2431: [HAOI2009] Reverse sequence

2431: [HAOI2009] Reverse to sequence time Limit:5 Sec Memory limit:128 MB
submit:954 solved:548
[Submit] [Status] Description

For a sequence {AI}, if there is i<j and Ai>aj, then we call the AI and AJ a pair of inverse pairs. For any sequence of 1~n natural numbers, it is easy to find out how many inverse logarithms are available. So how many of these natural number sequences with the inverse logarithm of k?

Input

The first behavior is two integers n,k.

Output

Write an integer that represents the number of numbers that meet the criteria, because this number can be large, you only need to output that number to 10000 after the remainder of the result.

Sample Input Example Inputs

4 1


Sample output Example outputs

3

Sample Description:

The following 3 sequence numbers are 1, respectively 1 2 4 3; 1 3 2 4; 2 1 3 4;



Test data range

30% of Data n<=12

100% of data N<=1000,k<=1000hint Source

Day1

Solving the puzzle: At first glance see the number of reverse order of such words, there are some small excitement, and then a look, incredibly is a DP (phile: hehe hansbug:tt)--Well, the recursive type is also very obvious--f[i,j]:=f[i-1,1]+f[i-1,2]+f[i-1,3] .... +F[I-1,J], so go directly (Note Note: Remember% 10000, and%10000 when you add a 10000 just in case)

1 var2 I,j,k,l,m,n:longint;3A:Array[0.. -,0.. -] ofLongint;4 begin5 readln (n,m);6Fillchar (A,sizeof (a),0);7       fori:=1  toN Do8A[i,0]:=1;9       fori:=2  toN DoTen          begin Onel:=a[i-1,0]; A                forj:=1  toM Do -                   begin -                        if  not(J<i) Thenl:=l-a[i-1, J-i]; thel:=l+a[i-1, j]; -a[i,j]:= (L +20000)MoD 10000; -                   End; -          End; + Writeln (a[n,m]); - End.

2431: [HAOI2009] Reverse sequence