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hdu 1023 Train Problem II

2018年05月02日 ⁄ 综合 ⁄ 共 1706字 ⁄ 字号小中大 ⁄ 评论关闭

Train Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5945 Accepted Submission(s): 3236

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

Sample Output

1
2
5
16796

Hint
The result will be very large, so you may not process it by 32-bit integers.

/*题解：
Catalan数应用中的凸多边形三角划分问题，直接将Catalan数输出即可
*/

Catalan模板：
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;
    }
}

AC代码：

#include<cstdio>
int a[102][102],b[102];
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()
{
	int i,n;
	catalan();
	while(scanf("%d",&n)!=EOF)
	{
		for(i=b[n]-1; i>=0; i--)
			printf("%d",a[n][i]);
		printf("\n");
	}
	return 0;
}

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