K6A editorial

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PROBLEM LINK:

Practice

Contest

Author: Chandan Boruah

Tester: Chandan Boruah

Editorialist: Chandan Boruah

DIFFICULTY:

EASY

PREREQUISITES:

Basic Maths

PROBLEM:

Given an array where elements can take only 2 values with some elements fixed, what is the number of ways in which the array can exist.

QUICK EXPLANATION:

Print 2^(number of elements not fixed). If all elements are fixed, which is a special case, print 1.

EXPLANATION:

The number of ways in which the elements can exist equals to 2^(number of elements that aren’t fixed), since there are only 2 possible values of each element. The problem was intended to confuse with bitmasks and also has a special case.

AUTHOR’S SOLUTION

using System;
class some
{
	public static void Main()
	{
		int t=int.Parse(Console.ReadLine());
		for(int l=0;l<t;l++)
		{
			int c=int.Parse(Console.ReadLine());
			string[]ss=Console.ReadLine().Split();
			int cc=0;
			foreach(string kk in ss)if(kk=="*")cc++;
			if(cc==0)Console.WriteLine(1);
			else
			Console.WriteLine(Math.Pow(2,cc));
		}
	}
}