407. Exchange

 

On Christmas, in the evening, there were three flowers in the window, from left to right: geranium, crocus and violet. Every morning Masha cleans the dust and changes position of the right flower with central one. In the afternoon, Tanya waters the flowers and swaps left and central flowers. Print the order of the flowers after k days, at night.

 

Input. The first line contains a number of test cases m (1 ≤ m ≤ 12). Each of the following m lines contains the number of days k (1 ≤ k ≤ 1000).

 

Output. Print for each test case in a separate line three Latin letters ‘G’, ‘C’ and ‘V’ (capital letters, no spaces), representing the order of flowers in k days (from left to right). Here G stands for geranium, C for crocus and V for violet.

 

Sample input

Sample output

2

1

5

VGC

CVG

 

 

SOLUTION

conditional statement

 

Algorithm analysis

Initially the flowers on the window have the order GCV (geranium, crocus and violet). After the first day the order of the flowers will be changed to VGC (violet, geranium and crocus). After the second day the flowers will be in the order CVG (crocus, violet and geranium). At the end of the third day the order of the flowers will be the same as it was at the start of the first day. Then the sequence of flowers permutations will be repeated. Thus the answer to the problem will be as follows:

·        If k is divisible by 3, the flowers will be in the order GCV.

·        If the remainder of the division of k by 3 equals to 1, the flowers will be in the order VGC.

·        If the remainder of the division of k by 3 equals to 2, the flowers will be in the order CVG.

 

Algorithm realization

Read input data and print the sequence of flowers depending on the remainder from division of number of the days k by 3.

 

scanf("%d",&n);

while(n--)

{

  scanf("%d",&a);

  if (a % 3 == 0) puts("GCV"); else

  if (a % 3 == 1) puts("VGC"); else puts("CVG");

}

 

Java realization

 

import java.util.*;

 

public class Main

{

  public static void main(String[] args)

  {

    Scanner con = new Scanner(System.in);

    int n = con.nextInt();

    while(n-- > 0)

    {

      int a = con.nextInt();

      if (a % 3 == 0) System.out.println("GCV"); else

      if (a % 3 == 1) System.out.println("VGC"); else

                      System.out.println("CVG");

    }

    con.close();

  }

}