hdu 1023

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

根据公式h(n)=h(n-1)*(4*n-2)/(n-1)打表，可以看出需要算一次大数乘法（大数*int型），一次大数除法（大数/int型），大数乘法从最低位开始用被乘数的每一位分别和乘数相乘，控制进位，大数除法被除数从最高位开始依次往下除以除数，记录本次的余数，乘10加到下次的被除数上（注意不可以写成h[i-1][j]加或者减去什么等于什么，这样就会改变上一个a[i][j]的值，对结果造成影响）
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int a[110][200];
void fun()
{
    int t,s,m,m1;
    a[1][110]=1;
    a[2][110]=2;
    a[3][110]=5;
    for(int i=4; i<=100; i++)
    {
        t=i*4-2;
        s=i+1,m=m1=0;
        for(int j=110; j>=0; j--)
        {

            a[i][j]+=a[i-1][j]*t;
            if(a[i][j]>9)
            {
                a[i][j-1]+=a[i][j]/10;
                a[i][j]=a[i][j]%10;
            }

        }
        for(int j=0; j<=110; j++)
        {
            m=a[i][j]+m1;
            a[i][j]=m/s;
            m1=m%s*10;
        }
    }
}
int main()
{
    fun();
    int n;
    while(~scanf("%d",&n))
    {
        int j=0;
        for(j=0; a[n][j]==0; j++);
        for(int i=j; i<=110; i++)
            printf("%d",a[n][i]);
        printf("\n");
    }
    return 0;
}