求一根阳线穿透3根均线公式

Question

求通达信一根阳线穿透3根均线选股公式

均线为5天.10天.20天

Answer 1

阳线穿透3根均线:C>MAX(MA(C,5),MAX(MA(C,10), MA(C,20))) AND O<MIN(MA(C,5),MIN(MA(C,10), MA(C,20)));

Answer 2

上面的答案很好很专业，只是有一点小小缺憾。
C>MAX(MA(C,5),MAX(MA(C,10), MA(C,20))) AND O<MIN(MA(C,5),MIN(MA(C,10), MA(C,20)));
这个答案忽略了当日高开且开盘价在三条均线以上的情况。
T1:=ref(C,1)<MIN(MA(C,5),MA(C,10), MA(C,20));
T2:=C>MAX(MA(C,5),MA(C,10), MA(C,20));
T1 and T2;
这个答案忽略了当日低开且前一日收盘价在三条均线上而当日开盘价在三条均线以下的情况。
两个答案结合在一起就完美了，以下为通达信公式：
T1:=C>MAX(MA(C,5),MAX(MA(C,10),MA(C,20))) AND O<MIN(MA(C,5),MIN(MA(C,10),MA(C,20)));
T2:=C>MAX(MA(C,5),MAX(MA(C,10),MA(C,20))) AND REF(C,1)<MIN(MA(C,5),MIN(MA(C,10),MA(C,20)));
T1 OR T2;

Answer 3

T1:=ref(C,1)<MIN(MA(C,5),MA(C,10), MA(C,20));
T2:=C>MAX(MA(C,5),MA(C,10), MA(C,20));
T1 and T2;

Answer 4

T1:=C>MAX(MA(C,5),MA(C,10), MA(C,20)) AND O<MIN(MA(C,5),MA(C,10), MA(C,10));
T2:EXIST(T1,10);