C# – CodeForces Round 368 – Problem 1

Posted on September 12, 2016 by

I have not been blogging for a while, because I got a little deeper into reading the SQL books, which are waiting for review and thus I have decided to blog about something today. Initially, the idea was to write about SELECT INTO or something like it, but I have considered that the web is full with better samples of SQL code so far. Thus, I have taken a look at the last CodeForces competition and I somehow noticed, that the first problem is unbelievably easy somehow. I have decided to solve it and within some 2 minutes I had a running answer. Thus, I have decided to dedicate an article about it 🙂

So, this is the problem:

Small, but very brave, mouse Brain was not accepted to summer school of young villains. He was upset and decided to postpone his plans of taking over the world, but to become a photographer instead.

As you may know, the coolest photos are on the film (because you can specify the hashtag #film for such).

Brain took a lot of colourful pictures on colored and black-and-white film. Then he developed and translated it into a digital form. But now, color and black-and-white photos are in one folder, and to sort them, one needs to spend more than one hour!

As soon as Brain is a photographer not programmer now, he asks you to help him determine for a single photo whether it is colored or black-and-white.

Photo can be represented as a matrix sized n × m, and each element of the matrix stores a symbol indicating corresponding pixel color. There are only 6 colors:

‘C’ (cyan)
‘M’ (magenta)
‘Y’ (yellow)
‘W’ (white)
‘G’ (grey)
‘B’ (black)

The photo is considered black-and-white if it has only white, black and grey pixels in it. If there are any of cyan, magenta or yellow pixels in the photo then it is considered colored.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 100) — the number of photo pixel matrix rows and columns respectively.

Then n lines describing matrix rows follow. Each of them contains m space-separated characters describing colors of pixels in a row. Each character in the line is one of the ‘C’, ‘M’, ‘Y’, ‘W’, ‘G’ or ‘B’.

Output

Print the “#Black&White” (without quotes), if the photo is black-and-white and “#Color” (without quotes), if it is colored, in the only line.

Examples

input

2 2
C M
Y Y

2 2

C M

Y Y

output

#Color

#Color

input

3 2
W W
W W
B B

3 2

W W

B B

output

#Black&amp;White

1	#Black&White

input

1 1
W

1 1

output

#Black&amp;White

1	#Black&White

My solution was just to read the first string of the input, then to put a for loop, reading the next and to put everything in one string. Then it is easy to say whether you have cyan, magenta or yellow at all, just by comparing with -1. Honestly, I did not believe that it is that easy, but it was. Thus, my code worked:

Here comes the code:

using System;

class check01
{
    static void Main()
    {
        string[] input = Console.ReadLine().Split();
        int i_row = int.Parse(input[0]);
        string str_result = "";

        for (int i = 0; i < i_row; i++)
        {
            str_result += Console.ReadLine();
        }

        if ((str_result.IndexOf('C') != -1) || (str_result.IndexOf('M') !=-1) || (str_result.IndexOf('Y') != -1))
        {
            Console.WriteLine("#Color");
        }
            else
        {
            Console.WriteLine("#Black&White");
        }
    }
}

using System;

class check01

{

static void Main()

{

string[] input = Console.ReadLine().Split();

int i_row = int.Parse(input[0]);

string str_result = "";

for (int i = 0; i < i_row; i++)

{

str_result += Console.ReadLine();

}

if ((str_result.IndexOf('C') != -1) || (str_result.IndexOf('M') !=-1) || (str_result.IndexOf('Y') != -1))

{

Console.WriteLine("#Color");

}

else

{

Console.WriteLine("#Black&White");

}

The funny part is that in C# I do not even need to care for the number of columns. In this tiny example C# produces better results than C++ and C 🙂

Enjoy it!

Tagged with: C#, CodeForces Round 368

C# – CodeForces Round 368 – Problem 1

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