ACM Number Theory Tour 3---Greatest common divisor gcd and least common multiple LCM (struck, rebuild (￣?￣))

GCD (A, B) is the greatest common divisor of A and B

LCM (A, b), which is the least common multiple of a and B

And then there's a formula.

A*b = gcd * LCM (GCD is gcd (A, B), (?∀?) shorthand do you understand?)

Explain (skip if you don't want to see) {

First, ask for a gcd, then ...

A/GCD and B/GCD These two numbers coprime, namely gcd (A/GCD, B/GCD) = 1, then ...

LCM = gcd * (A/GCD) * (B/GCD)

LCM = (A * b)/GCD

So.. A*b = GCD * LCM

}

So request LCM, ask GCD first

Spicy, the problem, gcd how to Beg

Euclidean method

While loop

1 ll GCD (ll A, ll b) {2    ll t; 3      while (b) {4         t = b; 5         b = a% b; 6         A = t; 7     }8     return  A; 9 }

and a recursive notation.

1 ll GCD (ll A, ll b) {2     if(b = =0)returnA;3     Else returnGCD (b, a%b);4 }5 6 ll GCD (ll A, ll b) {7     returnB? GCD (b, a%b): A;8 }9 //two kinds are available

Spicy, LCM = A * B/GCD

(Note that this may be wrong, because A * b may be too large to exceed int or exceed Longlong)

So the recommendation is written as: LCM = A/GCD * b

And then a few formulas to prove it yourself.

GCD (ka, KB) = k * GCD (A, B)

LCM (ka, KB) = k * LCM (A, B)

Last time I hit this formula.

LCM (s/a, s/b) = S/GCD (A, B)

S = 9,a = 4,b = 6, fractional does not LCM, had to retain fractional form to-pass numerator.

When I see the formula on the right ....

(╯°д°) ╯┻━┻

This TM I think of, to prove to me is a certificate. t_t

ACM number Theory Tour 3---Greatest common divisor gcd and least common multiple LCM (struck, rebuild (￣?￣))