CF1265E Solution

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题解在说啥？？？

期望步数不就是期望轮数乘上每轮的期望步数

期望轮数就是这轮结束的概率的倒数即 \(\dfrac1{\prod_{i=1}^np_i}\)

每轮期望步数根据期望的线性性就是 \(\sum_{i=1}^ni(1-p_i)\prod_{j=1}^{i-1}p_j\)

也就是步数乘上在这里停下来的概率，停下来的概率就是此处不通过，前面全通过的概率相乘。

期望步数还要加上一个 \(n\prod_{i=1}^np_i\)，就是到达终点的步数

#include <bits/stdc++.h>
#define LL long long
#define sl(n) strlen(n)
#define endline puts("")
#define pii pair<int , int>
#define pr_q priority_queue
#define DEBUG puts("DEBUG.")
using namespace std;
const int N = 2e5 + 10;
const int inf = ~0u >> 2;
const int p = 998244353;
int qpow(int a , int k)
{
	int res = 1;
	while(k)
	{
		if(k & 1)
			res = 1ll * res * a % p;
		a = 1ll * a * a % p;
		k >>= 1;
	}
	return res;
}
const int inv100 = qpow(100 , p - 2);
int n,pp[N],prd[N];
int main()
{
	cin >> n;
	prd[0] = 1;
	for(int i = 1;i <= n;i++)
		scanf("%d" , pp + i),pp[i] = 1ll * pp[i] * inv100 % p,prd[i] = 1ll * prd[i - 1] * pp[i] % p;
	int round = qpow(prd[n] , p - 2);
	int step = 0;
	for(int i = 1;i <= n;i++)
		step = (step + 1ll * i * prd[i - 1] % p * (1 - pp[i] + p) % p) % p;
	step = (step + 1ll * n * prd[n] % p) % p;
	cout << 1ll * round * step % p << endl;
	return 0;
}