I'm working on an assignment in Visual Studio Express 2012, and the challenge asks me to create an app that allows a user to enter three runners' names and finishing times. I'm working on the flowchart for the If...Then...Else portion for the first, second and third place winners. Can anyone help me or steer me in the right direction for creating the conditional statement portion?
Here's what I have:
If intRunner1 < intRunner2 And intRunner1 < intRunner3 Then
lblDisplay.Text = "First Place"
End If
If intRunner2 < intRunner1 And intRunner2 < intRunner3 Then
lblDisplay.Text = "First Place"
End If
If intRunner3 < intRunner1 And intRunner3 < intRunner2 Then
lblDisplay.Text = "First Place"
I also have this:
If intFinish1 < intFinish2 Then
If intFinish1 < intFinish3 Then
lblFirstPlace.Text = "Runner 1 finished in first place."
Else
lblSecondPlace.Text = "Runner 1 finished in second place."
Else
If intFinish1 < intFinish3 Then
lblFirstPlace.Text = "Runner 1 finished in second place."
Else
lblThirdPlace.Text = "Runner 1 finished in thirdplace"
End If
End If
If all data is input at the same time :
You should make 2d array of size equal to the number of runners. Insert the name of the runner in the first dimension, and runner time in the second. Then sort this array on the 2nd dimension to get winners. As the scope is very small and just for demonstration, you can just use a if then else condition to check for the sorting.
If you input the runner one at a time
Just insert the runners in the 2d array in the correct order and out put the array after all data has been inserted to get list of winners.
For example pseudocode for second method :
create 2d array A
repeat for each runner
input runner name
input runner time
if A has runners
check runner against each runner in A
insert runner at the correct position
else
insert runner in A
end repeat
If you are only using variables and not arrays : Let us assume u have runners A B & C
These are the possible scenarios : ABC, ACB, BAC, BCA, CAB, CBA
You have to check for each of them.
Take A
if A beats B (then he is definately second or first)
if A beats C (then he is definately first)
A is first
if B beats C
B is second, C is third
else
C is second, B is third
else
A is second
if B beats C
B is first, C is third
else
C is first, B is third
else ... if A beat C.... and so on
Project this to check for every possible scenario.
Related
This is a model solution code of the following problem;
"The input consists of T test cases. The first line of the input is given a T.
Each test case consists of three rows of integers separated by a single space, each consisting of three random points of x and y coordinates. The coordinates of the top left pixel in the browser viewport are (1, 1) and the coordinates of the bottom right pixel are (1000, 1000). All coordinates are located within the viewport and the positions of each point are different."
And here is the sample of the input.
2
5 5
5 7
7 5
30 20
10 10
10 20
7 7
30 10
The last two lines are the answers of the problem.
And here is my three questions.
1) What happened when we use cin statement in initialization?? It receives how many actions it will perform from the user in the initialization of for loop. I understand that this cin statement works properly. I cannot understand how this code knows how many times this for loop has to be repeated. This is because there is no action on T after initialization with a value of T from the user. There is no actrion in 'increment/decrement' also.
2) After googling, I understand when there is cin in condition, the loop ends when there is no more inputs or the inputs' type does not match the variables' type. But in this code, the for loop ends when the repeated time (T) is over. How could this happen???
3) Finally, the outcomes should be presented after all inputs are finished not by one-by-one. Then how could this for loop memorize the outcome of each set(3 inputs)??
I'm not English speaker T.T Thank you for reading my question.
#include<iostream>
int main()
{
int T,a,b,c,A,B,C;
for(std::cin>>T; std::cin>>a>>A>>b>>B>>c>>C; printf("%d %d\n",a^b^c,A^B^C));
}
What happened when we use cin statement in initialization??
That part of for() loop can contain any simple statement, not
just initialization statement. This statement is done only once. For
loop
for ( init condition ; iteration )
statement
is actually equivalent of this code:
{
init
while ( condition )
{
statement
iteration;
}
}
But in this code, the for loop ends when the repeated time (T) is
over. How could this happen???
The operator >> overloaded for streams return stream it acted on. Class
ios_base which is common parent of all streams, contains this
operator
std::ios_base::operator bool()
This operator is an equivalent of good() method. When >> fails to read and parse values from input stream, good() returns false,
loop breaks. T is not used in the provided code.
Then how could this for loop memorize the outcome of each set(3
inputs)??
It doesn't. It prints result after reading each set.
PS. People who read\proof-check code after programmer, would tend to have murderous intent toward those who write for() loops like that.
If we use a cin statement as our initialization, we execute this once. It will receive the input and place the value in the variable T. You are completely right; there is no action on T after initialization such as incrementing or decrementing its value.
This is not true. The code does not end when the it repeated T times. As long as (valid) input is given, this for-loop will continue. This is because the condition part of your for-loop consist of a cin statement. That is, as long as your cin statement succeeds, the for-loop will continue.
It cannot. Each time the loop runs, you overwrite the variables a, A, b, B, c and C. Hence, the old values are lost.
I am currently learning C++ at my school, and am making a word sleuth as part of a project that I have to submit. For this, I have already made the grid of alphabets and other necessary things (clues, rules, etc.). I am taking the input in the form of coordinates in an integer array whereby the user enters 4 values in the array, signifying the initial row and column number and the final row and column number, corresponding to which are the first and last alphabets of a particular word.
After doing this, I am now comparing the array input by the user with the array I have already defined that has the coordinates of that particular word. This is shown here :
cout<<"Enter the coordinates of starting and final characters : row1 col1 row2 col2 "<<endl;
for (z = 0; z < 4; z++) //first for loop
cin>>p[z]; //taking the input as an array 'p'
for (b = 0; b < 4; b++) //second for loop
{
if (p[b] == messi[b])
b+=0;
}
if (b == 4)
cout<<"Great!!!! You have answered the question correctly"<<"\n\n";
else
cout<<"You got this one wrong mate! Try again :)"<<"\n\n";
Here, messi[b] is the array which has the coordinates corresponding to the word 'MESSI' in the grid. Now, to my mind, the 'if' statement after the second for loop must contain the condition to check if b = 3. However, when I do that, the output always comes out to be what the 'else' statement says i.e. "You got this..." for every input. However, when I impose the condition to check if b = 4, the output comes out to be what the 'if' statement says i.e. "Great!!..." for every input.
What wrong am I doing? I hope I am clear enough in explaining the problem to you. I am using CodeBlocks 16.01.
It's a bit unclear what you are doing, as the program stands, b will always be equal to 4 after the second for-loop since the last time to condition was true, b < 4. So after the increment, it will be 4.
Inside the second for-loop you also have the NOP code b += 0; which does absolutely nothing to the code. What is the intention here?
How can I check if the value in cell #0 is equal to the value in cell #1? I am trying to write code equivalent to:
if(a == b)
{
//do stuff
}
else
{
//do something else
}
I have read Brainfuck compare 2 numbers as greater than or less than, and the second answer gave me a general idea of what I'd need to do, but I cannot figure it out. (That solution gives if a < b, else.)
I am thinking I need to do something along the lines of decrementing both values, and if they reach 0 at the same time, then they are true. But I keep getting stuck at the same exit point every time I think about it.
How can I check if two cells are equal in brainfuck?
I think I have it, I'm not a brainfuck expert but this question looked interesting. There might be a simpler way to do it, but I went with your method of decrementing values one by one.
In this case, if the two values in cell 0 and 1 are equal jump a ton forward, if they are not equal jump a little forward (second brackets is the not equal case, third brackets is the equal case)
Note that I'm using brainfucks while statements as a ghetto if (cell != 0)
+++++++++++++++++
>
+++++++++++++++++
>+<
[ - < - >] <[>>>>>] >> [>>>>>>>>>>>>>>>>>>>>>]
Try it online: http://fatiherikli.github.io/brainfuck-visualizer/#KysrKysrKysrKysrKysrKysKPgorKysrKysrKysrKysrKysrKwo+KzwKWyAtIDwgLSA+XSA8Wz4+Pj4+XSA+PiBbPj4+Pj4+Pj4+Pj4+Pj4+Pj4+Pj4+XQoKCg==
An example implementation, print T (true) if the two values are equal, F (false) if they are not equal
http://fatiherikli.github.io/brainfuck-visualizer/#KysrCj4KKysrKwo+KzwKWyAtIDwgLSA+XSA8Wz4+PgorKysrKysrKysrKysrKysrKysrKworKysrKysrKysrKysrKysrKysrKworKysrKysrKysrKysrKysrKysrKworKysrKysrKysrCi4KPgoKXSA+PiBbCisrKysrKysrKysrKysrKysrKysrCisrKysrKysrKysrKysrKysrKysrCisrKysrKysrKysrKysrKysrKysrCisrKysrKysrKysrKysrKysrKysrCisrKwouCj4KXQ==
+>>(a+++++)>(b+++++)>>+<<<
[[->]<<]
<
[>>>>>-<<<<<
a>b
]
>>
[->>-<
a<b
]
>>
[-
a=b
]
Pointer ends on the same pointer in the same state but the code within the appropriate brackets has been executed.
I came up with this for my bf compiler thing
basically it subtracts and then checks if the result is 0.
Can be easily changed to execute stuff in if/else-ish way
Layout:
[A] B
>[-<->]+<[>-<[-]]>
Output
0 [result]
Result is 1 if equal
I am trying to count how many times the unique words in a LinkedList appear using this code:
for(int i2 = 0; i2 < a.size(); i2++)
{
word2 = a.get(i2);
for(int j2 = 0; j2 < a.size(); j2++)
{
if(word2 == a.get(j2))
{
counter++;
}
}
System.out.println(word2 + " : " + counter);
counter = 0;
}
But the counter prints out:
Alphabet : 1
Alright : 1
Apple : 1
Alphabet : 1
Alright : 1
Apple : 1
Alphabet : 1
Alright : 1
Apple : 1
There is obviously more than one of the words, but the counter never gets higher then one. I think that the inner for loop is stopping when the if statement is satisfied, but I don't want it to. Any suggestions?
You should use
word2.equals(a.get(j2))
Using "==" you compare the references to those String which are not equal.
And by the way you will print the counter multiple times for the same word if you have repetitions. Let's say you have the word Apple on 2 positions in your list. When you will reach these 2 positions the counter will go up to 2 and you will print (Apple: 2) 2 times
This seems like java to me. You can't compare string with (==) operator.
if(word2.equals(a.get(j2)))
Just a new addition!!!
As suggested by #user3412998 its better to use Sets so that you can avoid duplication.
Now if you need to use ArrayList only then you need to take one element and check if there are multiple copies of that element using loop.The condition can be checked by .equals method if you are using string or my contains method incase you are inserting objects(dont forget to override your equals method in the corresponding class). This is a very tedious process and I do not recommend it.
I believe that the easiest way will to not to add duplicate elements in your array list. to accomplish that look at the code below..
if(!a.contains(word)){
a.add(word);
}
Note word is a String object.
Here I am checking weather the string is already contained in the ArrayList, before insertion.
make a bit of modifications so that you can use it, for arrays, or for directly inputting data etc..
I hope this helps..
In my c++ class, we got assigned pairs. Normally I can come up with an effective algorithm quite easily, this time I cannot figure out how to do this to save my life.
What I am looking for is someone to explain an algorithm (or just give me tips on what would work) in order to get this done. I'm still at the planning stage and want to get this code done on my own in order to learn. I just need a little help to get there.
We have to create histograms based on a 4 or 5 integer input. It is supposed to look something like this:
Calling histo(5, 4, 6, 2) should produce output that appears like:
*
* *
* * *
* * *
* * * *
* * * *
-------
A B C D
The formatting to this is just killing me. What makes it worse is that we cannot use any type of arrays or "advanced" sorting systems using other libraries.
At first I thought I could arrange the values from highest to lowest order. But then I realized I did not know how to do this without using the sort function and I was not sure how to go on from there.
Kudos for anyone who could help me get started on this assignment. :)
Try something along the lines of this:
Determine the largest number in the histogram
Using a loop like this to construct the histogram:
for(int i = largest; i >= 1; i--)
Inside the body of the loop, do steps 3 to 5 inclusive
If i <= value_of_column_a then print a *, otherwise print a space
Repeat step 3 for each column (or write a loop...)
Print a newline character
Print the horizontal line using -
Print the column labels
Maybe i'm mistaken on your q, but if you know how many items are in each column, it should be pretty easy to print them like your example:
Step 1: Find the Max of the numbers, store in variable, assign to column.
Step 2: Print spaces until you get to column with the max. Print star. Print remaining stars / spaces. Add a \n character.
Step 3: Find next max. Print stars in columns where the max is >= the max, otherwise print a space. Add newline. at end.
Step 4: Repeat step 3 (until stop condition below)
when you've printed the # of stars equal to the largest max, you've printed all of them.
Step 5: add the -------- line, and a \n
Step 6: add row headers and a \n
If I understood the problem correctly I think the problem can be solved like this:
a= <array of the numbers entered>
T=<number of numbers entered> = length(a) //This variable is used to
//determine if we have finished
//and it will change its value
Alph={A,B,C,D,E,F,G,..., Z} //A constant array containing the alphabet
//We will use it to print the bottom row
for (i=1 to T) {print Alph[i]+" "}; //Prints the letters (plus space),
//one for each number entered
for (i=1 to T) {print "--"}; //Prints the two dashes per letter above
//the letters, one for each
while (T!=0) do {
for (i=1 to N) do {
if (a[i]>0) {print "*"; a[i]--;} else {print " "; T--;};
};
if (T!=0) {T=N};
}
What this does is, for each non-zero entered number, it will print a * and then decrease the number entered. When one of the numbers becomes zero it stops putting *s for its column. When all numbers have become zero (notice that this will occur when the value of T comes out of the for as zero. This is what the variable T is for) then it stops.
I think the problem wasn't really about histograms. Notice it also doesn't require sorting or even knowing the