I'm trying to create a binary to decimal converter, and have got stuck on the code. I have researched forums for any help, but they all seam to use functions, which can not be used within a private sub. Please can anyone give me help on a solution to this problem?
I would use the positional notation method:
http://en.wikipedia.org/wiki/Positional_notation
http://www.wikihow.com/Convert-from-Binary-to-Decimal
So basically, without giving you the answer, you want to loop through binary place holders, filling up a variable as you go along. You would use an index to move from the least significant placeholder to the most.
For example : 10011011 in binary is 155 decimal.
So every place holder is a power with a base of two. Then you add the value for each one until your finished, like so:
placeholder 1 is: 2 pow 0 equals 1.
placeholder 2 is: 2 pow 1 equals 2.
placeholder 3 is: 2 pow 2 equals 4.
placeholder 4 is: 2 pow 3 equals 8.
placeholder 5 is: 2 pow 4 equals 16.
placeholder 6 is: 2 pow 5 equals 32.
placeholder 7 is: 2 pow 6 equals 64.
placeholder 8 is: 2 pow 7 equals 128.
Now we only add for the placeholders that have 1s.
128+16+8+2+1 = 155
What you will need:
A loop looping through indexes, and incrementing the exponent value as you go along, only adding the value if the index equals 1 in the binary number.
Hope my explanation makes sense. Good luck.
Related
I have read this: https://www.topcoder.com/community/competitive-programming/tutorials/binary-search.
I can't understand some parts==>
What we can call the main theorem states that binary search can be
used if and only if for all x in S, p(x) implies p(y) for all y > x.
This property is what we use when we discard the second half of the
search space. It is equivalent to saying that ¬p(x) implies ¬p(y) for
all y < x (the symbol ¬ denotes the logical not operator), which is
what we use when we discard the first half of the search space.
But I think this condition does not hold when we want to find an element(checking for equality only) in an array and this condition only holds when we're trying to find Inequality for example when we're searching for an element greater or equal to our target value.
Example: We are finding 5 in this array.
indexes=0 1 2 3 4 5 6 7 8
1 3 4 4 5 6 7 8 9
we define p(x)=>
if(a[x]==5) return true else return false
step one=>middle index = 8+1/2 = 9/2 = 4 ==> a[4]=5
and p(x) is correct for this and from the main theory, the result is that
p(x+1) ........ p(n) is true but its not.
So what is the problem?
We CAN use that theorem when looking for an exact value, because we
only use it when discarding one half. If we are looking for say 5,
and we find say 6 in the middle, the we can discard the upper half,
because we now know (due to the theorem) that all items in there are > 5
Also notice, that if we have a sorted sequence, and want to find any element
that satisfies an inequality, looking at the end elements is enough.
Could you please help me to understand this problem:
Convert the input sequence of N (1 ≤ N ≤ 20) input numbers so that
the subsequences of the same numbers are replaced with the first
numbers of the subsequences. Each input number is in the range [1, 2
000 000 000].
For example, the input sequence 1 2 2 3 1 1 1 4 4 is converted into
1 2 3 1 4.
Input: First, the number T of test cases is given. Each test case is
specified using two lines. The first one contains the number N and the
second one contains the numbers of the sequence.
Output: The converted sequence. The result for each test case should
be printed in a separate line.
For example, the input sequence 1 2 2 3 1 1 1 4 4 is converted into 1 2 3 1 4.
It looks like the idea is to remove duplicate numbers that occur adjacent to each other when creating the output.
You can do that by just keeping a state variable recording what the previous value was. When you get a new value, compare it to the state value. If it's the same, skip. If different, output it and update the state variable. Remember to initialize the state variable to a value not found in the input stream (e.g. -1 should work in this case).
I'm working on a homework assignment that asks us to create a type of Units class that can keep track of units and perform basic arithmetic on them. The problem description has this bit, which I don't completely understand:
Probably the easiest way to keep track of the units is to give Units a dictionary that maps symbols to integers. If you are dividing by a unit then it has a negative value in the dictionary. You add two Units together by adding the value together for each symbol in the dictionary. When it is zero, throw the symbol away!
For reference, this is also included in the description:
[...] you could write an expression 3 elephants / (1 sec sec) and it would return the right thing.
Could someone shed some light here? How can I use a dictionary to map these types of units? Am I making this way harder than it needs to be?
It sounds like your teacher is giving you a hint about how to wind up with the proper units at the end of the calculation.
When you're parsing the problem, as you encounter items that are obviously units, enter them into a dictionary. The dictionary would consist of a number and a string (the supposed "unit"). Then you'd use a set of rules to increase or decrease the integer count. The resultant integer value would help you to output the units correctly.
A count of 1 indicates it's a unit in the output.
A count of -1 indicates it's inverse is a unit in the output.
A count of 0 indicates that it doesn't appear in the output at all.
Similarly, a count of 2 would indicate that it's square appears as a unit in the output.
To wit:
5 Hippo + 10 Hippo = 15 Hippos
Parsing: Dictionary:
-------- -----------
5 Hippo Hippo:1
+
10 Hippo Hippo:1 (previous operation was addition or subtraction, and already have Hippo in dictionary
But consider this problem:
5 Hippo * 5 sec/Hippo = 25 sec
Parsing: Dictionary:
5 Hippo Hippo:1
*
5 sec Hippo:1, sec:1
/
Hippo Hippp:0, sec:1 (previous operation was division of Hippo, so decrement Hippo count)
Or perhaps:
10 feet / 5 sec = 2 feet/sec
Parsing: Dictionary:
10 feet feet:1
/
5 sec feet:1, sec:-1 (divided by sec, and second is not in dictionary, so second implicitly = 0. 0 + (-1) = -1.
In the example above, feet will be on the top of the bar because it's equal to 1, and sec will be below the bar because it's value is -1. If it's value had been -2, it would have been (feet/(sec*sec) or feet/(sec squared).
I'm writing a task in pascal.
Everything is ok, just my result is not right.
I'm summing some numbers
Example: 2.3 + 3.4+ 3.3 = 9
But output shows: 9.000000 + EEE or something like that.
So- how to convert, to be only 9, not this REAL variable.
To actually convert:
var
i: integer;
...
i := round(floatVar);
To output only the integer part:
writeln(floatVar:9:0);
Let's consider this quite simpler equation:
3.5 + 2.5
What do you expect? 6, right? Let's try this code
write(3.5 + 2.5);
Unfortunately, it's a floating-point number, so it would produce a number represented in a scientific way:
6.00000000000E+00
or 6.0000000000 x 100, or 6 x 10o. Whatever, you only care about 6, who need this weird useless long number? So the idea is to cut off the decimal part and output to the console only the integer part, which can be done with this line of code:
write(3.5 + 2.5 : 0 : 0);
Ok, now it outputs a beautiful number as expected
6
Seems like the problem is solved, but you say that:
I'm summing some numbers
Example: 2.3 + 3.4+ 3.3 = 9
Ohh so that the evenly, beautiful integer is just randomly appeared? Here the problem comes, how do you expect this equation would output?
3.6 + 2.5
It should be 6.1, right? Let's try it with the worked line of code:
write(3.6 + 2.5 : 0 : 0);
And the output is...
6
Unexpected, right? So how about rounding to some decimal places, like 1?
write(3.5 + 2.5 : 0 : 1);
write(3.6 + 2.5 : 0 : 1);
Then, 3.5 + 2.5 = 6.0 and 3.6 + 2.5 = 6.1. But 6.0 may look quite long, so how to make it output 6 for 6.0 and 6.1 for 6.1?
Actually, you can't make the program auto-detect if a real variable contains an integer value because the way a real var is stored is completely different from an integer var (how different they are, please contact Google; but you can do it manually by making a function to do the job).
So my solution is, to be easy, making the output rounded to some decimal places, and that's it.
For purpose of showing pretty output on the screen you can use something like this:
Writeln(result:0:2);
Result on screen would be this:
9.00
What this means someone would ask? Well first number 0 means how wide filed is. if you say it's 0 then Pascal writes it at the very left side of screen. If you said writeln(result:5:2) result would be:
9.00
In other words i would print form the right side and leave 5 chars to do so.
Second number 2, in this example means you want that result printed with 2 decimal places. You can place it only if you want to print on screen value that is real, single, double, extended and so on.You can round to any number of decimals, and if you do writeln(result:0:0) you would get ouput:
9
If you are printing integer and want to have some length of field, lets sat 5 you would do: writeln(int:5). If you added :2 to the end you would get compile time error.
This all also works for something like this: writeln(5/3.5+sqrt(3):0:3),
You should know that this does not round variable itself but just formats output. This is also legal:
program test;
var
a:real;
n,m:integer;
begin
readln(a,m,n);
writeln(a:m:n);
end.
What i did here is i asked user if on how many decimals and with what length of field he wants to write entered number a. This can be useful so i'm pointing it out. Thank you for reading. I hope i helped
You can convert to string, get the int part, e convert to int number!
Or Float to Str than Str to Int:
nPage := StrToInt(FloatToStr(Int(nReg / nTPages))) + 1;
I have a DFA question (Determinant Finite Automata) . We are using JFLAP to construct the automata. I cannot figure this question out to save my life! Here it is
"DFA to recognize the language of all strings that have an even number of zeros and an odd number of ones."
So the alphabet is {0,1} and only using 0,1. So I need to build an automata that recognizes an even number of zeros and an odd number of ones.
I don't know whether my understanding is right.
I could give you the description in Grail format that generate an even number of zeros and an odd number of ones.
START 1
1 1 2
2 1 1
1 0 3
3 0 4
4 0 3
FINAL 3