\(\mathcal{Description}\)

??Link.

??给定 \(n,p_a,p_b\)，初始有一个空串，每次操作有 \(\frac{p_a}{p_a+p_b}\) 的概率在其后添加字符 \(\texttt{‘a‘}\)，\(\frac{p_b}{p_a+p_b}\) 的概率添加字符 \(\texttt{‘b‘}\)，当子序列 \(\{\texttt{‘a‘},\texttt{‘b‘}\}\) 的个数不小于 \(n\) 时，结束操作。求子序列的期望个数，对 \(10^9+7\) 取模。

??\(n\le1000\)。

\(\mathcal{Solution}\)

??显然状态，\(f(i,j)\) 表示有 \(i\) 个 \(\texttt{‘a‘}\)，\(j\) 个 \(\{\texttt{‘a‘},\texttt{‘b‘}\}\) 的期望串长。为方便转移，令 \(p_a\) 为出现 \(\texttt{‘a‘}\) 的概率，\(p_b\) 同理。对于一般情况的转移：

??当 \(i+j\ge n\)，若再出现一个 \(\texttt{‘b‘}\)，操作必然停止。那么：

??而初始状态有：

??DP 就好了 w。

\(\mathcal{Code}\)

#include <cstdio>
#include <cstring>

typedef long long LL;

const int MAXN = 1000, MOD = 1e9 + 7;
int n, pa, pb, div, f[MAXN + 5][MAXN + 5];

inline int add ( int a, const int b ) { return ( a += b ) < MOD ? a : a - MOD; }
inline int mul ( LL a, const int b ) { return ( a *= b ) < MOD ? a : a % MOD; }
inline int sub ( int a, const int b ) { return ( a -= b ) < 0 ? a : a + MOD; }

inline int qkpow ( int a, int b ) {
	int ret = 1;
	for ( ; b; a = mul ( a, a ), b >>= 1 ) ret = mul ( ret, b & 1 ? a : 1 );
	return ret;
}

inline int solve ( const int a, const int k ) {
	if ( a + k >= n ) return add ( a, add ( k, div ) );
	if ( ~ f[a][k] ) return f[a][k];
	return f[a][k] = add ( mul ( pa, solve ( a + 1, k ) ), mul ( pb, solve ( a, a + k ) ) );
}

int main () {
	int ta, tb;
	scanf ( "%d %d %d", &n, &ta, &tb );
	memset ( f, -1, sizeof f );
	pa = mul ( ta, qkpow ( add ( ta, tb ), MOD - 2 ) );
	pb = mul ( tb, qkpow ( add ( ta, tb ), MOD - 2 ) );
	div = mul ( pa, qkpow ( pb, MOD - 2 ) );
	printf ( "%d\n", solve ( 1, 0 ) );
	return 0;
}

Solution -「CF 908D」New Year and Arbitrary Arrangement

原文：https://www.cnblogs.com/rainybunny/p/13428480.html