成功案例
指標源碼a1:=c/ref(c,1)>1.088 and count(c[ref(ma(v,135),1)*1.2,10)=0;]a2:=v>ma(v,135)*1.2 and v>ref(v,1) and ref(v,1)<> a3:=count(v>ref(ma(v,135),1)*1.05,6)<> a4:=c[ref(ma(v,135),1)*1.05;]a5:=count(a4,8)<2 and="" not(ref(c,1)="">ref(c,2) and ref(v,1)>ma(v,120)*1.05 and ref(c,2)>ref(c,3) and ref(v,2)>ma(v,135)*1.05); a1 and a2 and a3 and a5;
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