随笔分类 -  数论------gcd

摘要:洛谷P1072:https://www.luogu.org/problemnew/show/P1072 思路 gcd(x,a0)=a1 lcm(x,b0)=b1→b0*x=b1*gcd(x,b0) (由a*b=gcd(a,b)*lcm(a,b)) x=(b1/b0)*gcd(x,b0) 令i=gcd 阅读全文
posted @ 2018-09-16 16:34 Nanchtiy 阅读(184) 评论(0) 推荐(0)