<bits/stdc++.h>using namespace std;
const int N = 5005, MOD = 998244353;
int fac[N], inv[N];inline int qpow(int x, int n) {int res = 1;while (n) {if (n & 1) res = (1ll * res * x) % MOD;x = 1ll * x * x % MOD, n >>= 1;}return res;
}inline void A_init(const int n) {fac[0] = 1;for (int i = 1; i <= n; ++i)fac[i] = 1ll * fac[i - 1] * i % MOD;inv[n] = qpow(fac[n], MOD - 2);for (int i = n - 1; i >= 0; --i)inv[i] = 1ll * inv[i + 1] * (i + 1) % MOD;
}inline int A(int n, int m) {if (n < 0 || m < 0 || n < m) return 0;return 1ll * fac[n] * inv[n - m] % MOD;
}int lim[N], a[N], f[N], n;int main() {scanf("%d", &n), A_init(n);for (int i = 1; i <= n; i++)scanf("%d", &a[i]);sort(a + 1, a + 1 + n);for (int i = 1, j = 0; i <= n; i++) {while (2 * a[j + 1] <= a[i])j++;lim[i] = j;}if (lim[n] != n - 1)return puts("0"), 0;f[0] = 1, lim[0] = -1;for (int i = 1; i <= n; i++)for (int j = 0; j <= lim[i]; j++)f[i] = (f[i] + 1ll * f[j] * A(n - 2 - lim[j], lim[i] - lim[j] - 1) % MOD) % MOD;printf("%d\n", f[n]);return 0;
}

另外转移方程中的排列式子可作如下转化：
A(n−lim[j]−2,lim[i]−lim[j]−1)=(n−lim[j]−2)!((n−lim[j]−2)−(lim[i]−lim[j]−1))!=(n−lim[j]−2)!(n−lim[i]−1)!\begin{aligned} A(n-lim[j]-2,lim[i]-lim[j]-1) &= \frac{(n-lim[j]-2)!}{((n-lim[j]-2)-(lim[i]-lim[j]-1))!} \\ &= \frac{(n-lim[j]-2)!}{(n-lim[i]-1)!} \end{aligned} A(n−lim[j]−2,lim[i]−lim[j]−1)=((n−lim[j]−2)−(lim[i]−lim[j]−1))!(n−lim[j]−2)!=(n−lim[i]−1)!(n−lim[j]−2)!
其中式子一部分只和 iii 有关，另一部分只和 jjj 有关，因此只需维护 f[j]⋅(n−lim[j]−2)!f[j]\cdot (n-lim[j]-2)!f[j]⋅(n−lim[j]−2)! 的前缀和即可将 dp 复杂度降为 O(n)O(n)O(n) 。

#include<bits/stdc++.h>using namespace std;
const int N = 5005, MOD = 998244353;
int fac[N], inv[N];inline int qpow(int x, int n) {int res = 1;while (n) {if (n & 1) res = (1ll * res * x) % MOD;x = 1ll * x * x % MOD, n >>= 1;}return res;
}inline void A_init(const int n) {fac[0] = 1;for (int i = 1; i <= n; ++i)fac[i] = 1ll * fac[i - 1] * i % MOD;inv[n] = qpow(fac[n], MOD - 2);for (int i = n - 1; i >= 0; --i)inv[i] = 1ll * inv[i + 1] * (i + 1) % MOD;
}int lim[N], sum[N], a[N], f, n;int main() {scanf("%d", &n), A_init(n);for (int i = 1; i <= n; i++)scanf("%d", &a[i]);sort(a + 1, a + 1 + n);for (int i = 1, j = 0; i <= n; i++) {while (2 * a[j + 1] <= a[i])j++;lim[i] = j;}if (lim[n] != n - 1)return puts("0"), 0;sum[0] = fac[n - 1];for (int i = 1; i <= n; i++) {f = 1ll * sum[lim[i]] * inv[n - lim[i] - 1] % MOD;sum[i] = (sum[i - 1] + 1ll * f * fac[n - lim[i] - 2] % MOD) % MOD;}printf("%d\n", f);return 0;
}

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