答案:第\(n\)个卡特兰数
\(\large \displaystyle \frac{C_{2n}{n}}{n+1}\ mod \ p\)
因为\(p\)不是质数,不好求逆元所以用等价形式代替
\(\large \displaystyle (C_{2n}^{n}−C_{2n}^{n-1}) \ mod \ p\)
求法:先分解质因式\(p^{α1}*p^{α2}…p^{αk}\)+快速幂
实现代码
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 2000010;
int n, mod;
int primes[N], cnt;
bool st[N];
void get_primes(int n) {
for (int i = 2; i <= n; i++) {
if (!st[i]) primes[cnt++] = i;
for (int j = 0; primes[j] * i <= n; j++) {
st[i * primes[j]] = true;
if (i % primes[j] == 0) break;
}
}
}
int qmi(int a, int k) {
int res = 1;
while (k) {
if (k & 1) res = (LL)res * a % mod;
a = (LL)a * a % mod;
k >>= 1;
}
return res;
}
int get(int n, int p) {
int s = 0;
while (n) {
s += n / p;
n /= p;
}
return s;
}
int C(int a, int b) {
int res = 1;
for (int i = 0; i < cnt; i++) {
int p = primes[i];
int s = get(a, p) - get(b, p) - get(a - b, p);
res = (LL)res * qmi(p, s) % mod;
}
return res;
}
int main() {
//加快读入
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> mod;
get_primes(n * 2);
cout << (C(n * 2, n) - C(n * 2, n - 1) + mod) % mod << endl;
return 0;
}