飞狐可以支持,金字塔应怎样修改?
bz:=amount;
a:=c;
y:=a;
for i=2 to datacount do
y[i]:=(2*a[i]+(bz[i]-1)*y[i-1])/(bz[i]+1);
均线:y;
请耐心等待下个版本,或者
bz:=amount;
a:=c;
y:=a;
if not(islastbar) then
exit;
for i=2 to datacount do
y[i]:=(2*a[i]+(bz[i]-1)*y[i-1])/(bz[i]+1);
均线:y;