标题：高人帮忙翻译个飞狐公式变成金字塔的

 INPUT:N(10,1,100);
fc:=c;fo:=o;cc:0*c,LINETHICK;oo:0*c,LINETHICK;t1:=0*c;t2:=0*c; fc:=c;fo:=o;cc:0*c,LINETHICK;oo:0*c,LINETHICK;t1:=0*c;t2:=0*c;
for i=1 to datacount do begin for i=1 to datacount do begin
if i=1 then begin if i=1 then begin
cc[i]:=fc[i]; cc[i]:=fc[i];
oo[i]:=fc[i]; oo[i]:=fc[i];
end else begin end else begin
cc[i]:=cc[i-1]; cc[i]:=cc[i-1];
oo[i]:=oo[i-1]; oo[i]:=oo[i-1];
if fc[i]>cc[i]*(1+n/100) then begin if fc[i]>cc[i]*(1+n/100) then begin
cc[i]:=cc[i-1]*(1+n/100); cc[i]:=cc[i-1]*(1+n/100);
oo[i]:=cc[i-1]; oo[i]:=cc[i-1];
end; end;
if fc[i]<oo[i]/(1+n/100) then begin if fc[i]<oo[i]/(1+n/100) then begin
oo[i]:=oo[i-1]/(1+n/100); oo[i]:=oo[i-1]/(1+n/100);
cc[i]:=oo[i-1]; cc[i]:=oo[i-1];
end; end;
if oo[i]=oo[i-1] then if oo[i]=oo[i-1] then
t1[i]:=t1[i-1]+1; t1[i]:=t1[i-1]+1;
if cc[i]=cc[i-1] then if cc[i]=cc[i-1] then
t2[i]:=t2[i-1]+1; t2[i]:=t2[i-1]+1;
end; end;
end; end;
o1:=ref(oo,t1+1); o1:=ref(oo,t1+1);
c1:=ref(cc,t2+1); c1:=ref(cc,t2+1);
STICKLINE(c1=oo,cc,oo,8,1),COLORRED; STICKLINE(c1=oo,cc,oo,8,1),COLORRED;
STICKLINE(o1=cc,cc,oo,8,1),COLORCYAN; STICKLINE(o1=cc,cc,oo,8,1),COLORCYAN;

此多重未来数据数组的公式系统，无法在目前的架构的金字塔下运行，请等待日后升级版

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