I am trying to write the syntax to add up ten user-inputted numbers in a Console application, and then divide the final product by 10 in order to get the average number. So far I am able to allow the user to input the numbers properly, and I have the program set up to allow the user time to read the result, however I am slightly stuck on the syntax to add up the numbers overall. I know this is very simple, but the operation for creating this code is escaping me. I have tried finding the answer online already, but so far my only results have been overly-complex or just downright wrong.
Any and all help would be greatly appreciated.
There are lots of ways to acheive this, but the easiest is to just keep a running total. You'll need to cast the string input as a double using the double.Parse() method.
runningTotal = runningTotal + double.Parse(Console.ReadLine())
After the last input, simply divide runningTotal by 10 to display the result.
Exactly how depends on what you're using to store the numbers. A List(of Double) would be good for this because it'll store an arbitrary amount of numbers. Then to add them you can use a simple loop. Assuming your numbers are stored in a List called "numbers":
Dim total as Double = 0;
Dim average as Double = 0;
For Each number as Double in numbers
total += number
Next
average = total / numbers.Count()
What this does is goes through numbers, and for each number in it adds it to the total. At the end it divides the total by the count of numbers (however many you were given) to get the average. Note that you divide by the number of numbers to get the average, not necessarily 10. This one will still work if they only give you 7 numbers. :)
Related
When using forecast, you input a number and it should return a value based on the known X data and Known Y data.
However if you put in a time this does not work.
I need two things.
First of all I need the VBA equivalent of forecast. I suspect this to be application.forecast
Then how to use the date as a value for the forecast to work as it should
The formula is as follows:
=FORECAST(15:00:00,A10:A33,B10:B33)
Currently this equation flags up an error.
Any ideas to get this to work for time values?
I see two potential problem areas. The first is the time. Use the TIME function to get a precise time. Second, in D9:D12, the values are left-aligned. Typically, this means they are text, not true numbers. If you absolutely require the m suffix, use a Custom number Format of General\m in order that they retain their numeric status while displaying an m as an increment suffix. If you type the m in, they become text-that-look-like-numbers and are useless for any maths.
=FORECAST(TIME(15, 0, 0), B10:B33, A10:A33)
That returns 3.401666667 which is either 09:38 AM or 3.4 m (it's been a while since I played with the FORECAST function).
I am currently making a random math quiz which generates random addition, subtraction, and multiplication fine. Those 3 are simple. I would like to know just how to make a randomly generated division question that always equals a whole number, using 2 int variables, num1 and num2 but I just cannot figure it out Any help would be greatly appreciated!
It should be pretty simple. The logic may go like this:
Randomly generate a number and call it divisor
Multiply your divisor with a randomly generated number and call it dividend
Now you've got two integer numbers (dividend & divisor) that should solve into a whole number quotient.
I would construct the the two integers using a pool of random generated factors, making sure the two have at least one factor in common. Or to make it even simpler, pick a random integer n, then multiply it by another random integer m and show the division m*n/n.
I am making an anagram solver in Visual Basic that gives you every possible combination when you enter a string. I need to work out how many combinations there are depending on the amount of characters in the string and how many different characters there are.
E.G.
Sample string:
abc
Total characters: 3, Different Characters: 3
Possible combinations: 6
abc, acb, bac, bca, cab, cba
I need an equation (using the number of characters and different characters) to link this to a string that contains a different amount of characters.
I've been using trial and error to try and figure is out, but I can't quite get my head around it. So far I have:
((letters - 1) ^ (different letters - 1)) + (letters - 1)
which works for a few different letter counts but now for all.
Help please???
I'll lead you to the answer, but I'll try to explain along the way. Let's say you had 10 different letters. You'd have 10 choices for the first, 9 for the second, 8 for the third, etc. Ultimately, there would be 10*9*8*7*6...*2*1 = 10! possibilities. However, sometimes you'll have multiple instances of the same letter. For example, using that for the string "aaabcd" would overcount possibilities, because it counts each of the a's as distinct letters, even though they're not. To correct for that, you would have to divide by the factorial of the number of repeated letters. A good way to calculate the total number of possibilities would be (total number of letters factorial)/ (product of the factorials of the number of repeated instances of each letter).
For example:
There are 6!/(3!) ways to arrange the letters in "aaabcd"
There are 6! ways to arrange the letters is "abcdef"
There are 6!/(3!*2!) ways to arrange the letters in "aaabbc"
There are 10!/(5!*3!*2!) ways to arrange the letters in "aaaaabbbcc"
I hope this helps.
For the possible counting number, it's exactly the same as computing Multinomial Coefficient
A simple explanation is that, for no repeating characters,
It's simply permutation = n!
(It is easy to understand if you draw a tree diagram, with first character has n choices, second character has n-1choices...etc.)
However as you may have repeating characters, you will double count many of them.
Let's see an simple example: for aaa, how many possible arrangements IF WE COUNT EVEN THE OUTCOME IS THE SAME?
Answer is 3!(aaa,aaa,aaa,aaa,aaa,aaa)
This gives us an idea that, when we have a character appearing for m times, we will count m! instead of 1
So the counting is just n!(all possible arrangements, including same outcome) / m! (a character appear for m times)
Same for more characters repeating: n!/a!b!c!.. (first character appear a times, another appear for b times...)
If you understand the concept behind, then you will find that, actually for those "non-repeating" characters, it's just dividing an 1!. For eg, character (multi)set = {a,a,a,b,b,c}, #a = 3, #b = 2, #c = 1, so the answer (without repeating count) is (3+2+1)!/3!2!1! and fraction of this format is named multinomial coefficient as stated above.
In programming point of view, you can just pre-compute all factorials (with a pretty small n though as n~30 is already too large for a variable to store) with simple for loop
declare frac = array(n);
frac[0] = 1;
FOR i=1; i<=n;i++
frac[i] = i*frac[i-1]
For a larger n, you may just calculate double/float division on the fly in the loop to avoid overflow..you may face precision problem though.
If you further need to output the different strings, you may use DFS to backtrack all the possible outcomes. Or if you could use another language like C++, you can use built-in function like next_permutation() after sort the character set.
Well, it is a low-level question
Suppose I store a number (of course computer store number in binary format)
How can I print it in decimal format. It is obvious in high-level program, just print it and the library does it for you.
But how about in a very low-level situation where I don't have this library.
I can just tell what 'character' to output. How to convert the number into decimal characters?
I hope you understand my question. Thank You.
There are two ways of printing decimals - on CPUs with division/remainder instructions (modern CPUs are like that) and on CPUs where division is relatively slow (8-bit CPUs of 20+ years ago).
The first method is simple: int-divide the number by ten, and store the sequence of remainders in an array. Once you divided the number all the way to zero, start printing remainders starting from the back, adding the ASCII code of zero ('0') to each remainder.
The second method relies on the lookup table of powers of ten. You define an array of numbers like this:
int pow10 = {10000,1000,100,10,1}
Then you start with the largest power, and see if you can subtract it from the number at hand. If you can, keep subtracting it, and keep the count. Once you cannot subtract it without going negative, print the count plus the ASCII code of zero, and move on to the next smaller power of ten.
If integer, divide by ten, get both the result and the remainder. Repeat the process on the result until zero. The remainders will give you decimal digits from right to left. Add 48 for ASCII representation.
Basically, you want to tranform a number (stored in some arbitrary internal representation) into its decimal representation. You can do this with a few simple mathematical operations. Let's assume that we have a positive number, say 1234.
number mod 10 gives you a value between 0 and 9 (4 in our example), which you can map to a character¹. This is the rightmost digit.
Divide by 10, discarding the remainder (an operation commonly called "integer division"): 1234 → 123.
number mod 10 now yields 3, the second-to-rightmost digit.
continue until number is zero.
Footnotes:
¹ This can be done with a simple switch statement with 10 cases. Of course, if your character set has the characters 0..9 in consecutive order (like ASCII), '0' + number suffices.
It doesnt matter what the number system is, decimal, binary, octal. Say I have the decimal value 123 on a decimal computer, I would still need to convert that value to three characters to display them. Lets assume ASCII format. By looking at an ASCII table we know the answer we are looking for, 0x31,0x32,0x33.
If you divide 123 by 10 using integer math you get 12. Multiply 12*10 you get 120, the difference is 3, your least significant digit. we go back to the 12 and divide that by 10, giving a 1. 1 times 10 is 10, 12-10 is 2 our next digit. we take the 1 that is left over divide by 10 and get zero we know we are now done. the digits we found in order are 3, 2, 1. reverse the order 1, 2, 3. Add or OR 0x30 to each to convert them from integers to ascii.
change that to use a variable instead of 123 and use any numbering system you like so long as it has enough digits to do this kind of work
You can go the other way too, divide by 100...000, whatever the largest decimal you can store or intend to find, and work your way down. In this case the first non zero comes with a divide by 100 giving a 1. save the 1. 1 times 100 = 100, 123-100 = 23. now divide by 10, this gives a 2, save the 2, 2 times 10 is 20. 23 - 20 = 3. when you get to divide by 1 you are done save that value as your ones digit.
here is another given a number of seconds to convert to say hours and minutes and seconds, you can divide by 60, save the result a, subtract the original number - (a*60) giving your remainder which is seconds, save that. now take a and divide by 60, save that as b, this is your number of hours. subtract a - (b*60) this is the remainder which is minutes save that. done hours, minutes seconds. you can then divide the hours by 24 to get days if you want and days and then that by 7 if you want weeks.
A comment about divide instructions was brought up. Divides are very expensive and most processors do not have one. Expensive in that the divide, in a single clock, costs you gates and power. If you do the divide in many clocks you might as well just do a software divide and save the gates. Same reason most processors dont have an fpu, gates and power. (gates mean larger chips, more expensive chips, lower yield, etc). It is not a case of modern or old or 64 bit vs 8 bit or anything like that it is an engineering and business trade off. the 8088/86 has a divide with a remainder for example (it also has a bcd add). The gates/size if used might be better served than for a single instruction. Multiply falls into that category, not as bad but can be. If operand sizes are not done right you can make either instruction (family) not as useful to a programmer. Which brings up another point, I cant find the link right now but a way to avoid divides but convert from a number to a string of decimal digits is that you can multiply by .1 using fixed point. I also cant find the quote about real programmers not needing floating point related to keeping track of the decimal point yourself. its the slide rule vs calculator thing. I believe the link to the article on dividing by 10 using a multiply is somewhere on stack overflow.
I am relatively new to the world of programming and I was wondering if anybody could help me with a small project. I am trying to create two programs in VB.Net that each do one of the following individual actions:
Find the average grade given several user-inputted scores on assignments. The program should also provide the following feedback according to the final score (i.e. A, B, C, D, F).
Run two separate threads printing numbers (or words) in ascending and descending orders. (The numbers (or words) should be given by the user.)
I have a basic understanding of VB.Net, but I am having trouble when it comes to creating even remotely complex programs. I have a few ideas on how I may go about these, such as using an arraylist for the first question that takes user input to find the grades, and then uses a series of If-Then-Else statements to display the letter grade, and possibly using steps simply with dual threading that would result in numerical order being printed in ascending order and descending order. Any help or advice would be greatly appreciated.
P.S.
I will be adding the code I have so far for both of these programs shortly. In the meantime, if you can help me at all with the information I have given you, it would be helpful.
Here's some pseudocode to help you get started on #1:
get list of scores from user input (Console.Readline() if this is a console app)
assign scores to an array or list (List would be a good choice)
get the total score (assuming they're not weighted, if you have a List then you can just use the Sum() method)
get the number of grades (again, if you have a List then you can use Count() method)
divide total by count to get average (if you need a decimal average, you'll have to cast your values to double first, or if your average can be an int just leave them all as int)
use an if-then-else to compare the average to each grade cutoff (if > 90 then "A", else if > "80" then "B", etc)