text: uva10198,
  
  # Problem E: Counting
  
  ## The Problem
  
  Gustavo knows how to count, but he is now learning how write numbers. As he is a very good student, he already learned 1, 2, 3 and 4. But he didn't realize yet that 4 is different than 1, so he thinks that 4 is another way to write 1. Besides that, he is having fun with a little game he created himself: he make numbers (with those four digits) and sum their values. For instance:
  ```132 = 1 + 3 + 2 = 6
  112314 = 1 + 1 + 2 + 3 + 1 + 1 = 9 (remember that Gustavo thinks that 4 = 1)```
  After making a lot of numbers in this way, Gustavo now wants to know how much numbers he can create such that their sum is a number n. For instance, for n = 2 he noticed that he can make 5 numbers: 11, 14, 41, 44 and 2 (he knows how to count them up, but he doesn't know how to write five). However, he can't figure it out for n greater than 2. So, he asked you to help him.
  
  ## The Input
  
  Input will consist on an arbitrary number of sets. Each set will consist on an integer n such that 1 <= n <= 1000. You must read until you reach the end of file.
  
  ## The Output
  
  For each number read, you must output another number (on a line alone) stating how much numbers Gustavo can make such that the sum of their digits is equal to the given number.
  
  ```2
  3```
  
  ```5
  13```
  
  ## Solution:
  
  Deal with big number is much difficulty if using C++.
  The code is copied from Blog Link.
  The copyright is reserved by the owner of that blog.
  
  ## Source Code:
  
  //Wed Mar 30 12:31:14 CDT 2011
  import java.math.BigInteger;
  import java.util.Arrays;
  import java.util.Scanner;
  
  public class Main {
  
  public static void main(String[] args) {
  
  Scanner input = new Scanner(System.in);
  
  BigInteger dp[] = new BigInteger[1000 + 10];
  
  Arrays.fill(dp, BigInteger.ZERO);
  
  dp[0] = BigInteger.valueOf(1);
  dp[1] = BigInteger.valueOf(2);
  dp[2] = BigInteger.valueOf(5);
  dp[3] = BigInteger.valueOf(13);
  
  for (int i = 4; i <= 1000; i++) {
  dp[i] = dp[i].add(dp[i - 1]);
  dp[i] = dp[i].add(dp[i - 2]);
  dp[i] = dp[i].add(dp[i - 3]);
  dp[i] = dp[i].add(dp[i - 1]);
  }
  
  int n;
  
  while (input.hasNextInt()) {
  n = input.nextInt();
  System.out.println(dp[n]);
  }
  }
  }