| 非正经题解 | 统计acgo的出现次

#include<iostream>
#include<string>
using namespace std;
const int MOD=998244353;
int f(const string& S) {
    long long a=0,ac=0,acg=0,acgo=0;
    for (char c:S){
        if (c=='a'){
            a=(a+1)%MOD;
        }else if(c=='c'){
            ac=(ac+a)%MOD;
        }else if(c=='g'){
            acg=(acg+ac)%MOD;
        } else if (c=='o') {
            acgo=(acgo+acg)%MOD;
        }
    }

    return static_cast<int>(acgo);
}

int main() {
    int N;
    string S;
    cin>>N>>S;
    int result =f(S);
    cout<<result<<endl;
    return 0;
}