题目链接：https://codeforces.com/contest/1265/problem/E
题目大意：
有 (n) 个步骤，第 (i) 个步骤成功的概率是 (P_i) ，每一步只有成功了才会进入下一步，失败了会从第 (1) 步重新开始测。请问成功的期望步数是多少？
解题思路：
设期望步数是 (S) ，则有公式如下：
实现代码如下：
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 200020;
const ll MOD = 998244353LL;
void gcd(ll a , ll b , ll &d , ll &x , ll &y) {
if(!b) {d = a; x = 1; y = 0;}
else { gcd(b , a%b,d,y , x); y -= x * (a/b); }
}
ll inv(ll a , ll n) {
ll d , x , y;
gcd(a , n , d, x , y);
return d == 1 ? (x+n)%n : -1;
}
int n;
ll a[maxn], p[maxn], s[maxn];
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i ++) scanf("%lld", &a[i]);
for (int i = 1; i <= n; i ++) {
p[i] = a[i] * inv(100, MOD) % MOD;
}
s[0] = 1;
for (int i = 1; i <= n; i ++) s[i] = s[i-1] * p[i] % MOD;
ll x = 0, y = 0;
for (int i = 1; i <= n; i ++) {
x = (x + s[i-1] * (1 - p[i] + MOD) % MOD) % MOD;
y = (y + s[i-1] * (1 - p[i] + MOD) % MOD * i % MOD) % MOD;
}
y = (y + s[n] * n % MOD) % MOD;
x = (1 - x + MOD) % MOD;
ll res = y * inv(x, MOD) % MOD;
printf("%lld
", res);
return 0;
}