代码

#include <algorithm>
#include <iostream>

using namespace std;

typedef long long ll;

const int N = 1e5 + 10, mod = 1e9 + 7;

// 因为我们%的数是质数， 因此可以通过费马小定理来求逆元
int quick_power(int a, int k, int p)
{
    int res = 1 % p;
    while (k) {
        if (k & 1) {
            res = (ll)res * a % p;
        }
        k >>= 1;
        a = (ll)a * a % p;
    }
    return res;
}

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

    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=1;i<=b;i++){
         res = (ll) res *quick_power(i, mod - 2, mod)%mod;
    }

    res  = (ll ) res *quick_power(n+1, mod-2 , mod) %mod;

    cout<<res<<endl;

    return 0;
}