var1:=ema(winner(c)*70,3);var2:=ema(winner(c*1.1)-winner(c*0.9)*80,3);var3:=c/ma(c,40)*100;var4:=c/ma(c,60)*100;var5:=h>l*1.051;v**:=var5 and count(var5,5)>1;aab1:v** and(var3 or var4)and var2;
呵呵 你这个问题比较有针对性 不错 我再给你加个5 10 20均线成多头排列 这样会更好些 但是注意大盘的风险 a1:=ma(c,5);a2:=ma(c,10);a3:=ma(c,20);xg:a1>a2 and a2>a3 and ref(l,1)(c,2)>a2 and ref(c,3)>a1 and c>a1 and vol>ref(vol,1)and r...