Codeforces 1264D

Combinatorial Mathematics

Suppose the current enumeration to the location of the $ i $

$ a $: $ I $ before $ ($
$ B $: $ $ $ after I) $
$ $ C: before $ $ $ I $?
$ D $: $ I $ $ after $?

Then the current position to answer contribution

$$\sum{a + i \leqslant b + j}{\tbinom{c}{i} \tbinom{d}{j}}$$

Where i and j represent fill the $ i $ a $ ($, fill in the $ j $ th $) $

This formula is very clever, $ a + i $ represents the number of pairs of parentheses final answer, which is equivalent to calculate the current $ ($ contributions to answer this location

Then consider simplification

$$=\sum_{a + i \leqslant b + d - j}{\tbinom{c}{i} \tbinom{d}{d - j}}$$

$$=\sum_{x = 0}^{b + d - a}{\tbinom{c + d}{x}}$$

$ C + d $ Since there are only two values, can be pretreated

Complexity $ O (n) $

#include <bits/stdc++.h>
using namespace std;
const int P = 998244353;
int main() {
    string s;
    cin >> s;
    int n = s.size();
    vector<int> inv(n + 1), facinv(n + 1), fac(n + 1);
    fac[0] = inv[1] = facinv[0] = 1;
    for(int i = 1; i <= n; ++i) {
        if(i != 1) {
            inv[i] = 1LL * (P - P / i) * inv[P % i] % P;
        }
        facinv[i] = 1LL * facinv[i - 1] * inv[i] % P;
        fac[i] = 1LL * fac[i - 1] * i % P;
    }
    int a = 0, b = 0, c = 0, d = 0;
    for(auto e : s) {
        if(e == ')') {
            ++b;
        }
        if(e == '?') {
            ++d;
        }
    }
    map<int, vector<int> > dp;
    auto C = [&] (int n, int m) {
        if(n < m) return 0;
        if(dp.find(n) != dp.end()) {
            return dp[n][m];
        }
        auto &tmp = dp[n];
        int sum = 0;
        tmp.resize(n + 1);
        for(int i = 0; i <= n; ++i) {
            sum += 1LL * fac[n] * facinv[i] % P * facinv[n - i] % P;
            sum %= P;
            tmp[i] = sum;
        }
        return tmp[m];
    };
    auto calc = [&] (int a, int b, int c, int d) {
        int x = min(b + d - a, c + d);
        if(x < 0) return 0;
        return C(c + d, x);
    };
    int ans = 0;
    for(int i = 0; i < n; ++i) {
        if(s[i] == '(') {
            ++a;
        }
        if(s[i] == ')') {
            --b;
        }
        if(s[i] == '?') {
            ++c;
            --d;
        }
        if(s[i] == '(') {
            years + = calc (a, b, c, d); 
            years % = P; 
        } 
        If (s [i] == ' ? ' ) {
             ++ a;
            - c; 
            years + = calc (a, b, c, d); 
            years % = P;
            - a;
            ++ c; 
        } 
    } 
    Cout << years << ' \ n ' ;
    return  0 ; 
}

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