UVA Problem 12646 – Zero or One Solution

UVA Problem 12646 – Zero or One Solution:


Click here to go to this problem in uva Online Judge.

Solving Technique:

The input is all possible combination of 3 bits. If A, B, C are thought as bits then possible binary combination are,

000 \rightarrow * \\  001 \rightarrow C \\  010 \rightarrow B \\  011 \rightarrow A \\  ---- \\  100 \rightarrow A \\  101 \rightarrow B \\  110 \rightarrow C \\  111\rightarrow * \\

This problem can be solved in multiple ways such as using string, checking equality, performing xor equality etc. My code below is exactly same as equality checking but it perform xor operation between them.

XOR Table,

\begin{tabular}{l*{6}{c}r}  X & Y & X XOR Y \\ \hline  0 & 0 & 0 \\  0 & 1 & 1 \\  1 & 0 & 1 \\  1 & 1 & 0 \\  \end{tabular}
 

Important:  Be sure to add or print a new line after each output unless otherwise specified. The outputs should match exactly because sometimes even a space character causes the answer to be marked as wrong answer. Please compile with c++ compiler as some of my codes are in c and some in c++.


More Inputs of This Problem on uDebug.


Input:

1 1 0
0 0 0
1 0 0

 


Output:

C
*
A

Code:

/**
 * Author:    Asif Ahmed
 * Site:      https://quickgrid.wordpress.com
 * Problem:   UVA 12646 - zero or one
 * Technique: XOR Bits to check equality.
 */

#include<stdio.h>


int main(){

    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);


    int A, B, C;
    while( scanf("%d%d%d", &A, &B, &C) == 3 ){

        if( !( A ^ B ) && !( A ^ C ) )
            putchar('*');
        else{
            if( A ^ B ){
                if( B ^ C )
                    putchar('B');
                else
                    putchar('A');
            }
            else
                putchar('C');
        }
        putchar('\n');
    }


    return 0;
}
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