Codeforces Round #604 (Div. 2) E. Beautiful Mirrors

链接：

https://codeforces.com/contest/1265/problem/E

题意：

Creatnx has n mirrors, numbered from 1 to n. Every day, Creatnx asks exactly one mirror "Am I beautiful?". The i-th mirror will tell Creatnx that he is beautiful with probability pi100 for all 1≤i≤n.

Creatnx asks the mirrors one by one, starting from the 1-st mirror. Every day, if he asks i-th mirror, there are two possibilities:

The i-th mirror tells Creatnx that he is beautiful. In this case, if i=n Creatnx will stop and become happy, otherwise he will continue asking the i+1-th mirror next day;

In the other case, Creatnx will feel upset. The next day, Creatnx will start asking from the 1-st mirror again.

You need to calculate the expected number of days until Creatnx becomes happy.

This number should be found by modulo 998244353. Formally, let M=998244353. It can be shown that the answer can be expressed as an irreducible fraction pq, where p and q are integers and q≢0(modM). Output the integer equal to p⋅q−1modM. In other words, output such an integer x that 0≤x<M and x⋅q≡p(modM).

思路：

考虑期望Dp，Dp[i]是从第一天到第i天开心的期望天数。

正向推导\(Dp[i] = Dp[i-1]+ 1 * \frac{p_i}{100} + (1 - \frac{p_i}{100})*(Dp[i]+1)\)

化简得\(Dp[i] = \frac{100*(Dp[i-1]+1)}{p_i}\)

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MOD = 998244353;
const int MAXN = 2e5+10;

LL Dp[MAXN], inv;
int n;

LL ExGcd(LL a, LL b, LL &x, LL &y)
{
    if (b == 0)
    {
        x = 1, y = 0;
        return a;
    }
    LL d = ExGcd(b, a%b, x, y);
    LL tmp = y;
    y = x-(a/b)*y;
    x = tmp;
    return d;
}

LL GetInv(int a, int b)
{
    //a*x = 1 mod b
    LL d, x, y;
    d = ExGcd(a, b, x, y);
    if (d == 1)
        return (x%b+b)%b;
    return -1;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    cin >> n;
    Dp[0] = 0;
    int a;
    for (int i = 1;i <= n;i++)
    {
        cin >> a;
        inv = GetInv(a, MOD);
        Dp[i] = 100*(Dp[i-1]+1)%MOD*inv%MOD;
    }
    cout << Dp[n] << endl;

    return 0;
}