POJ 2456(二分哲学）

这题普通的二分会T…………

法一：只循环60遍，用ans记录答案(见标程）

法二：进行特判，若l+1==r 则 m=(l+r+1) shl 1 否则 m=(l+r) shl 1

Program P2456;
const
   maxd=1000000000;
   maxn=100000;
var
   n,c,i,j,k:longint;
   a:array[1..maxn] of longint;
procedure sort(l,r:longint);
var
   m,i,k,head,tot,ans:longint;
begin
   for k:=1 to 60 do
   begin
      m:=(l+r) shr 1;
      head:=1;tot:=0;
      for i:=2 to n do
      begin
         if a[i]-a[head]>=m then
         begin
            inc(tot);
            head:=i;
         end;
      end;
      if tot>=c-1 then begin ans:=m; l:=m+1 end else r:=m-1;
   end;
   writeln(ans);

end;
procedure qsort(l,r:longint);
var
   i,j,m,p:longint;
begin
   i:=l;
   j:=r;
   m:=a[(l+r) shr 1];
   repeat
      while a[i]<m do inc(i);
      while a[j]>m do dec(j);
      if i<=j then
      begin
         p:=a[i];a[i]:=a[j];a[j]:=p;
         inc(i);dec(j);
      end;
   until i>j;
   if l<j then qsort(l,j);
   if i<r then qsort(i,r);
end;
begin
   read(n,c);
   for i:=1 to n do read(a[i]);
   qsort(1,n);
   sort(1,maxd);

end.