VerySecureEncryption

Simply do what the statement says.

public class VerySecureEncryption{
  public String encrypt(String message, int[] key, int K){
    char[] current = message.toCharArray();
    char[] next = new char[message.length()];
    for(int times = 0; times < K; times++ ){
      for(int i = 0; i < key.length; i++ ){
        next[ key[i] ] = current[i];
      }
      System.arraycopy(next, 0, current, 0, next.length);
    }
    return String.valueOf(current);
  }
}

IsItASquare

Source

Try all permutations of the points (a total of $24$ ) and compare all distances. The diagonals should also be taken into account to make sure we found a square and not a rhombus.

public class IsItASquare{
  public int distSq(int x1, int y1, int x2, int y2){
    int dx = x2 - x1;
    int dy = y2 - y1;
    return dx * dx + dy * dy;
  }

  public String isSquare(int[] x, int[] y){
    for(int a = 0; a < 4; a++){
      for(int b = 0; b < 4 && (b != a); b++){
        for(int c = 0; c < 4 && (c != a) && (c != b); c++){
          int d = 0;
          while( (d == a) || (d == b) || (d == c) )
            d++;
          int ab = distSq( x[a], y[a], x[b], y[b] );
          int bc = distSq( x[b], y[b], x[c], y[c] );
          int cd = distSq( x[c], y[c], x[d], y[d] );
          int da = distSq( x[d], y[d], x[a], y[a] );
          int ac = distSq( x[a], y[a], x[c], y[c] ) ;
          int bd = distSq( x[b], y[b], x[d], y[d] ) ;

          if( (ab == bc) && (bc == cd) && (cd == da) && (ac == bd) )
            return "It's a square";
        }
      }
    }
    return "Not a square";
  }
}

GCD Table

Source

Let $m$ be the largest element in the table, since $gcd(x, x) = x$ , we can safely extract the $m$ from the table and add it to the answer. Now, let $n$ be the second largest element in the table. We can be sure that it belongs to the answer, however, we now must also remove $gcd(m,n)$ and $gcd(n,m)$ . Repeat this procedure until the original array is complete.

from fractions import gcd

n = int(input())
a = sorted(list(map(int, input().split())))
count = {}
for i in a:
    count[i] = count.get(i, 0) + 1

ans = []
i = n-1
while(len(ans) < n):
    top = a.pop()
    while(count[top] < 1):
        top = a.pop()
    count[top] -= 1
    ans.append(top)
    for i in range(len(ans)):
        count[gcd(ans[-1], ans[i])] -= 2
print(' '.join(str(x) for x in ans))

Conclusion

Never forget to copy your arrays instead of just assigning references to them!.