Codeforces Round #146 (Div. 2) C. LCM Challenge

http://codeforces.com/contest/236/problem/C

    觉得这题挺有意思，贴一下代码。if(a<=0)or(b<=0)or(c<=0)then continue；这句一开始没写，WA了。注意多个数的LCM求法。

program LCMchallenge;
uses math;
var i,j,k:longint;
    n,a,b,c,ans,temp:int64;
function GCD(a,b:int64):int64;
         var r:int64;
         begin
         while a mod b<>0 do begin
           r:=a mod b;
           a:=b;b:=r;
           end;
         GCD:=b;
         end;
function LCM(a,b:int64):int64;
begin  LCM:=a div GCD(a,b) *b;end;
begin
readln(n);ans:=0;
case n of
  1:ans:=1;
  2:ans:=2;
  3:ans:=6;
  else begin
       if(n and 1)=1
         then ans:=n*(n-1)*(n-2)
         else begin
              for i:=0 to 50 do
                for j:=i+1 to 50 do
                  for k:=j+1 to 50 do
                    begin
                    a:=n-i;b:=n-j;c:=n-k;
                    if(a<=0)or(b<=0)or(c<=0)then continue;
                    ans:=max(ans,LCM(LCM(a,b),c));
                    end;
              end;
       end;
  end;
writeln(ans); readln;
end.