简单数学(一）

GCD LCM

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

Sample Input

2

1 2

3 4

Sample Output

1 2

-1

//最小公倍数和最大公约数

#include<cstdio>
int main(){
	int T;
	scanf("%d",&T);
	int g,l;
	while(T--){
		scanf("%d%d",&g,&l);
		if(l%g==0)
		printf("%d %d\n",g,l);
		else printf("-1\n");
	}
}

Benefit

Recently Yaghoub is playing a new trick to sell some more. When somebody gives him A Tomans, he who never has appropriate changes, asks for B Tomans such that lowest common multiple of A and B equals to C and he will pay back a round bill. Or otherwise take some snack instead of the remaining of his money. He believes that finding such a number is hard enough that dissuades students from paying that.

You should write a program that help poor students giving the appropriate amount of money to Yaghoub. Of course if there are several answers you go for students’ benefit which is the lowest of them.

Input

The first line begin with an integer T (T ≤ 100000), the number of tests. Each test that comes in a separate line contains two integers A and C (1 ≤ A, C ≤ 107 ).

Output

Print the lowest integer B such that LCM(A, B) = C in a single line. If no such integer exists, print ‘NO SOLUTION’ instead. (Quotes for clarity)

Sample Input

3

2 6

32 1760

7 16

Sample Output

3

55

NO SOLUTION

a*b=gcd(a,b)*lcm(a,b);

令b1=lcm/a

不断枚举 得到b=b1*gcd(a,b);

#include<cstdio>
#include<cstring>
int gcd(int a,int b){
	if(b==0)
	return a;
	return gcd(b,a%b);
}
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		int a,c;
		scanf("%d%d",&a,&c);
		if(c%a!=0)
		printf("NO SOLUTION\n");
		else{
			int b,g;
			b=c/a;
			g=gcd(a,b);
			while(g!=1){
				b*=g;
				a/=g;
				g=gcd(a,b);
			}
			printf("%d\n",b);
		}
	}
}