ac1012.pas

来自「这是在网络上搜集到的在东京大学的ACM上面解决的一些题目的源码」· PAS 代码 · 共 28 行

program tju1012;
const
  maxn=99;
var
  code,pre,next:array[1..maxn]of word;
  n,m,i,p,j:byte;
begin
  repeat
    read(n,m);
    for i:=1 to n do begin
      read(code[i]);
      if i=1 then pre[i]:=n else pre[i]:=i-1;
      if i=n then next[i]:=1 else next[i]:=i+1;
    end;

    m:=m mod n;if m=0 then m:=n;write(m);p:=m;
    for i:=n-1 downto 1 do begin
      m:=code[p] mod i;pre[next[p]]:=pre[p];next[pre[p]]:=next[p];p:=pre[p];
      if m*2>i then
        for j:=1 to i-m do p:=pre[p]
      else
        for j:=1 to m do p:=next[p];
      write(' ',p);
    end;
    writeln;
  until seekeof;
end.