题面 结论:gcd(F[n],F[m])=F[gcd(n,m)]; F[n]=a和F[n+1]=b F[n+2]=a+b,F[n+3]=a+2b,…F[m]=F[m?n?1]a+F[m?n]b F[n]=a,F[n+1]=b,F[m]=F[m?n?1]a+F[m?n] F[m]=F[m?n?1]?F[n]+F[m?n]?F[n+1] gcd(F[n],F[m])=gcd(F[n],F[m?n?1]?F[n]+F[m?n]?F[n+1]) gcd(F[n],F[m])=gcd(F[n],F[m?n…