poj 1894 Alternative Scale of Notation

Alternative Scale of Notation

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions:2182		Accepted: 808

Description

One may define a map of strings over an alphabet Σ_B = { C1, C2, . . . C_B } of size B to non-negative integer numbers, using characters as digits C1 = 0, C2 = 1, . . . , C_B = B - 1 and interpreting the string as the representation of some number in a scale of notation with base B. Let us denote this map by U_B , for a string α[ 1...n ] of length n we put 
U_B(α)=Σ_0<=i<=n-1α[n-i]*Bⁱ
For example, U3(1001) = 1*27 + 0*9 + 0*3 + 1*1 = 28. 

However, this correspondence has one major drawback: it is not ont-to-one. For example, 
28 = U3(1001) = U3(01001) = U3(001001) = ... , 
infinitely many strings map to the number 28. 

In mathematical logic and computer science this may be unacceptable. To overcome this problem, the alternative interpretation is used. Let us interpret characters as digits, but in a slightly different way: C1 = 1, C2 = 2, . . . , C_B = B . Note that now we do not have 0 digit, but rather we have a rudiment B digit. Now we define the map V_B in a similar way, for each string α[ 1...n ] of length n we put 
V_B(α)=Σ_0<=i<=n-1α[n-i]*Bⁱ
For an empty string ε we put V_B(ε) = 0. 

This map looks very much like U_B , however, the set of digits is now different. So, for example, we have V3(1313) = 1*27 + 3*9 + 1*3 + 3*1 = 60. 

It can be easily proved that the correspondence defined by this map is one-to-one and onto. Such a map is called bijective, and it is well known that every bijective map has an inverse. Your task in this problem is to compute the inverse for the map V_B . That is, for a given integer number x you have to find the string α, such that V_B(α) = x. 

Input

The first line contains B (2 <= B <= 9) and the second line contains an integer number x given in a usual decimal scale of notation, 0 <= x <= 10¹⁰⁰.

Output

Output in one line such string α, consisting only of digits from the set { 1, 2, . . . , B }, that V_B(α) = x . 

Sample Input

3
60

Sample Output

1313

Source

import java.io.*;
import java.math.BigInteger;
import java.nio.Buffer;
import java.util.*;
public class Main
{
    public static void main(String args[]) throws Exception
    {
        Scanner cin=new Scanner(System.in);
        BigInteger n=cin.nextBigInteger();
        BigInteger a=cin.nextBigInteger();
        BigInteger x=BigInteger.valueOf(0);
        String s="";
        while (a.compareTo(BigInteger.valueOf(0))>0){
            if(a.remainder(n).equals(BigInteger.valueOf(0))){
                s+=n.toString();
                a=a.subtract(n);
                a=a.divide(n);
            }
            else{
                s+=a.remainder(n).toString();
                a=a.divide(n);
            }

        }

        s = new StringBuffer(s).reverse().toString();
        System.out.print(s);

    }
}