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[0..1500000] of longint;
procedure sort(h,l:longint);
 var i,j,m,temp:longint;
 begin
   i:=h;j:=l;m:=c[(i+j)>>1];
   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:=0;
  for i:=1 to n do
   begin
     readln(a[i]);inc(tot,a[i]);
   end;
  ave:=tot div n;
  c[0]:=0;
  for i:=1 to n-1 do c[i]:=c[i-1]+a[i]-ave;
  sort(0,n-1);
  ans:=0;x1:=c[n>>1];
  for i:=0 to n-1 do inc(ans,abs(x1-c[i]));
  writeln(ans);
 end;
begin
 main;
end.