1.​​题目链接​​。其实这也是卡特兰数的一个定义，也可以转化为进出栈问题。就是求一下Cantalan数的第n项。由于比较大，可以采用质因子分解统计贡献来计算。

#include<bits/stdc++.h>
#define N 2000003
#define LL long long
using namespace std;
LL prime[N], pd[N], num[N], mp[N];
int n, p;
void init()
{
  for (int i = 2; i <= 2000000; i++) {
    if (!pd[i]) prime[++prime[0]] = i, mp[i] = prime[0];
    for (int j = 1; j <= prime[0]; j++) {
      if (prime[j] * i > 2000000) break;
      pd[prime[j] * i] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}

void calc(int x, int val)
{
  int k = x;
  for (int i = 1; prime[i] * prime[i] <= k; i++)
    if (k % prime[i] == 0) {
      while (k % prime[i] == 0) {
        k /= prime[i]; num[i] += val;
      }
    }
  if (k > 1) num[mp[k]] += val;
}
LL qpow(LL a, LL b)
{
  LL res = 1;
  while (b)
  {
    if (b & 1)res = res * a % p;
    a = a * a % p;
    b >>= 1;
  }
  return res;
}
int main()
{
  scanf("%d%d", &n, &p);
  init();
  for (int i = n + 1; i <= 2 * n; i++) calc(i, 1);
  for (int i = 1; i <= n; i++) calc(i, -1);

  calc(n + 1, -1);
  LL ans = 1;
  for (int i = 1; i <= prime[0]; i++)
    ans = ans * qpow((LL)prime[i], num[i]) % p;
  printf("%lld\n", ans);
}