考虑g1=gcd(al,al+1,...,ar),g2=gcd(al,al+1,...,ar,ar+1)g_1=gcd(a_l,a_{l+1},...,a_{r}),g_2=gcd(a_l,a_{l+1},...,a_r,a_{r+1})g1​=gcd(al​,al+1​,...,ar​),g2​=gcd(al​,al+1​,...,ar​,ar+1​) 显然有g1g_1g1​为g2g_2g...