Day1 Solution
faebdc
1 lcm
1.1 g´
p°Ýêa†ÊÏêb¦lcm§éc"…c´bê
‡ÏLlcm˜5Ÿ{z¯K
1.2 ){
lcm(x, y) =
xy
gcd(x, y)
d
gcd(x, y) = gcd(x − y, y)
Υ
lcm(a, b) =
ab
gcd(a mod b, b)
duc´bê
lcm(a, b) =
ab
gcd(a mod c, b)
lcm(a, b) ≡
(a mod c)b
gcd((a mod c), b)
( mod c)
¤±Ñ\akéc§,†b¦lcm
2 irrev
2.1 µ
éu˜‡ÊÏS§Œ±ÏLlÑz¦SC¤1nü"
S/ÅÄ0§Ý´£ãS5Ÿž~^ëꃘ"
1