排列最小值 [期望]

$排列最小值$

---

$color{red}{正解部分}$

首先排列中第 $i$ 个元素作为前 $i$ 个元素的最小值的概率为 $frac{1}{i}$,

设 $f[i]$ 表示排列第 $i$ 个元素是否为前 $i$ 个元素的最小值, 值为 $0/1$,
根据题目要求, 期望为 $ans = left( sumlimits_{i=1}^n f[i] ight)^2 = sumlimits_{i=1}^n sumlimits_{j=1}^n f[i]f[j]$,

当 $i = j$ 时, $f[i]f[j] = 1$ 的概率为 $frac{1}{i}$, 在上式对期望的贡献为 $sumlimits_{i=1}^n frac{1}{i}$ .
当 $i != j$ 时, $f[i]f[j] = 1$ 的概率为 $frac{1}{i} imes frac{1}{j}$,
在上式对期望的贡献为 $sumlimits_{i=1}^n frac{1}{i}left( (sumlimits_{j=1}^n frac{1}{j}) - frac{1}{i} ight) = left(sumlimits_{i=1}^nfrac{1}{i} ight)^2 - sumlimits_{i=1}^n frac{1}{i^2}$ .

所以 $ans$ 的期望为 $sumlimits_{i=1}^n frac{1}{i}+left(sumlimits_{i=1}^nfrac{1}{i} ight)^2 - sumlimits_{i=1}^n frac{1}{i^2}$,

然后使用 $ans$ 乘上 $n!$ 即为题目中的答案 .

---

$color{red}{实现部分}$

#include<bits/stdc++.h>
#define reg register

int read(){
        char c;
        int s = 0, flag = 1;
        while((c=getchar()) && !isdigit(c))
                if(c == '-'){ flag = -1, c = getchar(); break ; }
        while(isdigit(c)) s = s*10 + c-'0', c = getchar();
        return s * flag;
}

const int maxn = 1e5 + 10;
const int mod = 998244353;

int N;
int Q_;
int inv[maxn];
int fac[maxn];
int sum_1[maxn];
int sum_2[maxn];

void Work(){
        N = read();
        int Ans = (sum_1[N] - sum_2[N] + mod) % mod;
        Ans = (1ll*Ans + (1ll*sum_1[N]*sum_1[N]%mod)) % mod;
        Ans = 1ll*Ans*fac[N] % mod;
        printf("%d
", Ans);
}

int main(){
        Q_ = read();
        fac[0] = 1; for(reg int i = 1; i < maxn; i ++) fac[i] = 1ll*fac[i-1]*i % mod;
        inv[1] = 1; for(reg int i = 2; i < maxn; i ++) inv[i] = ((-1ll*mod/i*inv[mod%i])%mod + mod) % mod;
        for(reg int i = 1; i < maxn; i ++){
                sum_1[i] = (sum_1[i-1] + inv[i]) % mod;
                sum_2[i] = (sum_2[i-1] + (1ll*inv[i]*inv[i]%mod)) % mod;
        }
        while(Q_ --) Work();
        return 0;
}