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Solution -「ARC 104E」Random LIS

\(\mathcal{Description}\)

  Link.

  给定整数序列 \(\{a_n\}\),对于整数序列 \(\{b_n\}\)\(b_i\)\([1,a_i]\) 中等概率随机。求 \(\{b_n\}\) 中 LIS(最长上升子序列)的期望长度。对 \(10^9+7\) 取模。

  \(n\le6\)\(a_i\le10^9\)

\(\mathcal{Solution}\)

  欺负这个 \(n\) 小得可爱,直接 \(\mathcal O(n!)\) 枚举 \((b_i,i)\) 的二维偏序关系,记排列 \(\{p_n\}\)\(p_i\) 表示第 \(i\) 小的二元组是 \((b_{p_i},p_i)\)。于是乎,\(\{b_n\}\) 就会满足:

\[b_{p_1}\le b_{p_2}\le \cdots \le b_{p_n} \]

  但其中有些地方是不能取等的。可以发现若 \(p_i<p_{i+1}\),则必须取 \(b_{p_i}<b_{p_{i+1}}\)。考虑把所有如此的小于全部变成小于等于:将所有 \(j>i\)\(b_{p_j}\leftarrow b_{p_j}-1\) 即可。

  为方便进一步思考,把 \(\{a_n\}\)\(\{b_n\}\)\(\{p_n\}\) 的位置排列。问题变成:求不降序列 \(\{b_n\}\) 的个数,满足 \(b_i\in[1,a_i]\)

  法一,「APIO 2016」划艇。尝试另一种方法——问题可转化为从 \((1,1)\) 走到 \((n+1,+\infty)\),只能向上或向右走,且横坐标为 \(x\) 时纵坐标不超过 \(a_x\),求方案数。令 \(f(i)\) 表示走到 \(x=i\) 时的合法方案数。转移时计算总方案减不合法方案。钦定不合法方案的第一个不合法位置为 \(x=j\),方案数即为 \(f(j)\binom{(a_i-a_j)+(i-j)-1}{i-j}\)(先走到 \((j,a_j+1)\) 保证不合法,再随便走)。所以转移:

\[f(i)=\binom{a_i+i-1}{i-1}-\sum_{j=1}^{i-1}f(j)\binom{a_i-a_j+i-j-1}{i-j} \]

  求组合数 \(\binom{a}{b}\)\(\mathcal O(b)\) 暴力求就行,因为 \(b\)\(\mathcal O(n)\) 的。两种方法的总复杂度均为 \(\mathcal O(n!n^3)\)

\(\mathcal{Code}\)

/* Clearink */

#include <cstdio>
#include <algorithm>

const int MAXN = 6, MOD = 1e9 + 7;
const int inv[] = { 0, 1, 500000004, 333333336, 250000002, 400000003, 166666668 };
int n, a[MAXN + 5], h[MAXN + 5], p[MAXN + 5], f[MAXN + 5];

inline void chkmax ( int& a, const int b ) { a < b ? a = b : 0; }
inline void chkmin ( int& a, const int b ) { b < a ? a = b : 0; }
inline void subeq ( int& a, const int b ) { ( a -= b ) < 0 ? a += MOD : 0; }
inline void addeq ( int& a, const int b ) { ( a += b ) < MOD ? 0 : a -= MOD; }

inline int calcLIS () {
	int f[MAXN + 5] {}, ret = 0;
	for ( int i = 1; i <= n; ++ i ) {
		for ( int j = i - 1; ~j; -- j ) {
			if ( p[i] > p[j] ) {
				chkmax ( f[i], f[j] + 1 );
			}
		}
		chkmax ( ret, f[i] );
	}
	return ret;
}

inline int qkpow ( int a, int b, const int p = MOD ) {
	int ret = 1;
	for ( ; b; a = 1ll * a * a % p, b >>= 1 ) ret = ret * ( b & 1 ? a : 1ll ) % MOD;
	return ret;
}

inline int comb ( const int n, const int m ) {
	int ret = 1;
	for ( int i = 1; i <= m; ++ i ) ret = ( n - i + 1ll ) * ret % MOD * inv[i] % MOD;
	return ret;
}

int main () {
	scanf ( "%d", &n );
	for ( int i = 1; i <= n; ++ i ) scanf ( "%d", &a[p[i] = i] );
	int ans = 0;
	do {
		int lis = calcLIS ();
		for ( int i = 1; i <= n; ++ i ) h[i] = a[p[i]] - 1;
		for ( int i = 1; i < n; ++ i ) {
			if ( p[i] > p[i + 1] ) continue;
			for ( int j = i + 1; j <= n; ++ j ) -- h[j];
		}
		for ( int i = n - 1; i; -- i ) chkmin ( h[i], h[i + 1] );
		int all = comb ( h[n] + n, n );
		for ( int i = 1; i <= n; ++ i ) {
			f[i] = comb ( h[i] + i - 1, i - 1 );
			for ( int j = 1; j < i; ++ j ) {
				subeq ( f[i], 1ll * f[j] * comb ( h[i] - h[j] + ( i - j - 1 ), i - j ) % MOD );
			}
			subeq ( all, 1ll * f[i] * comb ( h[n] - h[i] + n - i, n - i + 1 ) % MOD );
		}
		addeq ( ans, 1ll * all * lis % MOD );
	} while ( std::next_permutation ( p + 1, p + n + 1 ) );
	int s = 1;
	for ( int i = 1; i <= n; ++ i ) s = 1ll * s * a[i] % MOD;
	printf ( "%d\n", int ( 1ll * ans * qkpow ( s, MOD - 2 ) % MOD ) );
	return 0;
}
posted @   Rainybunny  阅读(169)  评论(0编辑  收藏  举报
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