House Numbering (KICTM UiTM Jasin 2015)

The government of Acmonia has decided that henceforth all house numbers should be given in binary instead of decimal notation.  Householders will now have to purchase 0 and 1 binary digits to display on their houses.  For reasons much too complicated to discuss here it seems that the cost to a householder of a 0 binary digit and of a 1 binary digit may well differ.  Your task is to write a program which will report to householders the cost of their new numbers.

Input 
The input text consists number of sets of problems. The first line of a set is of the form “COST a b”.
For that set:

• a and b are both integers, 0 ≤ a, b ≤ 1000 
• a 0 binary digit costs a dollars
• a 1 binary digit costs b dollars. 

The first line is followed by one or more lines each consisting of a single integer n.

• 0 ≤ n ≤ 2,000,000
• n indicates a house number, expressed as a standard decimal number. 

A single # on a line indicates the end of input.

Output
Each set of output data must begin with a single output line showing consisting of the word “Set”, followed by a space (“ ”), and the current set number (counted from 1).  This is followed by the cost of the binary digits for each house number, each cost being displayed as a decimal number on a separate line.

Samples Input
COST 1 1  
1 
34 
15 
COST 1 10
1 
34 
15 
COST 10 1
1
34
15
COST 0 5
1
16
# 

Samples Output
Set 1 
1 
6 
4 
Set 2 
10 
24 
40 
Set 3 
1 
42 
4 
Set 4 
5 
5 

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39

import java.util.*; public class Q4 { public static void main(String[] args) { Scanner scan = new Scanner(System.in); String line = System.getProperty("line.separator"); scan.useDelimiter(line); String ans = ""; int a = 0, b = 0, no = 0, ct = 1; String op = scan.next(); while (!op.equalsIgnoreCase("#")) { if (op.contains("COST")) { ans = ans + "Set " + ct + "\n"; String get[] = op.split(" "); a = Integer.parseInt(get[1]); b = Integer.parseInt(get[2]); ct++; } else { no = Integer.parseInt(op); String c = Integer.toBinaryString(no); int z = 0, s = 0; for (int x = 0; x < c.length(); x++) { if (c.charAt(x) == '1') s++; else z++; } int total = (z * a) + (s * b); ans = ans + total + "\n"; } op = scan.next(); } System.out.println(ans); } }