题目链接

E - Random Swaps of Balls (atcoder.jp)

Statements

Problem Statement

There are $N-1$ white balls and one black ball. These $N$ balls are arranged in a row, with the black ball initially at the leftmost position.

Takahashi will perform the following operation exactly $K$ times.

• Choose an integer uniformly at random between $1$ and $N$, inclusive, twice. Let $a$ and $b$ the chosen integers. If $a\ne b$, swap the $a$-th and $b$-th balls from the left.

After $K$ operations, let the black ball be at the $x$-th position from the left. Find the expected value of $x$, modulo $998244353$.

Solution

设 $d{p}_{i,j}$ 表示第 $i$ 轮黑球位于位置 $j$ 的概率

根据期望定义 ：$res={\sum }_{i=1}^{N}d{p}_{i,j}\ast i$

发现 2 ~ N 其实是没有区别的位置，即 ：

$d{p}_{k,2}=d{p}_{k,3}=\cdots =d{p}_{k,N}$

所以 ：
$res=d{p}_{k,2}\ast {\sum }_{i=2}^{N}i+d{p}_{k,1}$

求 $d{p}_{k,2}$ 可以递推 ：

设 $\lambda$ 为黑球保持不变的概率，多少我忘了

这一轮位于 x =上一轮位于 x * 保持 + 上一轮不位于 * 交换

公式我也懒得写了

Code

#include<bits/stdc++.h>
using namespace std;
#define int long longint const mod = 998244353;int ksm(int a, int k, int p){int res = 1;a %= p;while(k){if(k & 1) res = res * a % p;k >>= 1;a = a * a % p;}return res;
}
int inv(int x){x %= mod;return ksm(x, mod - 2, mod);
}
int n, k;
void solve(){cin >> n >> k;int p = 2LL * inv(n * n) % mod;for(int i = 2; i <= k; i ++){int t1 = 2LL * inv(n * n) % mod;int t2 = ((n * n % mod - 2 * n) % mod + mod) % mod;t2 = t2 * inv(n * n) % mod;p = (t1 + t2 * p % mod) % mod;}int res = 0; int tmp = ((n * (n + 1) % mod * inv(2LL) - 1) % mod + mod) % mod;res = tmp * p % mod; (p = (n * n % mod - 2 * n + 2) % mod + mod) %= mod;p = p * inv(n * n) % mod;for(int i = 2; i <= k; i ++){int t1 = 2LL * inv(n * n) % mod;int t2 = ((n * n % mod - 2 * n) % mod + mod) % mod;t2 = t2 * inv(n * n) % mod;p = (t1 + t2 * p % mod) % mod;} (res += p) %= mod;cout << res << '\n'; 
}signed main(){ios::sync_with_stdio(false);cin.tie(0), cout.tie(0); solve();return 0;
}