题目地址：

https://www.acwing.com/problem/content/891/

给定$n$个$0$和$n$个$1$，构造一个长$2n$的序列，要求对于任意前缀，$1$的个数都大于等于$0$的个数。答案模$10^9+7$输出。

数据范围：
$1\le n\le 10^5$

答案是Catalan数。参考https://blog.csdn.net/qq_46105170/article/details/104243540。有公式：$$C_n=\frac{1}{n+1}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$$这里还需要用到快速幂求逆元，参考https://blog.csdn.net/qq_46105170/article/details/113823892。代码如下：

#include <iostream>
using namespace std;

const int mod = 1e9 + 7;

int fast_pow(int a, int k, int p) {
    
    
    int res = 1;
    while (k) {
    
    
        if (k & 1) res = (long) res * a % p;
        a = (long) a * a % p;
        k >>= 1;
    }

    return res;
}

int main() {
    
    
    int n;
    cin >> n;

    int a = 2 * n, b = n;
    int res = 1;

    for (int i = a; i > a - b; i--) res = (long) res * i % mod;
    for (int i = 1; i <= b; i++) res = (long) res * fast_pow(i, mod - 2, mod) % mod;

    res = (long) res * fast_pow(n + 1, mod - 2, mod) % mod;
    
    cout << res << endl;

    return 0;
}

时间复杂度$O(n\log p)$，$p=10^9+7$，空间$O(1)$。