Ultraviolet A 11388 GCD lcm (number theory)

 

 

Ultraviolet A 11388 GCD lcm (number theory)

Question: Check whether there is a, B so that lcm (a, B) = L, gcd (a, B) = g, there is no output-1, there is output A, B, and a is as small as possible

Analysis: Forced violence is impossible. The data is huge. LlU is used. There are two ideas here.

Idea 1: a * B = g * l

If a is a multiple of a = g, enumerate
Then judge whether g * l can divide a, and finally judge whether B is a multiple of.
A from G to SQRT (G * l)

// Enter two integers, G, l // locate A and B so that gcd (a, B) = g lcm (A, B) = L // requires a to be as small as possible // train of thought:/* from question a * B = g * l enumerative A and B TLE enumerative, then, when determining whether B is to ensure that A is an appointment of G, judge whether g * l % A is zero, and then judge whether B % G is zero, prove it directly .... * // # Define bug # include <cstdio> # include <cstring> # include <algorithm> # include <map> # include <vector> # include <list> # include <queue> # include <ctime> # include <iostream> # include <cmath> # include <set> # include <string> using namespace STD; typedef long ll; typedef unsigned long ull; # define INF 0x7fffffff # define Max 0x3f3f3fvoid orz () {int t; ull T, G, L, A, B; scanf ("% d", & T); While (t --) {scanf ("% LlU", & G, & L); t = g * l; int flag = 0; for (a = g; A * A <= T; A + = g) {If (T % A = 0) {B = t/; if (B % G = 0) {flag = 1; break ;}} if (FLAG) printf ("% LlU \ n", a, B ); else printf ("-1 \ n") ;}} int main () {orz (); Return 0 ;}

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Train of Thought 2: directly prove that when l * g = a * B exists, a = G, B = L

If B %! = 0, there is no solution (not proof)
If B % A = 0, the solution exists, and the solution is a = G, B = L;
It can be thought that a should be as small as possible, then the minimum can only be the maximum public approx. g of two numbers
The larger the value of B is, and the maximum value is the least public multiple of two numbers.
A = G, B = L;

#include <cstdio>typedef unsigned long long ULL;void Orz(){    int T;    ULL G, L;    scanf("%d",&T);    while(T--)    {        scanf("%llu %llu",&G,&L);        if(L % G)            puts("-1");        else            printf("%llu %llu\n",G,L);    } }int main(){    Orz();    return 0;}

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Ultraviolet A 11388 GCD lcm (number theory)