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
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.
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An + Xn - X1 = G
解得
X1 = X1
X2 = A1 - G + X1
X3 = A1 + A2 - G - G + X1
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.
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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：
1 var i,n:longint; 2 x1,tot,ave,ans:int64; 3 a,c:array[0..1500000] of longint; 4 procedure sort(h,l:longint); 5 var i,j,m,temp:longint; 6 begin 7 i:=h;j:=l;m:=c[(i+j)>>1]; 8 repeat 9 while (c[i]<m) do inc(i); 10 while (c[j]>m) do dec(j); 11 if i<=j then 12 begin 13 temp:=c[i];c[i]:=c[j];c[j]:=temp; 14 inc(i);dec(j); 15 end; 16 until i>j ; 17 if i<l then sort(i,l); 18 if j>h then sort(h,j); 19 end; 20 procedure main; 21 begin 22 readln(n); 23 tot:=0; 24 for i:=1 to n do 25 begin 26 readln(a[i]);inc(tot,a[i]); 27 end; 28 ave:=tot div n; 29 c[0]:=0; 30 for i:=1 to n-1 do c[i]:=c[i-1]+a[i]-ave; 31 sort(0,n-1); 32 ans:=0;x1:=c[n>>1]; 33 for i:=0 to n-1 do inc(ans,abs(x1-c[i])); 34 writeln(ans); 35 end; 36 begin 37 main; 38 end.
