https://www.51nod.com/Challenge/Problem.html#!#problemId=1120

學到了卡特蘭數 這道題可以轉換成出棧次序問題

https://blog.csdn.net/wu_tongtong/article/details/78161211

https://baike.baidu.com/item/卡特蘭數/6125746?fr=aladdin

還有就是模數太小 求組合數得用盧卡斯定理 即c(n,k)%mod=(c(n/mod,k/mod)*c(n%mod,k%mod))%mod

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=1e4+7;
const int maxn=1e4+10;

ll pre[maxn];

void init()
{
    ll i;
    pre[0]=1;
    for(i=1;i<=10006;i++) pre[i]=(pre[i-1]*i)%mod;
}

ll quickpow(ll a,ll b)
{
    ll res;
    res=1;
    while(b>0){
        if(b%2) res=(res*a)%mod;
        a=(a*a)%mod,b/=2;
    }
    return res;
}

ll getcnk(ll n,ll k)
{
    return (pre[n]*quickpow(pre[k],mod-2)*quickpow(pre[n-k],mod-2))%mod;
}

ll lucas(ll n,ll k)
{
    ll res;
    res=getcnk(n%mod,k%mod);
    while(n/mod>=mod||k/mod>=mod){
        n/=mod,k/=mod;
        res=(res*getcnk(n%mod,k%mod))%mod;
    }
    res=(res*getcnk(n/mod,k/mod))%mod;
    return res;
}

int main()
{
    ll n;
    init();
    scanf("%lld",&n);
    n--;
    printf("%lld\n",(2ll*lucas(2*n,n)*quickpow(n+1,mod-2))%mod);
    return 0;
}