使用扩展欧几里得的方法解决此问题，时间上比 用快速幂求逆元快了一倍。

#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int  mod = 1e9 + 7;


/*int qmi(int a, int k, int p)
{
    int res = 1;
    while (k)
    {
        if (k & 1) res = (LL)res * a % p;
        a = (LL)a * a % p;
        k >>= 1;
    }
    return res;
}*/

int exgcd(int a, int b, int &x, int &y)
{
    if (!b)
    {
        x = 1, y = 0;
        return a;
    }
    int d = exgcd(b, a % b, y, x);

    y = y - (LL)a / b * x % mod;
    return d;
}

int main()
{
    int n;
    cin >> n;
    int x, y;
    int a = n * 2, b = n;
    int res = 1;
    for (int i = a; i > a - b; i -- )
    {
     res = (LL)res * i % mod;   
    }

    for (int i = 2; i <= b; i ++ )
    {
        exgcd(i, mod, x, y);
        // cout << x<< endl;
        // cout<< qmi(i, mod - 2, mod) << endl;
        //res = (LL)res * qmi(i, mod - 2, mod) % mod;
        res = (LL)res * x % mod;
    }
    exgcd(n+1, mod, x, y);

    //res = (LL)res * qmi(n + 1, mod - 2, mod) % mod;
    res = ((LL)res * x % mod +mod) % mod; 
    printf("%d",res);

    return 0;
}