Solution -「CF 908D」New Year&Arbitrary Arrangement
\(\mathcal{Description}\)
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：
\[f(i,j)=p_af(i+1,j)+p_bf(i,i+j)
\]
当 \(i+j\ge n\)，若再出现一个 \(\texttt{'b'}\)，操作必然停止。那么：
\[\begin{aligned}
f(i,j)&=p_b\sum_{k=0}^{+\infty}p_a^k(i+j+k)\\
p_af(i,j)&=p_b\sum_{k=1}^{+\infty}p_a^k(i+j+k-1)\\
(1-p_a)f(i,j)&=p_b\left(i+j+\sum_{k=1}^{+\infty}p_a^k\right)\\
p_bf(i,j)&=p_b\left(i+j+\frac{p_a}{p_b}\right)\\
f(i,j)&=i+j+\frac{p_a}{p_b}
\end{aligned}
\]
而初始状态有：
\[\begin{aligned}
f(0,0)&=p_af(1,0)+p_bf(0,0)\\
&=\frac{p_a}{1-p_b}f(1,0)\\
&=f(1,0)
\end{aligned}
\]
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;
}
