TopCoder Problem : Ameoba Division 2 Solution (C#) SRM 615


The Ameoba that eats :)

This one is a Division 2 250 the easiest from SRM 615. Here's the statement
Problem Statement
    
Monte-Carlo is an amoeba. Amoebas can feed on gel: whenever an amoeba encounters a piece of gel that is exactly as big as the amoeba, the amoeba will consume the gel and thus double its size.  Initially, the size of Monte-Carlo was A. During its lifetime, Monte-Carlo encountered several gels and consumed the ones it could.  You are given a int[] X and the int A. The elements of X are the sizes of gels Monte-Carlo encountered, in chronological order. Compute and return the final size of Monte-Carlo.
Definition
    
Class: AmebaDiv2
Method: simulate
Parameters: int[], int
Returns: int
Method signature: int simulate(int[] X, int A) (be sure your method is public)

Limits
Time Limit (s):2.000
Memory limit (MB):256
Constraints
-
X will contain between 1 and 200 integers, inclusive.
-
Each element of X will be between 1 and 1,000,000,000, inclusive.
-
A will be between 1 and 1,000,000,000, inclusive.

Examples
0) {2,1,3,1,2}
1
Returns: 4
Gel #0 is bigger than Monte-Carlo, nothing happens.
Monte-Carlo consumes gel #1. Its size is now 1+1 = 2.
Gel #2 is bigger than Monte-Carlo, nothing happens.
Gel #3 is smaller than Monte-Carlo, nothing happens.
Monte-Carlo consumes gel #4. Its size is now 2+2 = 4.

1){1,4,9,16,25,36,49}
10
Returns: 10
The size of Monte-Carlo doesn't change.

2){1,2,4,8,16,32,64,128,256,1024,2048}
1
Returns: 512

3){817,832,817,832,126,817,63,63,126,817,832,287,823,817,574}
63
Returns: 252

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And the Logic

All we have to do is to putup a loop that will go upto length(X) and whenever a term in X(food) is as big as size(Ameoba), add it to size of ameoba.

Then the code

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