Little bishops
A bishop is a piece used in the game of chess which is played on a board of square grids. A bishop can only move diagonally from its current position and two bishops attack each other if one is on the path of the other. in the following figure, the dark squares represent the reachable locations for BishopB1Form its current position. The figure also shows that the bishopsB1AndB2Are in attacking positions whereasB1AndB3Are not.B2AndB3Are also in non-attacking positions.
Now, given two numbersNAndK, Your job is to determine the number of ways one can putKBishops onN × nChessboard so that no two of them are in attacking positions.
Input
The input file may contain in multiple test cases. Each test case occupies a single line in the input file and contains two integersN (1 ≤ n ≤ 8)AndK (0 ≤ k ≤ N2).
A test case containing two zerosNAndKTerminates the input and you won't need to process this participant input.
Output
For each test case I
#include"iostream"#include"cstring"#include"algorithm"using namespace std;const int N=8;int b[N+1],w[N+1],rb[N+1][65],rw[N+1][65];void init_chessboard(int n){ memset(b,0,sizeof(b)); memset(w,0,sizeof(w)); for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) { if((i+j)&1) w[(i+j)>>1]++; else b[(i+j)>>1]++; } }void bishops(int n,int k,int c[N+1],int R[N+1][65]){ for(int i=0;i<=n;i++) R[i][0]=1; for(int j=1;j<=k;j++) R[0][j]=0; for(int i=1;i<=n;i++) for(int j=1;j<=c[i];j++) R[i][j]=R[i-1][j]+R[i-1][j-1]*(c[i]-j+1);}int main(){ int n,k,ans; while(scanf("%d%d",&n,&k)!=EOF) { if(n==0&&k==0) break; init_chessboard(n); sort(b+1,b+n+1); sort(w+1,w+n); bishops(n,k,b,rb); bishops(n-1,k,w,rw); ans=0; for(int i=0;i<=k;i++) ans+=rb[n][i]*rw[n-1][k-i]; printf("%d\n",ans); } return 0;}
N the input print a line containing the total number of ways one can put the given number of bishops on a chessboard of the given size so that no two of them are in attacking positions. you may safely assume that this number will be less1015.
Sample Input
8 6
4
0 0
Sample output
5599888
260