E - Critical Hit

https://atcoder.jp/contests/abc280/tasks/abc280_e

REFERENCE

https://blog.csdn.net/sophilex/article/details/128169335

dp[i]=(dp[i-2]+1)*p/100+(dp[i-1]+1)*(1-p/100)

 https://atcoder.jp/contests/abc280/submissions/36968112

求逆模

https://blog.csdn.net/qq_41897386/article/details/82289975

让除法运算改成乘法运算

费马小定理(Fermat's little theorem)

假如 p 是质数，那么 a ^ (p-1) ≡ 1 (mod p) 。

推论： b ^ (p - 2) % p 即为 b mod p 的乘法逆元。

Code

https://atcoder.jp/contests/abc280/submissions/37047001

#define ll long long
 
int n, p;
const int mod = 998244353;
ll in;
 
map<ll, ll> cache;
 
ll ksm(ll x,ll y)
{
    ll ans=1;
    while(y)
    {
        if(y&1) ans=ans*x%mod;
        x=x*x%mod;
        y>>=1;
    }
    return ans;
}
 
ll inv(ll x)
{
    return ksm(x,mod-2);
}
 
ll e(ll n){
    if(cache[n] != 0){
        return cache[n];
    }
    
    if (n<=0){
        return 0;
    } else if(n == 1){
        return 1;
    }
    
    ll big = e(n-2);
    ll small = e(n-1);
    
    big *= p;
    small *= 100 - p;
 
    big %= mod;
    small %= mod;
 
    big *= in;
    small *= in;
 
    big %= mod;
    small %= mod;
 
    ll av = big + small + 1;
    av %= mod;
    
    cache[n] = av;
    
    return av;
}
 
int main()
{
    cin >> n;
    cin >> p;
 
    in=inv(100);
 
    cout << e(n) << endl;
 
    return 0;
}