[HBCPC2024] Points on the Number Axis A

题目描述

Alice is playing a single-player game on the number axis.

There are $n$ points on the number axis. Each time, the player selects two points. The two points will be removed, and their midpoint will be added. When there is only one point on the axis, the game ends. Formally, if the two chosen points is $x_{i}$ and $x_{j}$ , then $\frac{x_{i} + x_{j}}{2}$ will be added after the operation.

In order to play this game happily, Alice will always select two points randomly.

Now Alice have a question: where is the expected position of the last point.

It can be proven that the answer can be expressed in the form $\frac{p}{q}$ , you only need to output the value of $p \cdot q^{- 1} mod 998 244 353$ .

输入格式

The first line contains one integer $n$ ( $1 \leq n \leq 10^{6}$ ).

The second line contains $n$ integers $x_{i}$ ( $0 \leq x_{1} \leq \dots \leq x_{n} < 998 244 353$ ), denoting the position of the $i$ -th point.

Note that two points may be in the same position.

输出格式

Output one integer, the answer modulo $998 244 353$ .

样例 #1

样例输入 #1

3
1 2 4

样例输出 #1

332748120

题解

本题其实就是求 $\sum_{i = x}^{n} \frac{x_{i}}{n}$ ；

记 $S = \sum_{i = x}^{n} x_{i}$ ， $m = 998244353$ ， $a$ 为 $n$ 在模 $m$ 下的逆元；

所以，仅求 $(S \times a) % m$ ；

因为 $a n \equiv 1 (m o d m)$ ；

所以 $a n \equiv n^{m - 1} (m o d m)$ （根据费马小定理 $a^{p - 1} \equiv 1 (m o d p)$ ）；

所以 $a \equiv m^{m - 2} (m o d m)$ 。

#include <iostream>

using namespace std;
 	`	1	1	1	1	1	1	1`
const int mod = 998244353;

//快速幂求逆元
long long qpow(int a, int b) {
	long long ans = 1;
	while (b) {
		if (b & 1)ans = ans * a % mod;

		a = (1ll * a * a) % mod;// 可能溢出，用long long防止 // 由于%的优先度高于*，所以先计算1ll*a*a再取模

		b >>= 1;

	}
	return ans;
}


int main()
{
	int n; cin >> n;

	long long sum = 0, x;
	for (int i = 0; i < n; i++)cin >> x, sum = (sum + x) % mod;

	cout << sum * qpow(n, mod - 2) % mod << endl;

	return 0;
}