卡特兰数模板

[1,1,2,5,14,42,132,429,1430,4862,16796,58786,
 208012,742900,2674440,9694845,35357670,129644790,
 477638700,1767263190,6564120420,24466267020,
 91482563640,343059613650,1289904147324,
 4861946401452,18367353072152,69533550916004,
 263747951750360,1002242216651368,3814986502092304]

const int C_maxn = 1e4 + 10;
LL CatalanNum[C_maxn];
LL inv[C_maxn];
inline void Catalan_Mod(int N, LL mod)
{
    inv[1] = 1;
    for(int i=2; i<=N+1; i++)///线性预处理 1 ~ N 关于 mod 的逆元
        inv[i] = (mod - mod / i) * inv[mod % i] % mod;

    CatalanNum[0] = CatalanNum[1] = 1;

    for(int i=2; i<=N; i++)
        CatalanNum[i] = CatalanNum[i-1] * (4 * i - 2) %mod * inv[i+1] %mod;
}

#include<bits/stdc++.h>
using namespace std;
const int C_maxn = 100 + 10;///项数

int Catalan_Num[C_maxn][1000];///保存卡特兰大数、第二维为具体每个数位的值
int NumLen[C_maxn];///每个大数的数长度、输出的时候需倒序输出

void catalan() //求卡特兰数
{
    int i, j, len, carry, temp;
    Catalan_Num[1][0] = NumLen[1] = 1;
    len = 1;
    for(i = 2; i < 100; i++)
    {
        for(j = 0; j < len; j++) //乘法
        Catalan_Num[i][j] = Catalan_Num[i-1][j]*(4*(i-1)+2);
        carry = 0;
        for(j = 0; j < len; j++) //处理相乘结果
        {
            temp = Catalan_Num[i][j] + carry;
            Catalan_Num[i][j] = temp % 10;
            carry = temp / 10;
        }
        while(carry) //进位处理
        {
            Catalan_Num[i][len++] = carry % 10;
            carry /= 10;
        }
        carry = 0;
        for(j = len-1; j >= 0; j--) //除法
        {
            temp = carry*10 + Catalan_Num[i][j];
            Catalan_Num[i][j] = temp/(i+1);
            carry = temp%(i+1);
        }
        while(!Catalan_Num[i][len-1]) //高位零处理
        len --;
        NumLen[i] = len;
    }
}

int main(void)
{
    catalan();
    for(int i=1; i<=30; i++){
        for(int j=NumLen[i]-1; j>=0; j--){
            printf("%d", Catalan_Num[i][j]);
        }puts("");
    }
    return 0;
}