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来函数检测结果如下：
该公式没有未来函数

M:=1;

N:=M*3;



BB:=REF(C,1)/REF(O,1)<1.04 AND REF(C,1)/REF(O,1)>0.9 AND REF(C,1)/REF(C,2)<1.03;

B1:=EVERY(BB,N);

B2:=EVERY(BB,N+1);

B3:=EVERY(BB,N+2);

BJ1:=(REF(C,N+2)+REF(L,N+1))/2;

BJ2:=(REF(C,N+3)+REF(L,N+2))/2;

BJ3:=(REF(C,N+4)+REF(L,N+3))/2;

BB1:=B1 AND EVERY(REF(H,1)<BJ1,N) AND REF(C,N+2)/REF(C,N+1)>1.08 AND C>=REF(O,N+1) AND C>=BJ1;

BB2:=B2 AND EVERY(REF(H,1)<BJ2,N+1) AND REF(C,N+3)/REF(C,N+2)>1.08 AND C>=REF(O,N+2) AND C>=BJ2;

BB3:=B3 AND EVERY(REF(H,1)<BJ3,N+2) AND REF(C,N+4)/REF(C,N+3)>1.08 AND C>=REF(O,N+3) AND C>=BJ3;

XG:BB1 OR BB2 OR BB3;

原理解析：
M赋值:1


N赋值:M*3




BB赋值:1日前的C/1日前的O<1.04 AND 1日前的C/1日前的O>0.9 AND 1日前的C/2日前的C<1.03


B1赋值:BB最近N日一直存在


B2赋值:BB最近N+1日一直存在


B3赋值:BB最近N+2日一直存在


BJ1赋值:(N+2日前的C+N+1日前的L)/2


BJ2赋值:(N+3日前的C+N+2日前的L)/2


BJ3赋值:(N+4日前的C+N+3日前的L)/2


BB1赋值:B1 AND EVERY(1日前的H1.08 AND C>=N+1日前的O AND C>=BJ1


BB2赋值:B2 AND EVERY(1日前的H1.08 AND C>=N+2日前的O AND C>=BJ2


BB3赋值:B3 AND EVERY(1日前的H1.08 AND C>=N+3日前的O AND C>=BJ3
输出

XG:BB1 OR BB2 OR BB3