[北大集训 2021] 随机游走 题解

前言

又是随机游走？

题目链接：洛谷。

题目分析

看到加边，可能性太多了。但是为了让步数最大化，我们可以贪心地想，肯定要往前面连，而且越前面要走的期望步数肯定越大。并且，我们不会浪费边在终点上。于是，题目转变成了 $1\sim n-1$ 连向起点 $1$ 连若干条边，使得随机游走到终点的期望步数最大。

那要如何分配这 $m$ 条边到 $1\sim n-1$ 个点呢？考虑假设已知第 $i$ 个点向 $1$ 连了 $d_i$ 条边，求期望步数。设 $f_i$ 为到了 $i$，还要期望多少步走到终点，显然 $f_n=0$。开始喜闻乐见的推式子环节：

$$f_i=\frac{1}{d_i+1}{f}_{i+1}+\frac{d_i}{d_i+1}{f}_1+1$$

从 $n-1$ 向前递推。

$$\begin{array}{rl}{f}_{n-1}& =\frac{1}{d_{n-1}+1}{f}_{n-1+1}+\frac{d_{n-1}}{d_{n-1}+1}{f}_1+1\\ & =\frac{d_{n-1}}{d_{n-1}+1}{f}_1+1\end{array}$$

推到 $n-2$。

$$\begin{array}{rl}{f}_{n-2}& =\frac{1}{d_{n-2}+1}{f}_{n-1}+\frac{d_{n-2}}{d_{n-2}+1}{f}_1+1\\ & =\frac{1}{d_{n-2}+1}\cdot (\frac{d_{n-1}}{d_{n-1}+1}{f}_1+1)+\frac{d_{n-2}}{d_{n-2}+1}{f}_1+1\\ & =\frac{1}{d_{n-2}+1}\cdot \frac{d_{n-1}}{d_{n-1}+1}{f}_1+\frac{d_{n-2}}{d_{n-2}+1}{f}_1+\frac{1}{d_{n-2}+1}+1\\ & =\frac{d_{n-1}+d_{n-2}\cdot (d_{n-1}+1)}{(d_{n-2}+1)\cdot (d_{n-1}+1)}{f}_1+\frac{(d_{n-2}+1)+1}{d_{n-2}+1}\\ & =\frac{(d_{n-2}+1)\cdot (d_{n-1}+1)-1}{(d_{n-2}+1)\cdot (d_{n-1}+1)}{f}_1+\frac{(d_{n-2}+1)+1}{d_{n-2}+1}\end{array}$$

再推到 $n-3$。

$$\begin{array}{rl}{f}_{n-3}=& \frac{1}{d_{n-3}+1}{f}_{n-2}+\frac{d_{n-3}}{d_{n-3}+1}{f}_1+1\\ =& \frac{1}{d_{n-3}+1}(\frac{(d_{n-2}+1)\cdot (d_{n-1}+1)-1}{(d_{n-2}+1)\cdot (d_{n-1}+1)}{f}_1+\frac{(d_{n-2}+1)+1}{d_{n-2}+1})+\frac{d_{n-3}}{d_{n-3}+1}{f}_1+1\\ =& \frac{(d_{n-2}+1)\cdot (d_{n-1}+1)-1}{(d_{n-3}+1)\cdot (d_{n-2}+1)\cdot (d_{n-1}+1)}{f}_1+\frac{(d_{n-2}+1)+1}{(d_{n-3}+1)\cdot (d_{n-2}+1)}+\frac{d_{n-3}}{d_{n-3}+1}{f}_1+1\\ =& \frac{d_{n-3}\cdot (d_{n-2}+1)\cdot (d_{n-1}+1)+(d_{n-2}+1)\cdot (d_{n-1}+1)-1}{(d_{n-3}+1)\cdot (d_{n-2}+1)\cdot (d_{n-1}+1)}{f}_1+\frac{(d_{n-3}+1)\cdot (d_{n-2}+1)+(d_{n-2}+1)+1}{(d_{n-3}+1)\cdot (d_{n-2}+1)}\\ =& \frac{(d_{n-3}+1)\cdot (d_{n-2}+1)\cdot (d_{n-1}+1)-1}{(d_{n-3}+1)\cdot (d_{n-2}+1)\cdot (d_{n-1}+1)}{f}_1+\frac{(d_{n-3}+1)\cdot (d_{n-2}+1)+(d_{n-2}+1)+1}{(d_{n-3}+1)\cdot (d_{n-2}+1)}\end{array}$$

找到一些规律，尝试去证明。假设对于 $i+1$ 满足：

$$f_{i+1}=\frac{\prod _{j=i+1}^{n-1}(d_j+1)-1}{\prod _{j=i+1}^{n-1}(d_j+1)}{f}_1+\frac{\sum _{j=i+1}^{n-2}\prod _{k=j}^{n-2}(d_k+1)+1}{\prod _{j=i+1}^{n-2}(d_j+1)}$$

显然该式对于 $n-1$ 成立。尝试用归纳法推到 $i$。

$$\begin{array}{rl}{f}_i& =\frac{1}{d_i+1}(\frac{\prod _{j=i+1}^{n-1}(d_j+1)-1}{\prod _{j=i+1}^{n-1}(d_j+1)}{f}_1+\frac{\sum _{j=i+1}^{n-2}\prod _{k=j}^{n-2}(d_k+1)+1}{\prod _{j=i+1}^{n-2}(d_j+1)})+\frac{d_i}{d_i+1}{f}_1+1\\ & =\frac{\prod _{j=i+1}^{n-1}(d_j+1)-1}{\prod _{j=i}^{n-1}(d_j+1)}{f}_1+\frac{\sum _{j=i+1}^{n-2}\prod _{k=j}^{n-2}(d_k+1)+1}{\prod _{j=i}^{n-2}(d_j+1)}+\frac{d_i}{d_i+1}{f}_1+1\\ & =\frac{\prod _{j=i}^{n-1}(d_j+1)-1}{\prod _{j=i}^{n-1}(d_j+1)}{f}_1+\frac{\sum _{j=i}^{n-2}\prod _{k=j}^{n-2}(d_k+1)+1}{\prod _{j=i}^{n-2}(d_j+1)}\end{array}$$

所以上式对于所有 $i$ 均成立。考虑边界，推到 $i=1$ 的时候是一个方程。

$$f_1=\frac{\prod _{j=1}^{n-1}(d_j+1)-1}{\prod _{j=1}^{n-1}(d_j+1)}{f}_1+\frac{\sum _{j=1}^{n-2}\prod _{k=j}^{n-2}(d_k+1)+1}{\prod _{j=1}^{n-2}(d_j+1)}$$

解方程。

$$(\prod _{j=1}^{n-1}(d_j+1))f_1=(\prod _{j=1}^{n-1}(d_j+1)-1)f_1+(d_{n-1}+1)\left(\sum _{j=1}^{n-2}\prod _{k=j}^{n-2}(d_k+1)+1\right)$$

$$f_1=\sum _{j=1}^{n-2}\prod _{k=j}^{n-1}(d_k+1)+(d_{n-1}+1)$$

$$f_1=\sum _{j=1}^{n-1}\prod _{k=j}^{n-1}(d_k+1)$$

于是我们可以很方便地求出期望步数，即 $f_1$。但是我们还是不知道最优的 $d$ 如何分配，考虑打表找规律。在此之前，我们不妨试着找一找 $d$ 的性质。

首先，$d$ 一定是单调不降的。因为显然，放在后面会给更多的 $\prod$ 提供贡献，从而使 $f_1$ 更大。

略证：

$\mathrm{\exists }t\in [1,n-2],d_t>d_{t+1}$，考虑原先的 $f_1$ 有

$$\begin{array}{rl}{f}_1& =\sum _{j=1}^{n-1}\prod _{k=j}^{n-1}(d_k+1)\\ & =\sum _{j=t+2}^{n-1}\prod _{k=j}^{n-1}(d_k+1)+(d_{t+1}+1)\prod _{k=t+2}^{n-1}(d_k+1)+\sum _{j=1}^t((d_t+1)(d_{t+1}+1)\prod _{k=j}^{t-1}(d_k+1)\prod _{k=t+2}^{n-1}(d_k+1))\end{array}$$

不妨交换 $d_t$ 和 $d_{t+1}$。

$$\begin{array}{rl}{f}_1& =\sum _{j=t+2}^{n-1}\prod _{k=j}^{n-1}(d_k+1)+(d_t+1)\prod _{k=t+2}^{n-1}(d_k+1)+\sum _{j=1}^t((d_{t+1}+1)(d_t+1)\prod _{k=j}^{t-1}(d_k+1)\prod _{k=t+2}^{n-1}(d_k+1))\end{array}$$

显然，因为 $d_t>d_{t+1}$，所以交换后的 $f_1$ 更优。我们可以不断进行这个过程，直到 $d$ 有序，即单调不降，$f_1$ 最大。

证毕。

其次……没其次了，打表吧。

while True:
    n, m = map(int, input().split())
 
    res = [0] * n
    ans = [0] * n
 
    maxx = 0
 
    def dfs(now, tot, x):
        global maxx, ans
        if now == n - 1:
            res[n - 1] = tot
          
            tmp = 0
            mul = 1
            for i in range(n - 1, 0, -1):
                mul *= res[i] + 1
                tmp += mul
          
            if tmp > maxx:
                maxx = tmp
                ans = res[::]
          
            return
        i = x
        while i * (n - now) <= tot:
            res[now] = i
            dfs(now + 1, tot - i, i)
            i += 1
 
    dfs(1, m, 0)
 
    print(' '.join(map(str, ans[1:])))

以上是打表程序。考虑 $n=20$ 时，不断加边对最优 $d$ 的分配造成的影响。

m = 1: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
m = 2: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1
......
m = 18: 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
m = 19: 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2
m = 20: 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2
......
m = 35: 0 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
m = 36: 0 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
m = 37: 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
m = 38: 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3
......
m = 54: 1 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
m = 55: 1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
m = 56: 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
m = 57: 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 4
......

发现，当 $m<n-1$ 时，加边是从 $n-1$ 开始放到 $2$，然后当 $m\ge n-1$ 时，从 $n-1$ 开始放到 $1$，如此往复，像是在不断从右往左往序列上刷。感性理解，具体证明交给读者。

Update on 2024.6.20 更新了进一步的证明

下证任意相邻两数之差不超过 $1$。

证明：

$\mathrm{\exists }t\in [1,n-2],d_t+2\le d_{t+1}$，证明 $d_t\leftarrow d_t+1,d_{t+1}\leftarrow d_{t+1}-1$ 后不劣。容易发现这样操作是合法的。

对于原先的 $f_1$：

$$\begin{array}{rl}{f}_1& =\sum _{j=1}^{n-1}\prod _{k=j}^{n-1}(d_k+1)\\ & =\sum _{j=t+2}^{n-1}\prod _{k=j}^{n-1}(d_k+1)+(d_{t+1}+1)\prod _{k=t+2}^{n-1}(d_k+1)+\sum _{j=1}^t((d_t+1)(d_{t+1}+1)\prod _{k=j}^{t-1}(d_k+1)\prod _{k=t+2}^{n-1}(d_k+1))\\ & =\sum _{j=t+2}^{n-1}\prod _{k=j}^{n-1}(d_k+1)+(d_{t+1}+1)\prod _{k=t+2}^{n-1}(d_k+1)+(d_t+1)(d_{t+1}+1)\sum _{j=1}^t\prod _{k=j}^{t-1}(d_k+1)\prod _{k=t+2}^{n-1}(d_k+1)\\ & =\sum _{j=t+2}^{n-1}\prod _{k=j}^{n-1}(d_k+1)+(d_{t+1}+1+(d_t+1)(d_{t+1}+1)\sum _{j=1}^t\prod _{k=j}^{t-1}(d_k+1))\prod _{k=t+2}^{n-1}(d_k+1)\end{array}$$

后来的 $f_1$，记作 $f_1^{\prime }$：

$$f_1^{\prime }=\sum _{j=t+2}^{n-1}\prod _{k=j}^{n-1}(d_k+1)+(d_{t+1}+(d_t+2)d_{t+1}\sum _{j=1}^t\prod _{k=j}^{t-1}(d_k+1))\prod _{k=t+2}^{n-1}(d_k+1)$$

那么考虑作差：

$$\begin{array}{rl}{f}_1^{\prime }-f_1=& ((d_{t+1}+(d_t+2)d_{t+1}\sum _{j=1}^t\prod _{k=j}^{t-1}(d_k+1))-\\ & (d_{t+1}+1+(d_t+1)(d_{t+1}+1)\sum _{j=1}^t\prod _{k=j}^{t-1}(d_k+1)))\prod _{k=t+2}^{n-1}(d_k+1)\\ =& (((d_t+2)d_{t+1}-(d_t+1)(d_{t+1}+1))\sum _{j=1}^t\prod _{k=j}^{t-1}(d_k+1)-1)\prod _{k=t+2}^{n-1}(d_k+1)\\ =& ((d_{t+1}-d_t-1)\sum _{j=1}^t\prod _{k=j}^{t-1}(d_k+1)-1)\prod _{k=t+2}^{n-1}(d_k+1)\end{array}$$

由于 $d_{t+1}-d_t\ge 2$，所以 $d_{t+1}-d_t-1\ge 1$。考虑到 $\sum _{j=1}^t$ 里至少有一个 $\prod _{k=t}^{t-1}=1$，则 $(d_{t+1}-d_t-1)\sum _{j=1}^t\prod _{k=j}^{t-1}(d_k+1)\ge 1$，那么有 $f_1^{\prime }-f_1\ge 0$，也就是 $f_1^{\prime }$ 是不劣于 $f_1$ 的。
如此操作，直至 $\mathrm{\forall }t\in [1,n-2],d_t+1\ge d_{t+1}$。

有了如上证明，发现 $\mathrm{\forall }t\in [1,n-2],d_{t+1}-d_t\in \{0,1\}$。有了这个性质，接下来的证明最优性就简单啦。交给读者自己证明。（中考考完再来补坑）

那么，我们分别考虑两种情况即可。

情况一

当 $m<n-1$ 时，$d_{n-m\sim n-1}=1$，其他位置 $d$ 均为 $0$。此时有：

$$\begin{array}{rl}{f}_1& =\sum _{j=1}^{n-1}\prod _{k=j}^{n-1}([k\ge n-m]+1)\\ & =\sum _{j=n-m}^{n-1}\prod _{k=j}^{n-1}2+(n-m-1)\prod _{k=n-m}^{n-1}2\\ & =\sum _{j=n-m}^{n-1}{2}^{n-j}+(n-m-1)2^m\\ & =2^{m+1}-2+(n-m-1)2^m\\ & =2^{m+1}-2^m-2+(n-m)2^m\\ & =2^m-2+(n-m)2^m\\ & =(n-m+1)2^m-2\end{array}$$

快速幂单次 $\mathrm{\Theta }(\log m)$ 解决。

情况二

当 $m\ge n-1$ 时，$d_1=\lfloor \frac{m-(n-2)}{n-1}\rfloor$，$m$ 除去第一次刷的 $n-2$，和 $d_1$ 次的刷完整个序列，还剩下 $m^{\prime }=m-(n-2)-(n-1)d_1$ 次从 $n-1$ 往左刷，故 $d_{2\sim n-1-m^{\prime }}=d_1+1$，$d_{n-m^{\prime }\sim n-1}=d_1+2$。然后推式子。

$$\begin{array}{rl}{f}_1& =\sum _{j=1}^{n-1}\prod _{k=j}^{n-1}(d_k+1)\\ & =\prod _{k=1}^{n-1}(d_k+1)+\sum _{j=2}^{n-1}\prod _{k=j}^{n-1}(d_k+1)\\ & =(d_1+1)\cdot ((d_1+1)+1{)}^{(n-1-m^{\prime })-2+1}\cdot ((d_1+2)+1{)}^{(n-1)-(n-m^{\prime })+1}+\sum _{j=2}^{n-1}\prod _{k=j}^{n-1}(d_k+1)\\ & =(d_1+1)\cdot (d_1+2{)}^{n-m^{\prime }-2}\cdot (d_1+3{)}^{m^{\prime }}+\sum _{j=2}^{n-1}\prod _{k=j}^{n-1}(d_k+1)\\ & =(d_1+1)\cdot (d_1+2{)}^{n-m^{\prime }-2}\cdot (d_1+3{)}^{m^{\prime }}+\sum _{j=2}^{n-1-m^{\prime }}\prod _{k=j}^{n-1}(d_k+1)+\sum _{j=n-m^{\prime }}^{n-1}\prod _{k=j}^{n-1}(d_k+1)\\ & =(d_1+1)\cdot (d_1+2{)}^{n-m^{\prime }-2}\cdot (d_1+3{)}^{m^{\prime }}+\sum _{j=2}^{n-1-m^{\prime }}((d_1+2{)}^{(n-1-m^{\prime })-j+1}\cdot (d_1+3{)}^{m^{\prime }})+\sum _{j=n-m^{\prime }}^{n-1}\prod _{k=j}^{n-1}(d_k+1)\\ & =(d_1+1)\cdot (d_1+2{)}^{n-m^{\prime }-2}\cdot (d_1+3{)}^{m^{\prime }}+(d_1+3{)}^{m^{\prime }}\cdot \sum _{j=2}^{n-1-m^{\prime }}(d_1+2{)}^{(n-1-m^{\prime })-j+1}+\sum _{j=n-m^{\prime }}^{n-1}\prod _{k=j}^{n-1}(d_k+1)\\ & =(d_1+1)\cdot (d_1+2{)}^{n-m^{\prime }-2}\cdot (d_1+3{)}^{m^{\prime }}+(d_1+3{)}^{m^{\prime }}\cdot \sum _{j=1}^{(n-1-m^{\prime })-2+1}(d_1+2{)}^j+\sum _{j=n-m^{\prime }}^{n-1}\prod _{k=j}^{n-1}(d_k+1)\\ & =(d_1+1)\cdot (d_1+2{)}^{n-m^{\prime }-2}\cdot (d_1+3{)}^{m^{\prime }}+(d_1+3{)}^{m^{\prime }}\frac{(d_1+2{)}^{n-1-m^{\prime }}-(d_1+2)}{d_1+1}+\sum _{j=n-m^{\prime }}^{n-1}\prod _{k=j}^{n-1}(d_k+1)\\ & =(d_1+1)\cdot (d_1+2{)}^{n-m^{\prime }-2}\cdot (d_1+3{)}^{m^{\prime }}+(d_1+3{)}^{m^{\prime }}\frac{(d_1+2{)}^{n-1-m^{\prime }}-(d_1+2)}{d_1+1}+\sum _{j=n-m^{\prime }}^{n-1}\prod _{k=j}^{n-1}(d_1+3)\\ & =(d_1+1)\cdot (d_1+2{)}^{n-m^{\prime }-2}\cdot (d_1+3{)}^{m^{\prime }}+(d_1+3{)}^{m^{\prime }}\frac{(d_1+2{)}^{n-1-m^{\prime }}-(d_1+2)}{d_1+1}+\sum _{j=n-m^{\prime }}^{n-1}(d_1+3{)}^{n-j}\\ & =(d_1+1)\cdot (d_1+2{)}^{n-m^{\prime }-2}\cdot (d_1+3{)}^{m^{\prime }}+(d_1+3{)}^{m^{\prime }}\frac{(d_1+2{)}^{n-1-m^{\prime }}-(d_1+2)}{d_1+1}+\sum _{j=1}^{m^{\prime }}(d_1+3{)}^j\\ & =(d_1+1)\cdot (d_1+2{)}^{n-m^{\prime }-2}\cdot (d_1+3{)}^{m^{\prime }}+(d_1+3{)}^{m^{\prime }}\frac{(d_1+2{)}^{n-1-m^{\prime }}-(d_1+2)}{d_1+1}+\frac{(d_1+3{)}^{m^{\prime }+1}-(d1+3)}{d_1+2}\end{array}$$

其中，$d_1$ 和 $m^{\prime }$ 在上文已经算出，故该式可以在 $\mathrm{\Theta }(\log m)$ 的时间内算出。

代码

注意特判 $n=1$ 的情况。

def fpow(a, b, mod):
    if b < 0:
        return 1
    res = 1
    base = a % mod
    while b:
        if b & 1:
            res = res * base % mod
        base = base * base % mod
        b >>= 1
    return res
 
def inv(x, mod):
    return fpow(x, mod - 2, mod)
 
def main():
    n, m, mod = map(int, input().split())
    if n == 1:
        print(0)
        return
    if m < n - 1:
        pr = fpow(2, m, mod)
        print((pr * (n - m + 1) + mod - 2) % mod)
        return
    d1 = (m - (n - 2)) // (n - 1)
    M = m - (n - 2) - (n - 1) * d1
    S1 = (d1 + 1) * fpow(d1 + 2, n - M - 2, mod) * fpow(d1 + 3, M, mod)
    S2 = fpow(d1 + 3, M, mod) * (fpow(d1 + 2, n - 1 - M, mod) - (d1 + 2) + mod) * inv(d1 + 1, mod)
    S3 = (fpow(d1 + 3, M + 1, mod) - (d1 + 3) + mod) * inv(d1 + 2, mod)
    S1 %= mod
    S2 %= mod
    S3 %= mod
    print((S1 + S2 + S3) % mod)
 
if __name__ == '__main__':
    t = int(input())
    for i in range(t):
        main()

后记

用 python 写的目的是因为教练讲题时用 python 打表唤起了我的回忆？