11388 - GCD LCM UVA

Problem D: GCD LCM
Input: standard input Output: standard output
The GCD of two positive integers is the largest integer that divides both the integers without any remainder. The LCM of two positive integers is the smallest positive integer that is divisible by both the integers. A positive integer can be the GCD of many pairs of numbers. Similarly, it can be the LCM of many pairs of numbers. In this problem, you will be given two positive integers. You have to output a pair of numbers whose GCD is the first number and LCM is the second number.
Input
The first line of input will consist of a positive integer T. T denotes the number of cases. Each of the next T lines will contain two positive integer, G and L.
Output
For each case of input, there will be one line of output. It will contain two positive integers a and b, a ≤ b, which has a GCD of G and LCM of L. In case there is more than one pair satisfying the condition, output the pair for which a is minimized. In case there is no such pair, output -1.
Constraints
-           T ≤ 100 -           Both G and L will be less than 231.
Sample Input	Output for Sample Input
2 1 2 3 4	1 2 -1

#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <string.h>
#include <stdlib.h>
using namespace std;
const int maxn = 10;
int main()
{
    int t;
    scanf("%d",&t);
    int i;
    for(i = 0;i < t;++i)
    {
        long long g,l;
        scanf("%lld %lld",&g,&l);
        long long  temp = g*l;
        if(g>l)
        printf("%d\n",-1);
        else
        for(long long j = g;j<=l;j+=g)
          if(l%g)
          {
          printf("%d\n",-1);
          break;
          }
          else
          if((g*l)%j==0&&(g*l)/j<=l)
          {
          printf("%lld %lld\n",j,(g*l)/j);
          break;
          }
    }
    return 0;
}
           

後來仔細一想發現上述的代碼寫的比較繁；

因為題目要求a需要最小，那麼滿足條件的a

就是g那麼對應的b也就隻能是l了！！！！！！

#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <string.h>
#include <stdlib.h>
using namespace std;
const int maxn = 10;
int main()
{
    int t;
    scanf("%d",&t);
    int i;
    for(i = 0;i < t;++i)
    {
        long long g,l;
        scanf("%lld %lld",&g,&l);
          if(l%g)
          {
          printf("%d\n",-1);
          }
          else
          printf("%lld %lld\n",g,l);

    }
    return 0;
}
           

兩個代碼都已ac