Solution:

? 直接计算不是很好搞，于是考虑将代价和变成代价的平均值乘以方案数( \(n!\) )

? 即 \(Ans = \text{删除物品的期望代价}\times n!\)

? 由期望的线性性质，考虑每一个 \(a_i\) 对期望代价的贡献，对于 \(a_i\) 与 \(a_j\)， 若 \(i\) 对 \(j\) 有贡献，必然是第一个删除 \(j\) ，这样 \(a_i\) 的值会被 \(j\) 统计一次，这样的概率为 \(\frac{1}{|j-i|+1}\)，于是有：
\[ Contribution(a_i) = a_i\sum_{j=1}^n\frac{1}{|j-i|+1} \]
答案为
\[ \sum_{i=1}^na_i\sum_{j=1}^n\frac{1}{|j-i|+1} \]
前缀和优化逆元。

Code

#include <vector>
#include <cmath>
#include <cstdio>
#include <cassert>
#include <cstring>
#include <iostream>
#include <algorithm>

typedef long long LL;
typedef unsigned long long uLL;

#define fir first
#define sec second
#define SZ(x) (int)x.size()
#define MP(x, y) std::make_pair(x, y)
#define PB(x) push_back(x)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define GO debug("GO\n")
#define rep(i, a, b) for (register int i = (a), i##end = (b); (i) <= i##end; ++ (i))
#define drep(i, a, b) for (register int i = (a), i##end = (b); (i) >= i##end; -- (i))
#define REP(i, a, b) for (register int i = (a), i##end = (b); (i) < i##end; ++ (i))

inline int read() {
    register int x = 0; register int f = 1; register char c;
    while (!isdigit(c = getchar())) if (c == '-') f = -1;
    while (x = (x << 1) + (x << 3) + (c xor 48), isdigit(c = getchar()));
    return x * f;
}
template<class T> inline void write(T x) {
    static char stk[30]; static int top = 0;
    if (x < 0) { x = -x, putchar('-'); }
    while (stk[++top] = x % 10 xor 48, x /= 10, x);
    while (putchar(stk[top--]), top);
}
template<typename T> inline bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }

#define int LL

using namespace std;

const int maxn = 1e5 + 2;
const int MOD = 1e9 + 7;

LL a[maxn], n;
LL inv[maxn], sum[maxn];

void Input() {
    n = read();
    rep (i, 1, n) a[i] = read();
}

void Init() {
    inv[0] = 0;
    inv[1] = 1;
    sum[1] = inv[1];
    rep (i, 2, n) {
        inv[i] = (MOD - (MOD / i) * inv[MOD % i] % MOD) % MOD;
        sum[i] = (sum[i - 1] + inv[i]) % MOD;
    }
}

int factor(int n) {
    int ans = 1;
    rep (i, 2, n) ans = ans * i % MOD;
    return ans;
}

void Solve() {
    LL ans(0);
    rep (i, 1, n) {
        ans = (ans + a[i] * ((sum[i] - sum[0] + sum[n - i + 1] - sum[1] + MOD) % MOD) % MOD ) % MOD;
    }
    cout << ans * factor(n) % MOD << endl;
}

signed main() 
{
    
    Input();

    Init();

    Solve();

    return 0;
}

[AGC028B]Removing Blocks(概率与期望)

原文：https://www.cnblogs.com/cnyali-Tea/p/11439872.html