C++14(g++5.4) 解法, 执行用时: 82ms, 内存消耗: 1532K, 提交时间: 2019-09-13 15:12:28

#include <bits/stdc++.h>
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define per(i, a, b) for (int i = a; i >= b; i--)
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 5;
const int mod = 998244353;

int n, a[maxn], suf[maxn], inv[maxn], ans;

int main() {
    cin >> n;
    rep(i, 1, n) cin >> a[i];
    inv[1] = 1;
    rep(i, 2, n + 1) inv[i] = mod - (ll)mod / i * inv[mod % i] % mod;
    per(i, n, 1) suf[i] = ((ll)a[i] + suf[i + 1]) % mod;
    rep(i, 1, n) ans = ((ll)ans + a[i] + (ll)inv[i + 1] * 2 * suf[i + 1] % mod) % mod;
    return !printf("%d\n", ans);
}

C++11(clang++ 3.9) 解法, 执行用时: 39ms, 内存消耗: 1256K, 提交时间: 2020-03-18 17:17:56

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const ll p=998244353;
ll n,sum,ans,a[500010];
ll ksm(ll x,ll y){
	ll xlh=1;
	while(y){
		if(y&1)xlh=xlh*x%p;
		x=x*x%p;
		y/=2;
	}
	return xlh;
}
int main(){
	ll i;
	scanf("%lld",&n);
	for(i=1;i<=n;i++){
		scanf("%lld",&a[i]);
		ans=(ans+(1+sum)*a[i])%p;
		sum=(sum+2*ksm(i+1,p-2)%p)%p;
	}
	printf("%lld",ans);
}