Bracket Sequence

题目链接
题意 :
给定括号的对数和括号的种类求出最后合法序列的数量
思路 :
如果只有一种括号序列就可以直接用卡特兰数
多种匹配的括号可以转换成同一种括号

最后结果为kNk^{N}kN x C2NNC_{2N}^{N}C2NN / (n + 1)
AC代码 :

#include <iostream>
#include <map>
#include <cstring>
#include <algorithm>
#include <math.h>#define IOS ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)using namespace std;
typedef long long LL;
const int N = 200010, mod = 1e9 + 7;int n,k;
int fact[N], infact[N];int ksm(int a, int k) {int res = 1;while (k) {if (k & 1) res = (LL)res * a % mod;a = (LL)a * a % mod;k >>= 1;}return res % mod;
}void init() {fact[0] = infact[0] = 1;for (int i = 1; i < N; i++) {fact[i] = (LL)fact[i - 1] * i % mod;infact[i] = (LL)infact[i - 1] * ksm(i, mod - 2) % mod;}
}int main() {init();IOS;cin >> n >> k;LL res = (LL)fact[2 * n] * infact[n] % mod * infact[n] % mod * ksm(n + 1, mod - 2) % mod;LL ans = ksm(k,n);cout << res * ans % mod << endl;return 0;
}

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