John and GCD list Hacker Rank solution in C

Problem:

https://www.hackerrank.com/challenges/john-and-gcd-list/problem

Explanation:

GCD(B[1],B[2]) =A[1] that is A[1] divides B[2]

GCD(B[2],B[3]) =A[2] that is A[2] also divides B[2]

so B[2] must be divisible by both A[1] and A[2].

as LCM of A[1],A[2] will be divisible by both A[1] and A[2].

Therefore B[2] must be a multiple of LCM(A[1],A[2])

Since in the question, they wanted minimum value of B[2]

So, B[2] = LCM(A[1],A[2])

#include<stdio.h>

int gcd(int,int);

int lcm(int,int);

int main()

{

    int t;

    scanf("%d",&t);

    while(t--)

    {

        int n,i;

        scanf("%d",&n);

        int a[n];

        for(i=0 ; i<n ; i++)

            scanf("%d",&a[i]);

        printf("%d ",a[0]);

        for(i=0 ; i<n-1 ; i++)

        {

            if(a[i+1] >a[i])

                printf("%d ",lcm(a[i+1],a[i]));

            else

                printf("%d ",lcm(a[i],a[i+1]));

        }

        printf("%d\n",a[n-1]);        

    }

    return 0;

}

int gcd(int a, int b)

{

    if (b == 0)

        return a;

    return gcd(b, a % b);

}

int lcm(int a, int b)

{

    return (a*b)/gcd(a,b);

}