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;
}
转载于:https://www.cnblogs.com/cnyali-Tea/p/11439872.html
