牛客多校round9 E Music Game

题目：https://www.nowcoder.com/acm/contest/147/E

Niuniu likes to play OSU!
We simplify the game OSU to the following problem.

Given n and m, there are n clicks. Each click may success or fail.
For a continuous success sequence with length X, the player can score X^m.
The probability that the i-th click success is p[i]/100.
We want to know the expectation of score.
As the result might be very large (and not integral), you only need to output the result mod 1000000007.
输入描述:
The first line contains two integers, which are n and m.
The second line contains n integers. The i-th integer is p[i].

1 <= n <= 1000
1 <= m <= 1000
0 <= p[i] <= 100
输出描述:
You should output an integer, which is the answer.

sol:

这个题有 $O(n^2)$ 的做法，这里不多加讨论。说下从bls学到的用第二类 $Stirling$ 数优化计算 $x^m$ 的方法。

1.考虑朴素的dp

设 $x$ 为最长连续成功后缀的长度。设 $a[i][j]$ 为考虑前i次点击， $x^j$ 的期望。
由二项式定理可知

(x + 1)^{m} = \sum_{i = 0}^{m} (\binom{m}{i}) x^{i}

$(x+1)^m = \sum_{i=0} ^ {m} \binom{m} {i} x^i$
从而得到转移方程

a [i] [j] = p [i] * \sum_{k = 0}^{j} (\binom{j}{k}) a [i - 1] [k]

$a[i][j] = p[i] * \sum_{k=0} ^{j} \binom{j} {k} a[i-1][k]$
设

b [i] [j]

$b[i][j]$ 为

x^{j}

$x^j$ 的得分期望，由

(x + 1)^{m} - x^{m} = \sum_{i = 0}^{m - 1} (\binom{m}{i}) x^{i}

$(x+1)^m - x^m = \sum_{i=0}^{m-1} \binom{m} {i} x^i$
则有

b [i] [j] = {\begin{aligned} b [i - 1] [j] & (f a i l) \\ b [i - 1] [j] + \sum_{k = 0}^{m - 1} (\binom{m}{k}) a [i - 1] [k] & (s u c e e s s) \end{aligned}

$b[i][j]=\left\{ \begin{aligned} &b[i-1][j] &(fail)\\ &b[i-1][j]+\sum_{k=0}^{m-1} \binom{m} {k} a[i-1][k] &(suceess) \end{aligned} \right.$

合并起来就是

b [i] [j] = b [i - 1] [j] + p [i] * \sum_{k = 0}^{m - 1} (\binom{m}{k}) a [i - 1] [k]

$b[i][j] = b[i-1][j] + p[i] * \sum_{k=0}^{m-1} \binom{m} {k} a[i-1][k]$

这样做的复杂度是 $O(nm^2)$

2.用Stirling数优化

先放一个结论

n^{p} = \sum_{k = 0}^{p} S (p, k) [n]_{k}

$n^p = \sum_{k=0}^{p} S(p,k)[n]_k$
也可以写为

n^{p} = \sum_{k = 0}^{p} S (p, k) * i! * (\binom{n}{k})

$n^p = \sum_{k=0}^{p} S(p,k)*i!* \binom{n} {k}$

这个可以在组合数学的教材里找到。

$[n]_k$ 表示n的下阶乘，等价于 $A(n,k)$ 。 $S(p,k)$ 表示第二类 $stirling$ 数

那么我们可以将维护 $n^p$ 的期望转换到维护 $\binom{n} {k}$ 的期望上去。

同样的，我们设最长成功后缀的长度为 $x$ ,设 $a[i][j]$ 为考虑到第i 位后 $\binom {x} {j}$ 的期望，设 $b[i][j]$ 为所有 $\binom {x} {j}$ 的期望。

由杨辉三角：

(\binom{n}{i}) = (\binom{n - 1}{i}) + (\binom{n - 1}{i - 1})

$\binom {n} {i} = \binom {n-1} {i} + \binom{n-1} {i-1}$
可得

a [i] [j] = p [i] * (a [i - 1] [j] + a [i - 1] [j - 1])

$a[i][j] = p[i]*(a[i-1][j] + a[i-1][j-1])$
同理可得

\begin{aligned} b [i] [j] & = b [i - 1] [j] + p [i] * (a [i - 1] [j - 1]) \end{aligned}

$\begin{aligned} b[i][j] &= b[i-1][j] + p[i]*(a[i-1][j-1]) \end{aligned}$

至此,这道题就可以能够在 $O(nm)$ 的时间内完成.

code:

#include <bits/stdc++.h>
using namespace std;
int n, m, mod = 1000000007;
long long p[1020];
long long a[1020][1020];
long long b[1020][1020];
long long s[1020][1020];
long long f[1020];

typedef long long ll;

int main() {
    scanf("%d%d", &n, &m);
    s[0][0] = 1;
    for (int i = 0; i <= m; i++) {
        for (int j = 1; j <= i; j++) {
            s[i][j] = (s[i - 1][j - 1] + j * s[i - 1][j]) % mod;
        }
    }
    for (int i = f[0] = 1; i <= m; i++) {
        f[i] = f[i - 1] * i % mod;
    }
    a[0][0] = 1;
    for (int i = 1; i <= n; i++) {
        scanf("%lld", &p[i]);
        p[i] = p[i] * 570000004 % mod;
        a[i][0] = 1;
        for (int j = 1; j <= m; j++) {
            a[i][j] = a[i-1][j] + a[i-1][j - 1];
            a[i][j] *= p[i];
            a[i][j] %= mod;
            b[i][j] = p[i] * a[i-1][j-1]; 
            b[i][j] %= mod;
            b[i][j] += b[i-1][j];
            b[i][j] %= mod;
        }
    }
    long long z = 0;
    for (int i = 1; i <= m; i++) {
        z = (z + b[n][i] * f[i] % mod * s[m][i]) % mod;
    }
    printf("%lld\n", z);
    return 0;
}