1 #include <iostream>
2 #include <string.h>
3 #include <string>
4 #include <fstream>
5 #include <algorithm>
6 #include <stdio.h>
7 #include <vector>
8 #include <queue>
9 #include <set>
10 #include <cmath>
11 using namespace std;
12 const double eps = 1e-8;
13 const double pi=acos(-1.0);
14 const int INF=0x7fffffff;
15 unsigned long long uINF = ~0LL;
16 #define MAXN 1007
17 #define mod 1000000007
18 typedef long long LL;
19
20 LL d1[33][1000],d2[33][1000];
21 LL len1[33],len2[33];
22 int n,k;
23 void init()
24 {
25 len1[1]=len1[2]=1;
26 len2[1]=len2[2]=2;
27 for(int i=3;i<=30;i+=2)
28 {
29 len1[i]=len1[i+1]=len1[i-1]+2;
30 len2[i]=len2[i+1]=len2[i-1]+2;
31 }
32 }
33 //D[i,j] = (Len[i]-j+1)*D[i-1,j-1] + D[i-1,j]
34 int main()
35 {
36 //freopen("0.in","r",stdin);
37 init();
38 while(scanf("%d%d",&n,&k),n+k!=0)
39 {
40 memset(d1,0,sizeof(d1));
41 memset(d2,0,sizeof(d2));
42
43 for(int i=0;i<=n;i++)
44 {
45 d1[i][0]=1;d2[i][0]=1;
46 }
47
48 for(int i=1;i<=n;i++)
49 for(int j=1;j<=k;j++)
50 {
51 d1[i][j]=(len1[i]-j+1)*d1[i-1][j-1]+d1[i-1][j];
52 if(i!=n)d2[i][j]=(len2[i]-j+1)*d2[i-1][j-1]+d2[i-1][j];
53 }
54
55 LL ans=0;
56 for(int i=0;i<=k;i++)
57 ans+=d1[n][i]*d2[n-1][k-i];
58 printf("%lld\n",ans);
59 }
60
61 return 0;
62 }
D[i,j] = (Len[i]-j+1)*D[i-1,j-1] + D[i-1,j]
对棋盘染色 跟国际象棋的棋盘一样 分成两部分 旋转45° 后 分别dp
ans=d[n][i]+d[n-1][k-i];
