CODEWITHSUNDEEP
c++
CocoHot
CocoHot

Reputation: 1226

UVA 11959 Dice wrong answer

So I was trying to solve uva 11959 Dice. But the third example the question gave give me the wrong answer. Then I found out if I change

cin >> d1 >> d2;

to 

scanf("%lx %lx", &d1, &d2);

and it works but I don't know why. However when I submit my code it shows that my answer is wrong. Can anyone help me with both questions?

My code:

#include <iostream>
#include <cstdio>
using namespace std;

//rotate and flip the dice
#define R(d) ( ((d) & 0x000F00) << 12 | ((d) & 0x00000F) << 16 | ((d) & 0x00F0F0)
  | ((d) & 0x0F0000) >> 8 | ((d) & 0xF00000) >> 20  )
#define F(d) (((d) & 0x000F0F) | ((d) & 0x00F000) << 8 | ((d) & 0x0000F0) << 12 
  | ((d) & 0x0F0000) >> 4 | ((d) & 0xF00000) >> 16)

bool E(long d1, long d2)
{
    return ( (d1) == (d2) || (d1) == R(d2) || (d1) == R(R(d2)) || (d1) == R(R(R(d2))) );
}


int main()
{
    long d1, d2;
    long counter;
    bool equal;
    cin >> counter; 
    for( int i = 0 ; i < counter ; i++ )
    {
        scanf("%lx %lx", &d1, &d2);
        //  cin >> d1 >> d2;
        equal =  E(d1, d2) || E(d1, F(d2)) || E(d1, F(F(d2))) 
             || E(d1,F(F(F(d2)))) ;

    if(equal)
        cout << "Equal";
    else
        cout << "Not Equal";
    cout << endl;
    }

    return 0;
}

ADD: I found out in my for loop I need to add two more condition to determine "equal". So it should looks like this:

equal = E(d1, d2) || E(d1, F(d2)) || E(d1, F(F(d2))) || E(d1, F(F(F(d2)))) || E(d1, F(R(d2))) || E(d1, F(R(R(R(d2)))));

But I still dont know why I need to add these two conditions. Isn't it already been covered?

Upvotes: 1

Views: 282

Answers (2)

Joel Lee
Joel Lee

Reputation: 3844

You need the two additional conditions because a six-sided die (cube) can be oriented one of 24 different ways. Your original algorithm was only checking 16 orientations (4 positions in main times 4 positions in E = 16) , and would therefore sometimes fail to detect equality. With the two new conditions you have 6 positions in main times 4 positions in E, for a total of 24. Note, however, that it isn't immediately obvious that the 24 orientations you generate are all distinct. (I think they probably are, but I didn't take the time to verify it.)

Upvotes: 2

Jerry Coffin
Jerry Coffin

Reputation: 490048

cin >> d1 >> d2; reads the input as decimal. scanf("%lx %lx", &d1, &d2); reads the input as hexadecimal.

If you want to read the input as hexadecimal with iostream, you probably want to use the std::hex manipulator, something like: cin >> hex >> d1 >> d2;

Note that the hex bit is "sticky", so it'll remain set until you specify another base. That means the one hex affects both when you read d1 and d2 above -- but it also means if you read something else afterwards, it'll still read it as hex.

Upvotes: 1

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