CQOI2011分金币&HAOI2008糖果传递

双倍经验……

没想到白书上竟然有……我还看过……还忘了……

抄份题解：

A1 + X1 - X2 = G

A2 + X2 - X3 = G

.

.

.

An + Xn - X1 = G

解得

X1 = X1

X2 = A1 - G + X1

X3 = A1 + A2 - G - G + X1

.

.

.

Xn = sigma(1, n)A - n * G + X1

注意到ans = abs (X1 - lamda1) + abs (X2 - lamda2) + ... + abs (Xn - lamdan)

于是sort 取中位数

代码：

 var i,n:longint;
     x1,tot,ave,ans:int64;
     a,c:array[..] of longint;
 procedure sort(h,l:longint);
  var i,j,m,temp:longint;
  begin
    i:=h;j:=l;m:=c[(i+j)>>];
    repeat
      while (c[i]<m) do inc(i);
      while (c[j]>m) do dec(j);
      if i<=j then
       begin
         temp:=c[i];c[i]:=c[j];c[j]:=temp;
         inc(i);dec(j);
       end;
    until i>j ;
    if i<l then sort(i,l);
    if j>h then sort(h,j);
  end;
 procedure main;
  begin
   readln(n);
   tot:=;
   for i:= to n do
    begin
      readln(a[i]);inc(tot,a[i]);
    end;
   ave:=tot div n;
   c[]:=;
   for i:= to n- do c[i]:=c[i-]+a[i]-ave;
   sort(,n-);
   ans:=;x1:=c[n>>];
   for i:= to n- do inc(ans,abs(x1-c[i]));
   writeln(ans);
  end;
 begin
  main;
 end.