C solution

int t;
scanf("%d", &t);

while(t--){
    int a_count;
    scanf("%d", &a_count);
    int a[a_count];
    for(int j = 0 ; j < a_count ; j++){
        scanf("%d", &a[j]);
    }
    
    int gcd[a_count+1];
    int result_count = 0;
    
    gcd[result_count++] = a[0];
    for(int i = 0 ; i < a_count-1 ; i++){
        int y = a[i];
        while(y % a[i+1] != 0 || (y % a[i] != 0 && a[i] % y != 0)){
            y++;
        }
        gcd[result_count++] = y;
    }
    gcd[result_count++] = a[a_count-1];
    
    for(int i = 0 ; i <= a_count ; i++){
        printf("%d ", gcd[i]);
    }
    printf("\n");
}   

1. def gcd(a,b):

   if b==0 :

   return a
   

   • return gcd(b,a%b)![https://)

def ppcm(a,b): return a*b//gcd(a,b)

def solve(a): # Write your code here L=[a[0]] for i in range(len(a)-1): L.append(ppcm(a[i],a[i+1])) L.append(a[-1]) return L

include

int gcd(int,int); int lcm(int,int); 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); } int main() { int t; scanf("%d",&t); while(t--) { int n,i; scanf("%d",&n); int a[n]; for(i=0 ; ia[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; }

C# - the only problem, calculate min Multiple for 2 numbers

private static int GCD(int a, int b)
{
    return a == 0 ? b : GCD(b % a, a);
}
private static List<int> CalcNext(ref int prev, List<int> lst, int cur)
{
    int gcd = GCD(prev,cur);
    lst.Add((prev/gcd)*(cur/gcd)*gcd);
    prev = cur;
    return lst;
}
public static List<int> solve(List<int> a)
{
    int prev = a.ElementAt(0), prevB = prev;
    List<int> ret = a.Skip(1).Aggregate(new List<int>{prev}, (lst,cur)=>CalcNext(ref prev, lst, cur)).ToList();
    ret.Add(a.Last());
    return ret;
}

...not the shortest code though

Some hint:

Suppose the input is [A,B,C,,...]

The first value of the list should be one number X such that : A = gcd(X,??) In other words: X should be multiple of A.

Because the sum have to be minimal, lets try the lowest multiple of A (X = A).

From here, whats the second number Y of the list? We have two restrictions of Y: A = gcd(X,Y) B = gcd(Y, ??) In other words Y should be a multiple of A and a multiple of B. Again, because the sum have to be minimal, get the lowest multiple of A and B ( Least Common Multiple)

Iterate from there...