HDU 1023 Train ProblemII(カトラン数)未解決未解決未解決未解決未解決
Train Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7876 Accepted Submission(s): 4223
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
カトラン数は、カタラン数とも呼ばれ、数学の組み合わせでよく使われるカウント数列です.
h(0)=1,h(1)=1,catalan数を繰返し式を満たす
h(n)= h(0)*h(n-1)+h(1)*h(n-2) + ... + h(n-1)h(0) (n>=2)
例えば、h(2)=h(0)*h(1)+h(1)*h(0)=1*1+1*1=2
h(3)=h(0)*h(2)+h(1)*h(1)+h(2)*h(0)=1*2+1*1+2*1=5
別のプッシュ
h(n)=h(n-1)*(4*n-2)/(n+1);
繰返し関係の解は次のとおりです.
h(n)=C(2n,n)/(n+1) (n=0,1,2,...)
繰返し関係の別の解釈は、次のとおりです.
h(n)=c(2n,n)-c(2n,n-1)(n=0,1,2,...)
(from baike)
待ってくれ
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7876 Accepted Submission(s): 4223
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
Hint
The result will be very large, so you may not process it by 32-bit integers.
カトラン数は、カタラン数とも呼ばれ、数学の組み合わせでよく使われるカウント数列です.
h(0)=1,h(1)=1,catalan数を繰返し式を満たす
h(n)= h(0)*h(n-1)+h(1)*h(n-2) + ... + h(n-1)h(0) (n>=2)
例えば、h(2)=h(0)*h(1)+h(1)*h(0)=1*1+1*1=2
h(3)=h(0)*h(2)+h(1)*h(1)+h(2)*h(0)=1*2+1*1+2*1=5
別のプッシュ
h(n)=h(n-1)*(4*n-2)/(n+1);
繰返し関係の解は次のとおりです.
h(n)=C(2n,n)/(n+1) (n=0,1,2,...)
繰返し関係の別の解釈は、次のとおりです.
h(n)=c(2n,n)-c(2n,n-1)(n=0,1,2,...)
(from baike)
待ってくれ