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I I U C O N L I N E C O N T E S T 2 0 0 8 |
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Problem D: GCD LCM |
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Input: standard input |
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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. |
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Input |
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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. |
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Output |
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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. |
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Constraints |
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- T ≤ 100 - Both G and L will be less than 231. |
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Sample Input |
Output for Sample Input |
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2 1 2 3 4 |
1 2 -1 |
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Problem setter: Shamim Hafiz 题意 ：给出两个数G,L，问是否存在一对数a,b。使得gcd(a,b)==G,lcm(a,b)==L; 能够这么想：当gcd(G,L)==G(a),lcm(G,L)==L(b)时。此时G==a,L==b,满足上述条件。否则不成立。 #include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <list>
#define ll long long
using namespace std;
const int INF = 0x3f3f3f3f;
ll gcd(ll a,ll b)
{
if(b==0) return a;
else return gcd(b,a%b);
}
ll lcm(ll a,ll b)
{
return a*b/gcd(a,b);
}
int main()
{
int t;ll a,b;
scanf("%d",&t);
while(t--)
{
scanf("%lld%lld",&a,&b);
ll G=gcd(a,b),L=lcm(a,b);
if(G==a&&L==b)
printf("%lld %lld
",a,b);
else
puts("-1");
}
return 0;
}
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