bzoj3503

显然知道第一行就可以只道整个矩阵
但n<=40，搜索是不行的，我们设第一行为x1~xm
可以由轻易由第一行未知数推出第n+1行，这一步我们可以压成二进制位（因为只和奇偶有关）
显然n+1行必须是0，由此可以列方程高斯消元即可

 var a,c:array[..,..] of longint;
     b:array[..,..] of int64;
     n,m,i,j:longint;

 procedure swap(var a,b:longint);
   var c:longint;
   begin
     c:=a;
     a:=b;
     b:=c;
   end;

 procedure gauss;
   var i,j,k,p:longint;
   begin
     k:=;
     for i:= to m do
     begin
       j:=k;
       while (j<=m) and (a[j,i]=) do inc(j);
       if (j>m) then continue;
       if j<>k then
       begin
         for p:=i to m+ do
          swap(a[k,p],a[j,p]);
       end;
       for j:=k+ to m do
         if a[j,i]= then
         begin
           for p:=i to m+ do
             a[j,p]:=a[j,p] xor a[k,p];
         end;
       inc(k);
     end;
     for i:=m downto  do
     begin
       c[,i]:=a[i,m+];
       if a[i,i]= then
       begin
         c[,i]:=;
         continue;
       end;
       for j:=i+ to m do
         if a[i,j]= then c[,i]:=c[,i] xor c[,j];
     end;
   end;

 begin
   readln(n,m);
   for i:= to m do
     b[,i]:=int64() shl int64(i-);

   for i:= to n+ do
     for j:= to m do
       b[i,j]:=b[i-,j] xor b[i-,j+] xor b[i-,j-] xor b[i-,j];

   for i:= to m do
     for j:= to m do
       if b[n+,i] and (int64() shl int64(j-))<> then a[i,j]:=;

   gauss;
   for i:= to n do
     for j:= to m do
       c[i,j]:=c[i-,j] xor c[i-,j+] xor c[i-,j-] xor c[i-,j];

   for i:= to n do
   begin
     for j:= to m- do
       write(c[i,j],' ');
     writeln(c[i,m]);
   end;
 end.