python/TangDou/AcWing/Math/ZhuHeJiShu/CatalanNumber.cpp

#include <bits/stdc++.h>

using namespace std;
typedef long long LL;
const int mod = 1e9 + 7;
const int N = 110;
int f[N];
int qmi(int a, int k, int p) {
    int res = 1;
    while (k) {
        if (k & 1) res = (LL)res * a % p;
        a = (LL)a * a % p;
        k >>= 1;
    }
    return res;
}

int C(int a, int b, int p) {
    if (a < b) return 0;
    int down = 1, up = 1;

    for (int i = a, j = 1; j <= b; i--, j++) {
        up = (LL)up * i % p;
        down = (LL)down * j % p;
    }
    return (LL)up * qmi(down, p - 2, p) % p;
}

//卡特兰数的前几项为：1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786,208012
int main() {
    //目标：求出卡特兰数的前12位

    //公式法计算I
    //方法1:f(n)=C(n,2n)-C(n-1,2n)
    for (int i = 1; i <= 12; i++)
        cout << C(2 * i, i, mod) - C(2 * i, i - 1, mod) << " ";
    cout << endl;

    //公式法计算II
    //方法2：C(n,2n)/(n+1)
    for (int i = 1; i <= 12; i++)
        cout << C(2 * i, i, mod) / (i + 1) << " ";
    cout << endl;

    //递推法计算I
    //方法3：f(n)=(4n-2)/(n+1)*f(n-1)
    f[0] = 1;
    for (int i = 1; i <= 12; i++) {
        f[i] = (LL)f[i - 1] * (4 * i - 2) / (i + 1); //注意：这里要先乘再除，否则可能丢失精度
        cout << f[i] << " ";
    }
    cout << endl;

    //递推法计算II
    //方法4： f(n)=\sum_{i=0}^{n-1}f(i)*f(n-i+1)
    //要是每次都这么推，不是效率太差了吗？
    memset(f, 0, sizeof f);
    f[0] = 1;
    for (int i = 1; i <= 12; i++) {
        for (int j = 1; j <= i; j++)
            f[i] += f[j - 1] * f[i - j];
        cout << f[i] << " ";
    }
    return 0;
}