HDU-1023 Train Problem II（高精度+卡特兰数）

HDU-1023 Train Problem II（高精度+卡特兰数）

Train Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13764 Accepted Submission(s): 7339

Problem Description
As we all know the Train Problem I, the boss of the Ignatius Train Station want to know if all the trains come in strict-increasing order, how many orders that all the trains can get out of the railway.

Input
The input contains several test cases. Each test cases consists of a number N(1<=N<=100). The input is terminated by the end of file.

Output
For each test case, you should output how many ways that all the trains can get out of the railway.

Sample Input
1
2
3
10

Sample Output
1
2
5
16796

思路：
顺序入栈求出栈方案为典型的卡特兰数问题，比赛开始我第一个看到这题，这不是很简单嘛，卡特兰数题，可一看是高精度，我头一回面对这么大数的卡特兰数，没写过，连忙翻看资料，以应对如此大的数据。
对于卡特兰数的应用可看我另一篇博客：
https://blog.youkuaiyun.com/zuzhiang/article/details/77966726
接下来就是套板子时刻了。
看代码

//经典卡特兰数问题
#include<iostream>
#include<string>
#include<algorithm>

using namespace std;

typedef long long LL;
int a[101][101],b[101];
const int N = 200010, mod = 1e9 + 7;

int n;
int fact[N], infact[N];

int ksm(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;
}

void init() {
    fact[0] = infact[0] = 1;
    for (int i = 1; i < N; i++) {
        fact[i] = (LL)fact[i - 1] * i % mod;
        infact[i] = (LL)infact[i - 1] * ksm(i, mod - 2) % mod;
    }
}

 
void catalan() 
{
    int i, j, len, carry, temp;
    a[1][0] = b[1] = 1;
    len = 1;
    for(i = 2; i <= 100; i++)
    {
        for(j = 0; j < len; j++) 
        a[i][j] = a[i-1][j]*(4*(i-1)+2);
        carry = 0;
        for(j = 0; j < len; j++)
        {
            temp = a[i][j] + carry;
            a[i][j] = temp % 10;
            carry = temp / 10;
        }
        while(carry) 
        {
            a[i][len++] = carry % 10;
            carry /= 10;
        }
        carry = 0;
        for(j = len-1; j >= 0; j--) 
        {
            temp = carry*10 + a[i][j];
            a[i][j] = temp/(i+1);
            carry = temp%(i+1);
        }
        while(!a[i][len-1])
        len --;
        b[i] = len;
    }
}
int main() {
    // init();
    catalan();
    while(cin>>n){

        for(int i=b[n];i>0;i--)
            cout<<a[n][i-1];
        cout<<endl;
    }
   
    // int res = (LL)fact[2 * n] * infact[n] % mod * infact[n] % mod * ksm(n + 1, mod - 2) % mod;
    
    return 0;
}