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AA:=(MA(C,1)+MA(C,3)+MA(C,5)+MA(C,7)+MA(C,9))/5;
BB:=(MA(C,2)+MA(C,4)+MA(C,6)+MA(C,8)+MA(C,10))/5;
CC:=EMA(AA,2);
DD:=EMA(BB,5);
TJ0:=CROSS(CC,DD);
上:=EMA(C,3);
中:=EMA(C,5);
下:=EMA(中,5);
TJ1:=CROSS(上,中);
TJ2:=上>REF(上,1);
X1:=(C+L+H)/3;
BB5:=ATAN(EMA(X1,5)-REF(EMA(X1,5),1))*3.1416*10;
EMA3:=EMA(C,3);
TJ3:=BB5>1 AND EMA(X1,5)>REF(EMA(X1,3),1);
TJ1 AND TJ2 AND TJ0 AND TJ3 OR
TJ1 AND TJ2 AND TJ0 AND TJ3;
这样的公式，竟然也有人发上来，
p20124242352646853.jpg
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