源码中a1:=abs((c-ref(c,1))/ref(c,1))<=0.05
and abs((o-ref(c,1))/ref(c,1))<=0.05
and c>=ref(c,1) and l<=ref(h,1);
a2:=ref(c,1)>ref(o,1) and (ref(c,1)/ref(o,1))>=1+m/100
and ref(c,1)/ref(c,2)>=1.04 and ref(c,1)>ref(c,2)
and ref(c,1)>ref(c,3) and ref(c,1)>ref(c,4)
and ref(c,1)>ref(c,5) and ref(c,1)>ref(c,6)
and ref(c,1)>ref(c,7) and ref(c,1)>ref(c,8)
and ref(c,1)>ref(c,9);
a3:=if(a1 and a2,3,0);
a4:=barslast(a1 and a2);
a8:=ref(c,a4+1);
a5:=abs((c-ref(c,1))/ref(c,1))<=0.03
and abs((o-ref(c,1))/ref(c,1))<=0.03;
a6:=(c>=a8 and c/a8<=(1+n/100));
a4=count(a5 and a6,a4)
红字部分and连接的两个条件中，第二个：ref(c,1)>ref(c,2)
是否多余？前个条件似乎已经包虑它了。
是的，确认无疑。放心服用