HDOJ_ACM_小兔的棋盘

Problem Description

小兔的叔叔从外面旅游回来给她带来了一个礼物，小兔高兴地跑回自己的房间，拆开一看是一个棋盘，小兔有所失望。不过没过几天发现了棋盘的好玩之处。从起点(0，0)走到终点(n,n)的最短路径数是C(2n,n),现在小兔又想如果不穿越对角线(但可接触对角线上的格点)，这样的路径数有多少?小兔想了很长时间都没想出来，现在想请你帮助小兔解决这个问题，对于你来说应该不难吧!

 

Input

每次输入一个数n(1<=n<=35)，当n等于－1时结束输入。

 

Output

对于每个输入数据输出路径数，具体格式看Sample。

 

Sample Input

1
3
12
-1

 

Sample Output

1 1 2
2 3 10
3 12 416024

//h(n) = h(n-1)*(4n-2)/(n+1)
#include <stdio.h>
int a[40][100];
void catalan()
{
    //initializing the begin of the array
    //circulate from 2 to 35, for each, multiple (4n-2) and divide(n+1)
    //care about initializing the carry and length, and don't forget make a[][0] = length
    int carry, remainder,length, i, j;
    a[1][0] = 1;
    a[1][1] = 2;
    length = 1;
    for (i = 2; i <= 35; i++)
    {
        //multipling (4n-2)
        carry = 0;
        for (j = 1; j <= length; j++)
        {
            carry += a[i - 1][j] * (4 * i - 2);
            a[i][j] = carry % 10;
            carry /= 10;
        }
        while (carry)
        {
            a[i][++length] = carry % 10;
            carry /= 10;
        }
        //dividing (n+1)
        remainder = 0;
        for (j = length; j >= 1; j--)
        {
            remainder = remainder * 10 + a[i][j];
            a[i][j] = remainder / (i + 1);
            remainder = remainder % (i + 1);
        }
        while (!a[i][length])
        {
            length--;
        }
        a[i][0] = length;
    }
}
int main()
{
    int n, count, i;
    catalan();
    count = 1;
    while (scanf("%d", &n) != EOF && n != -1)
    {
        //formatting printing
        printf("%d %d ", count++, n);
        for (i = a[n][0]; i >= 1;i--)
            printf("%d", a[n][i]);
        puts("");
    }
    return 0;
}

Key points

firstly, it said that the number of 'up' is always larger than the number of 'right', or always smaller. Consequestly, it's double of the catalan. you will find that just change the begin number, then you will get the right answer. 

secondly, the above program is designed by getting the results in advance, then I finish other program which is designed by getting the result according the input. For this question, I can't find any difference. The code is as follow.

//h(n) = h(n-1)*(4n-2)/(n+1)
#include <stdio.h>
int a[40][100];
int main()
{
    int carry, remainder,length, i, j, count, n;
    count = 1;
    a[1][0] = 1;
    a[1][1] = 2;
    length = 1;
    while (scanf("%d", &n) != EOF && n != -1)
    {
        a[1][0] = 1;
        a[1][1] = 2;
        length = 1;
        for (i = 2; i <= n; i++)
        {
            //multipling (4n-2)
            carry = 0;
            for (j = 1; j <= length; j++)
            {
                carry += a[i - 1][j] * (4 * i - 2);
                a[i][j] = carry % 10;
                carry /= 10;
            }
            while (carry)
            {
                a[i][++length] = carry % 10;
                carry /= 10;
            }
            //dividing (n+1)
            remainder = 0;
            for (j = length; j >= 1; j--)
            {
                remainder = remainder * 10 + a[i][j];
                a[i][j] = remainder / (i + 1);
                remainder = remainder % (i + 1);
            }
            while (!a[i][length])
            {
                length--;
            }
            a[i][0] = length;
        }
        printf("%d %d ", count++, n);
        for (i = a[n][0]; i >= 1;i--)
            printf("%d", a[n][i]);
        puts("");
    }
    return 0;
}