原题链接

题解

非交叉匹配

code

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const ll mod = 1e9+7;

ll qpow(ll a, ll b) {
    ll res = 1;
    while (b) {
        if (b & 1) res = res * a % mod;
        a = a * a % mod;
        b >>= 1;
    }
    return res;
}

ll inv(ll x) {
    return qpow(x, mod - 2);  // 使用费马小定理计算逆元
}

ll C(ll n, ll m) {
    if (m > n) return 0;
    ll res = 1;
    for (ll i = 1; i <= m; i++) {
        res = res * (n - i + 1) % mod;
        res = res * inv(i) % mod;
    }
    return res;
}

void solve() {
    ll n;
    cin >> n;
    if(n & 1) {
        cout << 0 << endl;
        return;
    }
    ll ans = C(n, n/2) * inv(n/2 + 1) % mod;  // 计算 Catalan 数
    cout << ans << endl;
}

int main() {
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    ll TT = 1;
    //cin >> TT;
    while (TT--) solve();
    return 0;
}