> 程序在FREE PASCAL 上运行正常，应该没错。
> 
> var a:array[1..20,1..20,0..1]of int64;
>     b:array[1..20]of byte;
>     c,total:int64;
>     n,i,k,fence:integer;
>     t:byte;
> procedure init;
> var i,j,k:integer;
> begin
>   a[1,1,0]:=1;a[1,1,1]:=1;
>   for i:=2 to n do
>   for j:=1 to i do
>   begin
>     a[i,j,0]:=0;
>     for k:=1 to j-1 do
>       a[i,j,0]:=a[i,j,0]+a[i-1,k,1];
>     a[i,j,1]:=0;
>     for k:=j to i-1 do
>       a[i,j,1]:=a[i,j,1]+a[i-1,k,0];
>   end;
>   for i:=1 to n do
>     b[i]:=i;
> end;
> 
> procedure check(m:integer);
> var i,j,ii:integer; sum:int64;
> begin
>   if m=1 then begin writeln(b[1]); exit; end;
>   if m=n
>   then begin
>     i:=0;sum:=0;
>     while sum<c do
>     begin
>       inc(i);
>       ii:=(i+i mod 2)div 2;
>       sum:=sum+a[m,ii,1-i mod 2];
>     end;
>     write(b[ii],' ');
>     for j:=ii to m-1 do
>       b[j]:=b[j+1];
>     c:=c+a[m,ii,1-i mod 2]-sum;
>     t:=i mod 2;
>     fence:=ii;
>     check(m-1);
>   end
>   else begin
>     i:=0;
>     sum:=0;
>     while sum<c do
>     begin
>       inc(i);
>       if ((b[i]>fence)and (t=0))or ((b[i]<fence)and (t=1))
>       then sum:=sum+a[m,i,t];
>     end;
>     write(b[i],' ');
>     fence:=b[i];
>     for j:=i to m-1 do
>       b[j]:=b[j+1];
>     c:=c+a[m,i,t]-sum;
>     t:=1-t;
>     check(m-1);
>   end;
> end;
> 
> begin
>   readln(k);
>   for i:=1 to k do
>   begin
>     readln(n,c);
>     init;
>     check(n);
>   end;
> end.

• 哭吧，当年是virtual pascal.. (0) frkstyc 2008-04-09 12:23:20
  • ~~~~>.<~~~~ 五年了 你才rejudge啊 (108) cuiaoxiang 2008-04-09 15:23:35
    • virtual pascal....听起来好古老... (110) scat 2008-04-09 20:05:18
      • Orz (0) alpc16 2008-04-09 21:33:52
    • Re:~~~~>.<~~~~ 五年了 你才rejudge啊 (110) ghy21 2019-04-23 11:43:41
• Re:终于ac乐 (3) kantianfadai 2008-06-24 08:51:39
• Re:终于ac乐 (5) Yudus 2010-04-13 20:00:19