5205. Скобки (Brackets)

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

#define int long long
#define pb push_back
#define mp make_pair
#define FOR(i,a,b) for(int i = (a) ; i < (b) ; i++)
#define ALL(A) A.begin(),A.end()

typedef vector<int> VI;
typedef vector<VI> VVI;

const int INF = (int)1e9;
const int MOD = (int)301907;

void solve_case()
{
    string s;
    cin >> s;

    int n = s.size();

    vector<int> a(n+3,0);  /// Creating vector for odd positions
    vector<int> b(n+3,0);  /// Creating vector for even positions

    a[1] = 1;

    if(n&1)  /// if n % 2 != 0, then we cannot even make 1 case to answer, so cout << 0
    {
        cout << "0";
        return;
    }

    for(int i = 1 ; i <= n ; i++)   /// Here do DP for i as position in string and j as amount of openned brackets, a[j] or b[j]
                                    /// will be number of ways to go to this position with j openned brackets.
    {
        for(int j=2;j<=n;j+=2)
        {
            if(s[i - 1] == '(')
                b[j] = a[j-1] % MOD;
            else if(s[i - 1] == ')')
                b[j] = a[j+1] % MOD;
            else
                b[j] = (a[j-1] + a[j+1]) % MOD;
        }

        i++;

        for(int j=1;j<=n;j+=2)
        {
            if(s[i - 1] == '(')
                a[j] = b[j-1] % MOD;
            else if(s[i - 1] == ')')
                a[j] = b[j+1] % MOD;
            else
                a[j] = (b[j-1] + b[j+1]) % MOD;
        }
    }

    cout << (a[1] + b[1]) % MOD;    /// Actually i didn`t want to think hard, so I cout sum of a[1] and b[1], but one of this parts
                                    /// will be zero

    return;
}

signed main()
{
    ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);

//    freopen("boxes.in","r",stdin);
//    freopen("boxes.out","w",stdout);

    int t = 1;
//    cin >> t;

    while(t--)
    {
        solve_case();
    }

    return 0;
}