题目链接：hdu 4828 Grids

题目大意：略。

解题思路：将上一行看成是入栈，下一行看成是出栈，那么执着的方案就是卡特兰数，用递推的方式求解。

#include <cstdio>
#include <cstring>

typedef long long ll;
const int N = 1000005;
const ll MOD = 1e9+7;

ll dp[N];

ll extendGcd(ll a, ll b, ll& x, ll& y) {
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }

    ll d = extendGcd(b, a%b, y, x);
    y -= a / b * x;
    return d;
}

ll solve (ll n) {
    ll x, y;
    ll tmp = extendGcd(n + 1, MOD, x, y);
    x = (x % MOD + MOD) % MOD;
    return x;
}

void init () {
    dp[1] = 1;
    dp[2] = 2;
    for (ll i = 3; i < N; i++)
        dp[i] = (dp[i-1] * (4 * i - 2) % MOD * solve(i)) % MOD;
}

int main () {
    int cas, n;
    scanf("%d", &cas);
    init();
    for (int i = 1; i <= cas; i++) {
        scanf("%d", &n);
        printf("Case #%d:\n%lld\n", i, dp[n]);
    }
    return 0;
}