复制内容到剪贴板
代码:
fc:=c;
for i1=13 to datacount do begin
s1:=0;
for j1=2 to 13 do s1:=s1+fc[i1-j1];//循环取得第2根K线到第13根K线的值
ma1[i1]:s1/10.8;
end;
for i2=25 to datacount do begin
s2:=0;
for j2=12 to 25 do s2:=s2+fc[i2-j2];//循环取得第12根K线到第25根K线的值
ma2[i2]:s2/15.2;
end;
for i3=58 to datacount do begin
s3:=0;
for j3=35 to 58 do s3:=s3+fc[i3-j3];//循环取得第35根K线到第58根K线的值
ma3[i3]:s3/24;
end;