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CF1426F Number of Subsequences

Description

一个字符串中包含(a,b,c,?) 这4种字符，(?)可变成(a,b,c)
求所有可能的串中有多少个子串是(abc)

Code

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#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
 
const int N = 2e5 + 10;
const int mod = 1e9 + 7;
 
int n;
ll ans, Fa[N], Fb[N], Fc[N];
char ch[N];
 
int main() {
    scanf("%d", &n);
    scanf("%s", ch + 1);
    ll pow_3 = 1;
    for (int i = 1; i <= n; ++i) {
        if (ch[i] == 'a') {
            Fa[i] = (Fa[i - 1] + pow_3) % mod;
            Fb[i] = Fb[i - 1];
            Fc[i] = Fc[i - 1];
        }
        if (ch[i] == 'b') {
            Fa[i] = Fa[i - 1];
            Fb[i] = (Fb[i - 1] + Fa[i - 1]) % mod;
            Fc[i] = Fc[i - 1]; 
        }
        if (ch[i] == 'c') {
            Fa[i] = Fa[i - 1];
            Fb[i] = Fb[i - 1];
            Fc[i] = (Fc[i - 1] + Fb[i - 1]) % mod;
        }
        if (ch[i] == '?') {
            Fa[i] = Fa[i - 1] * 3 % mod;
            Fb[i] = Fb[i - 1] * 3 % mod;
            Fc[i] = Fc[i - 1] * 3 % mod;
            Fa[i] = (Fa[i] + pow_3) % mod;
            Fb[i] = (Fb[i] + Fa[i - 1]) % mod;
            Fc[i] = (Fc[i] + Fb[i - 1]) % mod;
            pow_3 = pow_3 * 3 % mod;
        }
    } 
    printf("%lld
", Fc[n]);
}