[CF785D] Anton and School

[CF785D] Anton and School - 2

Description

给定一个长度≤2×10^5由(和)组成的字符串，问有多少个子串（可以不连续），前半部分是由(组成后半部分由)组成.

Solution

枚举最后一个 ( 的位置，枚举选了多少个括号，此时的贡献是 C(ai-1,j)C(bi,j+1)

这个东西可以转化为 C(ai-1,j) C(bi,bi-j-1)，对 j 求和，其实就是 C(ai+bi-1,bi-1)=C(ai+bi-1,ai)

于是我们暴力枚举 i 然后算组合数即可

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

#define int long long

const int mod = 1e9 + 7;

namespace math_mod
{
    int c__[5005][5005], fac__[3000005];

    int qpow(int p, int q)
    {
        return (q & 1 ? p : 1) * (q ? qpow(p * p % mod, q / 2) : 1) % mod;
    }

    int inv(int p)
    {
        return qpow(p, mod - 2);
    }

    int fac(int p)
    {
        if (p <= 3000000)
            return fac__[p];
        if (p == 0)
            return 1;
        return p * fac(p - 1) % mod;
    }

    int __fac(int p)
    {
        return fac(p);
    }

    int ncr(int n, int r)
    {
        if (r < 0 || r > n)
            return 0;
        return fac(n) * inv(fac(r)) % mod * inv(fac(n - r)) % mod;
    }

    void math_presolve()
    {
        fac__[0] = 1;
        for (int i = 1; i <= 3000000; i++)
        {
            fac__[i] = fac__[i - 1] * i % mod;
        }
        for (int i = 0; i <= 5000; i++)
        {
            c__[i][0] = c__[i][i] = 1;
            for (int j = 1; j < i; j++)
                c__[i][j] = c__[i - 1][j] + c__[i - 1][j - 1], c__[i][j] %= mod;
        }
    }

    int __c(int n, int r)
    {
        if (r < 0 || r > n)
            return 0;
        if (n > 5000)
            return ncr(n, r);
        return c__[n][r];
    }
}

using namespace math_mod;

const int N = 1e6 + 5;
string s;
int a[N], b[N];

signed main()
{
    ios::sync_with_stdio(false);
    cin >> s;
    s = ' ' + s;
    int n = s.length() - 1;
    for (int i = 1; i <= n; i++)
        a[i] = a[i - 1] + (s[i] == '(');
    for (int i = n; i >= 1; i--)
        b[i] = b[i + 1] + (s[i] == ')');
    math_presolve();
    int ans = 0;
    for (int i = 1; i <= n; i++)
        if (s[i] == '(')
            ans = (ans + __c(a[i] + b[i] - 1, a[i])) % mod;
    cout << ans << endl;
}