CodeForces - 1426 F. Number of Subsequences

题目

每一个?会使得原串的数量扩大为原来的三倍，困难在于如何统计a的数量

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int N = 2e5 + 10, mod = 1e9 + 7;

ll f[N], g[N], k[N]; //在由于？的出现所分裂产生的所有前缀串[1, i]中，abc, ab, a的总数

void solve()
{
    int n;
    string s;
    cin >> n >> s;
    s = "#" + s;
    
    ll cnt = 1;
    for(int i = 1; i <= n; i ++ )
    {
    	if(s[i] == 'a') k[i] = (k[i - 1] + cnt) % mod;
    	else if(s[i] == '?') k[i] = (3 * k[i - 1] % mod + cnt) % mod, cnt = cnt * 3 % mod;
    	else k[i] = k[i - 1];
	}
	
	for(int i = 1; i <= n; i ++ )
	{
		if(s[i] == 'c')
		{
			f[i] = (f[i - 1] + g[i - 1]) % mod;
			g[i] = g[i - 1];
		}
		else if(s[i] == 'b')
		{
			f[i] = f[i - 1];
			g[i] = (g[i - 1] + k[i - 1]) % mod;
		}
		else if(s[i] == '?')
		{
			f[i] = (3 * f[i - 1] % mod + g[i - 1]) % mod;
			g[i] = (3 * g[i - 1] % mod + k[i - 1]) % mod;
		}
		else 
		{
			f[i] = f[i - 1];
			g[i] = g[i - 1];
		}
	}
	cout << f[n] % mod << endl;
	return;
}
 
signed main()
{
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	int t = 1;
	//cin >> t;
	while(t --) solve();
	return 0;
}