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namespace SquareStars
{
class Program
{
static void Main(string[] args)
{
var n = int.Parse(Console.ReadLine());
for (int i = 0; i <n-1; i++)
{
Console.Write("*");
}
for (int i = 0; i < n-1; i++)
{
Console.WriteLine("*");
}
for (int i = 0; i < n; i++)
{
Console.Write("*");
}
}
}
}
My exercises is to do Square of Stars like this:
depends of "n".I try to use for loops,but I can do right side of square.
I do this:
***
*
***
but I want this:
*** ****
* * * *
*** or * *
****
Can sameone help me to this code????
The following should do:
public static string CreateASCIISquare(int squareSideLength, char c = '*')
{
if (squareSideLength < 1)
return "";
if (squareSideLength == 1)
return c.ToString();
var horizontalOuterRow = new String(c, squareSideLength);
var horizontalInnerRow = $"{c}{new string(' ', squareSideLength - 2)}{c}";
var squareBuilder = new StringBuilder();
squareBuilder.AppendLine(horizontalOuterRow);
for (int i = 0; i < squareSideLength - 2; i++)
{
squareBuilder.AppendLine(horizontalInnerRow);
}
squareBuilder.Append(horizontalOuterRow);
return squareBuilder.ToString();
}
Ok, so lets explain the code a little bit:
Always validate the data going into your method. In your case the user can specify the length of the square's side. Are there any invalid values? Well clearly yes, -2 doesn't seem to be a valid choice.
You have many options here. You can tell the user the value is not valid, you can simply ignore it, do nothing and let your application crash and die miserably, or you can return an empty square. All are valid choices, but consciously choose one and design according to your decision. In my case I chose to return an empty square:
if (squareSideLength < 0)
return "";
Are there any trivial cases I can manage easily without for loops and Console.Writes etc.? Yes, lengths 0 and 1 seem pretty straightforward. It stands to reason that if I've returned the empty square for negative values, I do the same for 0 sized squares. So I change the previous code to:
if (squareSideLength < 1)
return "";
Good, now what about 1? Well that's pretty easy too isn't it?
if (squareSideLength == 1)
return c.ToString();
Moral of the story: take care of invalid data or trivial cases first. Many times trivial cases can also be corner cases that can complicate your general solution, get them out of the way fast!
Ok, now lets think about what a square looks like:
**
**
***
* *
***
****
* *
* *
****
Well the pattern seems pretty obvious. A square is made up of two rows containing the specified number of stars, and 0 or more rows containing only 2 stars and squareSideLength - 2 spaces in between. Well, its only two types of rows, lets build them up front:
var horizontalOuterRow = new String(c, squareSideLength);
var horizontalInnerRow = $"{c}{new string(' ', squareSideLength - 2)}{c}";
Great! We got our building blocks, now lets build our square:
So how does that go? Well we simply start by adding a horizontalOuterRow then we add the squareSideLength - 2 horizontalInnerRows and we finish up adding one more horizontalOuterRow.
And voilá, we got ourselves our nice little ASCII square:
squareBuilder.AppendLine(horizontalOuterRow);
for (int i = 0; i < squareSideLength - 2; i++)
{
squareBuilder.AppendLine(horizontalInnerRow);
}
squareBuilder.Append(horizontalOuterRow);
The fact that I've used a StringBuilder is not really germane to the question, but get into the habit of using this tool when constructing dynamically built strings you don't know the length of beforehand. Concatenating strings can give you pretty bad performance so it's best to avoid it if possible.
Now we proudly return our beautiful ASCII square back to our extatic user and bask under the general round of applause:
return squareBuilder.ToString();
Hope this little answer helps you.
Remember, to code well, break everything up into very small tasks, and take care of them one at a time. Think how you'd do it by hand and write it that way. Once you have it working there'll be time to see if you need to optimize, refactor, etc. your code. But the key is getting it to work with code as clear as possible.
You can use the string(char, int) constructor to create strings of repeating characters. This lets you simplify the code to use a single for loop:
var n = int.Parse(Console.ReadLine());
Console.WriteLine(new string('*', n));
for (int i = 0; i < n - 2; i++)
{
Console.WriteLine("*" + new string(' ', n - 2) + "*");
}
Console.WriteLine(new string('*', n));
class Program
{
static void Main(string[] args)
{
var n = int.Parse(Console.ReadLine());
for (int row = 1; row <= n; row++)
{
for (int col = 1; col <= n; col++)
{
if (row == 1 || row == n)
{
Console.Write("*");
}
else
{
if (col == 1 || col == n)
{
Console.Write("*");
}
else
{
Console.Write(" ");
}
}
}
Console.WriteLine();
}
Console.ReadKey();
}
}
You can change Console.WriteLine in 2nd loop like this :
class Program
{
static void Main(string[] args)
{
var n = int.Parse(Console.ReadLine());
for (int i = 0; i < n - 1; i++)
{
Console.Write("*");
}
for (int i = 0; i < n - 1; i++)
{
Console.WriteLine(i == 0 ? "*" : "*" + string.Empty.PadLeft(n - 2) + "*");
}
for (int i = 0; i < n; i++)
{
Console.Write("*");
}
}
}
Related
I have been trying to fix this issue for some days now but can't find the error. It seems as simple as an if statement for the code to print commas for all but the last number. It worked for me putting random numbers but when i put the specific numbers (24,7,35,2,27,7,89)it prints the comma at the end.
Print even numbers with commas
This is my code, but i tried multiple other ways.
using System.Collections.Generic;
using System.Text;
using System.Transactions;
namespace ArrayExercises
{
class TaskFour
{
public static void FindEvenNumbers()
{
int[] input = new int[7];
int count = 1;
string comma = ",";
Console.WriteLine("[== Please Enter 7 numbers ==]");
Console.WriteLine();
for (int i = 0; i < input.Length; i++)
{
Console.WriteLine($"Enter number {count}:");
input[i] = int.Parse(Console.ReadLine());
count++;
}
Console.WriteLine("The even numbers in this array are: ");
for (int i = 0; i < input.Length; i++)
{
if (input[i] % 2 == 0)
{
if (i < input.Length)
{
Console.Write(input[i] + comma);
}
else
{
Console.Write(input[i]);
}
}
}
}
}
}
Thanks in advance :)
You can use below code without introducing any extra space using for loop.
for (int i = 0; i < input.Length; i++)
{
if (input[i] % 2 == 0)
{
if (i != 0) Console.Write(comma);
Console.Write(input[i]);
}
}
Or use the inbuilt string.Join function.
Console.WriteLine(string.Join(',', input.Where(i => i % 2 == 0)));
This answer assumes that the above is homework/tutorial, there are more efficient means that what I am posting here
The problem is this segment:
if (i < input.Length)
{
Console.Write(input[i] + comma);
}
The problem is that you are always outputting a comma, regardless of what is to follow. I think that the easiest approach would be to add the comma before hand, meaning, while you are going through the second loop, if you have already printed a number before, you pre-pend a comma and print the number you have, if you did not print any numbers before (that is, you are about to print the first number) then you do not prepend the comma.
i will always be less than the length the input due to the loop's condition.
Instead of implementing this yourself, you could let string.join do the heavy lifting for you:
string result = string.Join(comma, input.Where(i => i % 2 == 0));
Console.WriteLine(result);
Your else condition will never execute. If you are rookie in C# then I would suggest get all even numbers first and store it in the list. Iterate over again to print with comma and just skip the last comma.
//Your existing source code
List<int> evenNumber = new List<int>();
for (int i = 0; i < input.Length; i++)
{
if (input[i] % 2 == 0)
{
evenNumber.Add(input[i]);
}
}
for(int i = 0; i < evenNumber.Length; i++)
{
//Print just number if it is last, otherwise print number with comma
if(i == evenNumber.Length - 1)
Console.Write(evenNumber[i]);
else
Console.Write(evenNumber[i]+comma);
}
If you know string.Join() then use linq to get list of even numbers and print list of even numbers with ',' as a delimiter
var evenNumbers = input.Where(x => x % 2 ==0); //Filter all even numbers
Console.WriteLine(string.Join(",", evenNumbers); //Print using string.Join
Yet another version:
string result = "";
for (int i = 0; i < input.Length; i++)
{
if (input[i] % 2 == 0)
{
result += $",{input[i]}";
}
}
Console.WriteLine( result.Length > 0 ? result.Substring(1).ToString() : "");
I want to print a rectangle like this :
&#*#
#*#&
*#&#
#&#*
But problem is that i can't find the algorithm to print this.
I only know how to print a simple rectangle/square
public static void Main(string[] args)
{
Console.Out.Write("Saisir la taille : ");
int taille = int.Parse(Console.In.ReadLine());
int i;
int j;
for(i = 0; i < taille; i++){
for(j = 0; j < taille; j++){
Console.Write("*");
}
Console.WriteLine("");
}
}
Thank you !
First things first unless you need your iterators outside of your loop, just declare them in the for declaration
public static void Main(string[] args)
{
Console.Out.Write("Saisir la taille : ");
int taille = int.Parse(Console.In.ReadLine());
for(int i = 0; i < taille; i++){
for(int j = 0; j < taille; j++){
Console.Write("*");
}
Console.WriteLine("");
}
}
Second you'll need a list of the characters you want to use, given your example
char[] chars = { '&', `#`, `*`, '#' };
and we'll need a way to know which character we want to use at any given time, say an iterator we can call characterIndex for simplicity. We will increment it each iteration. If incrementing it puts it out of the range of our character array, if characterIndex == 4, we set it back to zero.
int characterIndex;
To get the scrolling effect you have, before each line we must select a characterIndex that is offset by the row
characterIndex = i % chars.Length;
Tying it all together
public static void Main(string[] args)
{
char[] chars = { '&', `#`, `*`, '#' };
int characterIndex;
Console.Out.Write("Saisir la taille : ");
int taille = int.Parse(Console.In.ReadLine());
for(int i = 0; i < taille; i++){
characterIndex = i % chars.Length;
for(int j = 0; j < taille; j++){
Console.Write(chars[characterIndex]);
characterIndex++;
if(characterIndex == chars.Length)
characterIndex = 0;
}
Console.WriteLine("");
}
}
Getting the permutations by nesting for loops will only work if you know exactly how many elements there will be. Basically you need to write a for-loop for every element after the 1st.
The proper way to deal with this is Recursion. While there are cases where Recursion and nested for-loops are interchangeable. And in cases where they are, for loops have a potential speed advantage. While normally the speed rant applies to such differences, with the sheer amount of data both Recursion and Loops might have to deal with, it often maters - so best to prefer loops where possible.
Permutations is AFAIK not a case where loops and recursion are interchangeable. Recurions seems to be mandatory. Some problem as simply inherently recursive. As the recursion version is fairly well known, I will not post any example code.
You should defiitely use Recursion. With your example code I basically asume you are:
In a learning environment
You just learned recursion
A input variant recurions can effortless solve (like a 6 or 20 size input), is the next assignment
I am a beginner learning c#, I have this code
static void Main(string[] args)
{
int a = 0;
while (a < 100)
{
a = a + 1;
if ((a % 10) == 0)
{
Console.WriteLine(a);
}
else
{
Console.Write(a);
Console.Write(",");
}
}
}
Is there a more efficient way of writing this code? I feel there might be a better way of doing this in c#. This is my very first code. I will appreciate a response. Thanks
the short version would look like this:
int stepSize = 10;
for (int i = 1; i < 100; i+=stepSize )
{
Console.WriteLine(String.Join(",", Enumerable.Range(i, stepSize)));
}
Explanation:
You walk in steps of 10 through your for-loop. at each step the
Enumerable.Range method creates an array which holds numbers enumerated from a start value (i) until the count value (10).
The String.Join method takes each element of this array and combines them into a string separated by a ,
Since it looks like homework:
You should research how to useString.Format. This way you could arrange elements in one line at certain positions.
For iterations with a counter variable a for-loop is preferable, because it is exactly made for it with a clearly readable head signature.
You actually wrote a very readable code, which in my opinion is efficient. A shortening of codelines does not make it necessarily more efficient or faster or more readable. Sometimes the only advantage is that it looks a little more elegant ;) that's all
EDIT:
You can even get it down to one line:
Console.WriteLine(Enumerable.Range(0, 10).Select(x => String.Join(",", Enumerable.Range(x * 10 + 1, 10))));
it is short, but it is horrible to read and understand :)
First step would be using a foor loop instead.
for(int i = 0; i <= 100; i++)
{
if ((i % 10) == 0)
{
Console.WriteLine(i);
}
else
{
Console.Write(i);
Console.Write(",");
}
}
You could replace
Console.Write(i);
Console.Write(",");
with
Console.Write(string.Format("{0},", i));
or even better with
Console.Write($"{i},");
Just yet another approach
for (var i = 1; i <= 100; i++)
{
Console.Write(i);
Console.Write(i % 10 == 0 ? Environment.NewLine : ",");
}
If you have a for loop:
for(i=0;i<10;i++){}
Now, when i==5, how can I exit the for loop completely without using break, return, or if?
The best I could come up with was this:
for (int i = 0; i < 10; i++)
{
i = (i == 5) ? 10 : i;
Trace.WriteLine("i = " + i.ToString());
}
...which will cause the loop to run six times (i=0..5) and display this..
i = 0
i = 1
i = 2
i = 3
i = 4
i = 10
The alternative way to "exit the loop" (in a particularly nasty fashion) would be to do this...
for (int i = 0; i < 10; i++)
{
int a = 3 / ((i == 5) ? 0 : 1);
Trace.WriteLine("i = " + i.ToString());
}
..which crashes out, errr, successfully exits the loop without using the break, return or if commands.
i = 0
i = 1
i = 2
i = 3
i = 4
A first chance exception of type 'System.DivideByZeroException' occurred in MikesProgram.dll
language is C# .it was an interview question actually..curious
Do I get the job ?
I would need to check your health & dental plans, and I have to leave early on Thursdays to collect my daughters from school.
;-)
for (int n = 0; n < 10; n++)
{
n += (n / 5) * 5;
}
well here's another way if you want to break the processing exactly when i = 5 without using break, return, or if
for (int lowsetLimit = 0, highestLimit = 10, i = lowsetLimit; i < highestLimit; i++)
{
//normal code which process before i gets eqaul to 5 goes here...
i = (i < 5) ? i : highestLimit; //and here is the pivot point.
}
for(i=0;i<10;i++){
(i==5) ? goto Outer : //do something;
}
Outer:
//do something
The question says that loop should end when i=5, It says nothing about start, So this should be valid (ternary operator solution is better, but if we are not allowed to use any conditional operator)
for (int i = 0; i < 10; i++)
{
i=i-4;
Console.WriteLine("i = " + i.ToString());
i=i+4;
}
this Starts at -4 and ends at 5.
The real answer of course would be the following:
for (i=0; i!=5; i++)
{
// do something
}
But let's make it a bit more generic: stop if (expression) becomes true.
The second argument of the for loop is a boolean expression which determines whether to continue the loop with the next element or not.
So if you want to stop looping because of any condition:
for (i=0; !(expression) && i<10; i++)
{
// do something
}
This works using while and goto:
for (int i = 0; i < 10; i++)
{
while (i < 5)
{
Console.Write(i + " ");
goto OutsideWhile;
}
OutsideWhile:
continue;
}
// 0 1 2 3 4
can you help me with the following exercise pls? (it's not homework, just an exercise in the book I'm using.)
"An integer is said to be a perfect number if its factors, including one (but not the number itself), sum to the number. For example, 6 is a perfect number, because 6 = 1 + 2 + 3. Write method Perfect that determines whether parameter value is a perfect number. Use this method in an app that determines and displays all the perfect numbers between 2 and 1000. Display the factors of each perfect number to confirm that the number is indeed perfect."
so here's what i got so far:
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace Perfect_Numbers2
{
class Program
{
static bool IsItPerfect(int value)
{
int x = 0;
int counter = 0;
bool IsPerfect = false;
List<int> myList = new List<int>();
for (int i = value; i <= value; i++)
{
for (int j = 1; j < value; j++)
{
// if the remainder of i divided by j is zero, then j is a factor of i
if (i%j == 0) {
myList[counter] = j; //add j to the list
counter++;
}
for (int k = 0; k < counter; k++)
{
// add all the numbers in the list together, then
x = myList[k] + myList[k + 1];
}
// test if the sum of the factors equals the number itself (in which case it is a perfect number)
if (x == i) {
IsPerfect = true;
}
}
Console.WriteLine(i);
}
return IsPerfect;
}
static void Main(string[] args)
{
bool IsItAPerfectNum = false;
for (int i = 2; i < 1001; i++)
{
IsItAPerfectNum = IsItPerfect(i);
}
}
}
}
how would you do it? is my code fixable? how would you fix it? thanks!
im getting an error at line myList[counter] = j; (index was out of range) and besides it's not displaying the perfect numbers like it's supposed to....
EDIT = I made some changes;
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
namespace Perfect_Numbers2
{
class Program
{
static bool IsItPerfect(int value)
{
int x = 0;
int counter = 0;
bool IsPerfect = false;
List<int> myList = new List<int>();
for (int i = value; i <= value; i++)
{
for (int j = 1; j < i; j++)
{
if (i%j == 0) // if the remainder of i divided by j is zero, then j is a factor of i
{
myList.Add(j); //add j to the list
}
x = myList.Sum();
if (x == i) // test if the sum of the factors equals the number itself (in which case it is a perfect number)
{
IsPerfect = true;
}
}
Console.WriteLine(i);
}
return IsPerfect;
}
static void Main(string[] args)
{
bool IsItAPerfectNum = false;
for (int i = 2; i < 1001; i++)
{
IsItAPerfectNum = IsItPerfect(i);
Console.WriteLine(IsItAPerfectNum);
Console.ReadKey(true);
}
}
}
}
now i can cycle through all the numbers until 1000 and it displays if it's perfect or not (true or false) [which isn't what the exercise called for, but it's a step in the right direction (the exercise says that it should display only the perfect numbers)].
In any case, what's strange is that it says true at number 24, which isn't a perfect number.... http://en.wikipedia.org/wiki/Perfect_numbers#Examples
why is 24 different?
thanks very much
can you help me with the following exercise please?
Yes. Rather than showing you where your error is, I'll teach you how to find your error. Even better, the same technique will lower the chances of you causing the error in the first place.
The key here is to break the problem down into small parts where each small part can be tested independently. You have already started to do this! You have two methods: Main and IsItPerfect. You should have at least three more methods. The methods you should have are:
IsDivisor -- takes two integers, returns true if the first divides the second.
GetAllDivisors -- takes an integer, returns a list of all the divisors
Sum -- takes a list of integers, returns the sum
Your method IsPerfect should be calling GetAllDivisors and Sum and comparing the sum to the original number, and that's all it should be doing. Your method GetAllDivisors should be calling IsDivisor, and so on.
You can't find the bug easily because your method is doing too much. If you're not getting the correct result out and you have four methods instead of one then you can test each method independently to make sure that it works, or fix it if it does not.
Your first for loop will be executed exactly once.
for (int i = value; i <= value; i++)
For example for value = 6
for (int i = 6; i <= 6; i++)
Some help with the 24 issue you are having: 24 is returning true as you are actually checking if it is perfect on every additional factor. So 24 gets flipped to true here:
Factors of 24 | Total so far
1 1
2 3
3 6
4 10
6 16
8 24 <-- returns true
12 36 <-- should be false, but flag is never reset
I have just now completed the same exercise which is from a really great book called visual c# 2012 by Mr Deitel.
The way i started to tackle is, i started off with figuring out how to work out the factorials of numbers and then slowly kept building on from there.
Since you are following the same book, i would suggest you not to use things that are not covered up to that chapters exercise, like list collections which you have used, As this will make the exercise unnecessarily difficult. and negates the learning methodology set out by of the author.
here is my code which i hope can help you in some way.
class Program
{
static int factorTotal = 1;
static void Main(string[] args)
{
int count = 1;
while (count <= 10000)
{
bool isPerfect = IsPerfectNumber(count);
if (isPerfect && (factorTotal >1))
{
Console.WriteLine("Is Perfect: {0}", factorTotal);
}
factorTotal = 1;
count++;
}
} // end main
static bool IsPerfectNumber(int n)
{
int temp;
int counter = 2;
bool IsPerfect = false;
while (counter <= (n - 1))
{
temp = n % counter;
if (temp == 0) // if true than factor found
{
factorTotal = factorTotal + counter;
}
counter++;
}
if ((factorTotal) == n)
IsPerfect = true;
else
IsPerfect = false;
return IsPerfect;
}
}//end class
under the Main method of you console application copy and paste below code.
I explained few things at the end of the code...
=====================================================================
{
Console.WriteLine("perfect numbers/n");
Console.Write("Enter upper limit: ");
int iUpperLimit = int.Parse(Console.ReadLine());
string sNumbers = "";
List<int> lstFactor = new List<int>();
for(int i = 1;i<=iUpperLimit;i++)
{
for(int k = 1;k<i;k++)
{
if (i % k == 0)
{
lstFactor.Add(k); //this collect all factors
}
if (k == i-1)
{
if (lstFactor.Sum() == i) //explain1
{
sNumbers += " " + i;
lstFactor.Clear(); //explain2
break;
}
else
{
lstFactor.Clear(); //explain2
}
}
}
}
Console.WriteLine("\nperfect numbers are: " + sNumbers);
Console.ReadKey();
}
}
=======================================================================
note that i is a number that we test and k is its factors.
explain1 => we add all factors collected and check if they are equal to i (we simply check if i is perfect number)
explain2 => we have to clear our list before we can check if the next number i is a perfect number or not so that factors of the previous number does not interfere with factors of the current number.
int start=1;
int end=50;
for(int a=end ; a > start ;a--)
{
int b=1;
int c=0;
bool x=false;
for(int i=1 ; i < a ;i++)
{
b=a/i;
if(b*i==a)
{
c+=i;
}
if(c==a & i==a/2)
{
x=true;
}
}
if(x==true)
Console.Write("{0} is : {1}",a,x);
}