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  • Write a C++ program to encrypt and decrypt certain codes.

    - by Amber
    Step 1: Write a function int GetText(char[],int); which fills a character array from a requested file. That is, the function should prompt the user to input the filename, and then read up to the number of characters given as the second argument, terminating when the number has been reached or when the end of file is encountered. The file should then be closed. The number of characters placed in the array is then returned as the value of the function. Every character in the file should be transferred to the array. Whitespace should not be removed. When testing, assume that no more than 5000 characters will be read. The function should be placed in a file called coding.cpp while the main will be in ass5.cpp. To enable the prototypes to be accessible, the file coding.h contains the prototypes for all the functions that are to be written in coding.cpp for this assignment. (You may write other functions. If they are called from any of the functions in coding.h, they must appear in coding.cpp where their prototypes should also appear. Do not alter coding.h. Any other functions written for this assignment should be placed, along with their prototypes, with the main function.) Step 2: Write a function int SimplifyText(char[],int); which simplifies the text in the first argument, an array containing the number of characters as given in the second argument, by converting all alphabetic characters to lower case, removing all non-alpha characters, and replacing multiple whitespace by one blank. Any leading whitespace at the beginning of the array should be removed completely. The resulting number of characters should be returned as the value of the function. Note that another array cannot appear in the function (as the file does not contain one). For example, if the array contained the 29 characters "The 39 Steps" by John Buchan (with the " appearing in the array), the simplified text would be the steps by john buchan of length 24. The array should not contain a null character at the end. Step 3: Using the file test.txt, test your program so far. You will need to write a function void PrintText(const char[],int,int); that prints out the contents of the array, whose length is the second argument, breaking the lines to exactly the number of characters in the third argument. Be warned that, if the array contains newlines (as it would when read from a file), lines will be broken earlier than the specified length. Step 4: Write a function void Caesar(const char[],int,char[],int); which takes the first argument array, with length given by the second argument and codes it into the third argument array, using the shift given in the fourth argument. The shift must be performed cyclicly and must also be able to handle negative shifts. Shifts exceeding 26 can be reduced by modulo arithmetic. (Is C++'s modulo operations on negative numbers a problem here?) Demonstrate that the test file, as simplified, can be coded and decoded using a given shift by listing the original input text, the simplified text (indicating the new length), the coded text and finally the decoded text. Step 5: The permutation cypher does not limit the character substitution to just a shift. In fact, each of the 26 characters is coded to one of the others in an arbitrary way. So, for example, a might become f, b become q, c become d, but a letter never remains the same. How the letters are rearranged can be specified using a seed to the random number generator. The code can then be decoded, if the decoder has the same random number generator and knows the seed. Write the function void Permute(const char[],int,char[],unsigned long); with the same first three arguments as Caesar above, with the fourth argument being the seed. The function will have to make up a permutation table as follows: To find what a is coded as, generate a random number from 1 to 25. Add that to a to get the coded letter. Mark that letter as used. For b, generate 1 to 24, then step that many letters after b, ignoring the used letter if encountered. For c, generate 1 to 23, ignoring a or b's codes if encountered. Wrap around at z. Here's an example, for only the 6 letters a, b, c, d, e, f. For the letter a, generate, from 1-5, a 2. Then a - c. c is marked as used. For the letter b, generate, from 1-4, a 3. So count 3 from b, skipping c (since it is marked as used) yielding the coding of b - f. Mark f as used. For c, generate, from 1-3, a 3. So count 3 from c, skipping f, giving a. Note the wrap at the last letter back to the first. And so on, yielding a - c b - f c - a d - b (it got a 2) e - d f - e Thus, for a given seed, a translation table is required. To decode a piece of text, we need the table generated to be re-arranged so that the right hand column is in order. In fact you can just store the table in the reverse way (e.g., if a gets encoded to c, put a opposite c is the table). Write a function called void DePermute(const char[],int,char[], unsigned long); to reverse the permutation cypher. Again, test your functions using the test file. At this point, any main program used to test these functions will not be required as part of the assignment. The remainder of the assignment uses some of these functions, and needs its own main function. When submitted, all the above functions will be tested by the marker's own main function. Step 6: If the seed number is unknown, decoding is difficult. Write a main program which: (i) reads in a piece of text using GetText; (ii) simplifies the text using SimplifyText; (iii) prints the text using PrintText; (iv) requests two letters to swap. If we think 'a' in the text should be 'q' we would type aq as input. The text would be modified by swapping the a's and q's, and the text reprinted. Repeat this last step until the user considers the text is decoded, when the input of the same letter twice (requesting a letter to be swapped with itself) terminates the program. Step 7: If we have a large enough sample of coded text, we can use knowledge of English to aid in finding the permutation. The first clue is in the frequency of occurrence of each letter. Write a function void LetterFreq(const char[],int,freq[]); which takes the piece of text given as the first two arguments (same as above) and returns in the 26 long array of structs (the third argument), the table of the frequency of the 26 letters. This frequency table should be in decreasing order of popularity. A simple Selection Sort will suffice. (This will be described in lectures.) When printed, this summary would look something like v x r s z j p t n c l h u o i b w d g e a q y k f m 168106 68 66 59 54 48 45 44 35 26 24 22 20 20 20 17 13 12 12 4 4 1 0 0 0 The formatting will require the use of input/output manipulators. See the header file for the definition of the struct called freq. Modify the program so that, before each swap is requested, the current frequency of the letters is printed. This does not require further calls to LetterFreq, however. You may use the traditional order of regular letter frequencies (E T A I O N S H R D L U) as a guide when deciding what characters to exchange. Step 8: The decoding process can be made more difficult if blank is also coded. That is, consider the alphabet to be 27 letters. Rewrite LetterFreq and your main program to handle blank as another character to code. In the above frequency order, space usually comes first.

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  • Find the shortest path in a graph which visits certain nodes.

    - by dmd
    I have a undirected graph with about 100 nodes and about 200 edges. One node is labelled 'start', one is 'end', and there's about a dozen labelled 'mustpass'. I need to find the shortest path through this graph that starts at 'start', ends at 'end', and passes through all of the 'mustpass' nodes (in any order). ( http://3e.org/local/maize-graph.png / http://3e.org/local/maize-graph.dot.txt is the graph in question - it represents a corn maze in Lancaster, PA)

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  • an x86 question

    - by wide
    i'm working for my exam. i didn't resolved this question. does anyone help me? assume that there are two block, BLOCK1 AND BLOCK2. every block has 50 bytes. write a program to add BLOCK1 with BLOCK2 , and store result to BLOCK2 using LODS, STOS and LOOP etc. assembly commands?

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  • Convert Java program to C

    - by imicrothinking
    I need a bit of guidance with writing a C program...a bit of quick background as to my level, I've programmed in Java previously, but this is my first time programming in C, and we've been tasked to translate a word count program from Java to C that consists of the following: Read a file from memory Count the words in the file For each occurrence of a unique word, keep a word counter variable Print out the top ten most frequent words and their corresponding occurrences Here's the source program in Java: package lab0; import java.io.File; import java.io.FileReader; import java.util.ArrayList; import java.util.Calendar; import java.util.Collections; public class WordCount { private ArrayList<WordCountNode> outputlist = null; public WordCount(){ this.outputlist = new ArrayList<WordCountNode>(); } /** * Read the file into memory. * * @param filename name of the file. * @return content of the file. * @throws Exception if the file is too large or other file related exception. */ public char[] readFile(String filename) throws Exception{ char [] result = null; File file = new File(filename); long size = file.length(); if (size > Integer.MAX_VALUE){ throw new Exception("File is too large"); } result = new char[(int)size]; FileReader reader = new FileReader(file); int len, offset = 0, size2read = (int)size; while(size2read > 0){ len = reader.read(result, offset, size2read); if(len == -1) break; size2read -= len; offset += len; } return result; } /** * Make article word by word. * * @param article the content of file to be counted. * @return string contains only letters and "'". */ private enum SPLIT_STATE {IN_WORD, NOT_IN_WORD}; /** * Go through article, find all the words and add to output list * with their count. * * @param article the content of the file to be counted. * @return words in the file and their counts. */ public ArrayList<WordCountNode> countWords(char[] article){ SPLIT_STATE state = SPLIT_STATE.NOT_IN_WORD; if(null == article) return null; char curr_ltr; int curr_start = 0; for(int i = 0; i < article.length; i++){ curr_ltr = Character.toUpperCase( article[i]); if(state == SPLIT_STATE.IN_WORD){ article[i] = curr_ltr; if ((curr_ltr < 'A' || curr_ltr > 'Z') && curr_ltr != '\'') { article[i] = ' '; //printf("\nthe word is %s\n\n",curr_start); if(i - curr_start < 0){ System.out.println("i = " + i + " curr_start = " + curr_start); } addWord(new String(article, curr_start, i-curr_start)); state = SPLIT_STATE.NOT_IN_WORD; } }else{ if (curr_ltr >= 'A' && curr_ltr <= 'Z') { curr_start = i; article[i] = curr_ltr; state = SPLIT_STATE.IN_WORD; } } } return outputlist; } /** * Add the word to output list. */ public void addWord(String word){ int pos = dobsearch(word); if(pos >= outputlist.size()){ outputlist.add(new WordCountNode(1L, word)); }else{ WordCountNode tmp = outputlist.get(pos); if(tmp.getWord().compareTo(word) == 0){ tmp.setCount(tmp.getCount() + 1); }else{ outputlist.add(pos, new WordCountNode(1L, word)); } } } /** * Search the output list and return the position to put word. * @param word is the word to be put into output list. * @return position in the output list to insert the word. */ public int dobsearch(String word){ int cmp, high = outputlist.size(), low = -1, next; // Binary search the array to find the key while (high - low > 1) { next = (high + low) / 2; // all in upper case cmp = word.compareTo((outputlist.get(next)).getWord()); if (cmp == 0) return next; else if (cmp < 0) high = next; else low = next; } return high; } public static void main(String args[]){ // handle input if (args.length == 0){ System.out.println("USAGE: WordCount <filename> [Top # of results to display]\n"); System.exit(1); } String filename = args[0]; int dispnum; try{ dispnum = Integer.parseInt(args[1]); }catch(Exception e){ dispnum = 10; } long start_time = Calendar.getInstance().getTimeInMillis(); WordCount wordcount = new WordCount(); System.out.println("Wordcount: Running..."); // read file char[] input = null; try { input = wordcount.readFile(filename); } catch (Exception e) { // TODO Auto-generated catch block e.printStackTrace(); System.exit(1); } // count all word ArrayList<WordCountNode> result = wordcount.countWords(input); long end_time = Calendar.getInstance().getTimeInMillis(); System.out.println("wordcount: completed " + (end_time - start_time)/1000000 + "." + (end_time - start_time)%1000000 + "(s)"); System.out.println("wordsort: running ..."); start_time = Calendar.getInstance().getTimeInMillis(); Collections.sort(result); end_time = Calendar.getInstance().getTimeInMillis(); System.out.println("wordsort: completed " + (end_time - start_time)/1000000 + "." + (end_time - start_time)%1000000 + "(s)"); Collections.reverse(result); System.out.println("\nresults (TOP "+ dispnum +" from "+ result.size() +"):\n" ); // print out result String str ; for (int i = 0; i < result.size() && i < dispnum; i++){ if(result.get(i).getWord().length() > 15) str = result.get(i).getWord().substring(0, 14); else str = result.get(i).getWord(); System.out.println(str + " - " + result.get(i).getCount()); } } public class WordCountNode implements Comparable{ private String word; private long count; public WordCountNode(long count, String word){ this.count = count; this.word = word; } public String getWord() { return word; } public void setWord(String word) { this.word = word; } public long getCount() { return count; } public void setCount(long count) { this.count = count; } public int compareTo(Object arg0) { // TODO Auto-generated method stub WordCountNode obj = (WordCountNode)arg0; if( count - obj.getCount() < 0) return -1; else if( count - obj.getCount() == 0) return 0; else return 1; } } } Here's my attempt (so far) in C: #include <stdio.h> #include <stdlib.h> #include <stdbool.h> #include <string.h> // Read in a file FILE *readFile (char filename[]) { FILE *inputFile; inputFile = fopen (filename, "r"); if (inputFile == NULL) { printf ("File could not be opened.\n"); exit (EXIT_FAILURE); } return inputFile; } // Return number of words in an array int wordCount (FILE *filePointer, char filename[]) {//, char *words[]) { // count words int count = 0; char temp; while ((temp = getc(filePointer)) != EOF) { //printf ("%c", temp); if ((temp == ' ' || temp == '\n') && (temp != '\'')) count++; } count += 1; // counting method uses space AFTER last character in word - the last space // of the last character isn't counted - off by one error // close file fclose (filePointer); return count; } // Print out the frequencies of the 10 most frequent words in the console int main (int argc, char *argv[]) { /* Step 1: Read in file and check for errors */ FILE *filePointer; filePointer = readFile (argv[1]); /* Step 2: Do a word count to prep for array size */ int count = wordCount (filePointer, argv[1]); printf ("Number of words is: %i\n", count); /* Step 3: Create a 2D array to store words in the file */ // open file to reset marker to beginning of file filePointer = fopen (argv[1], "r"); // store words in character array (each element in array = consecutive word) char allWords[count][100]; // 100 is an arbitrary size - max length of word int i,j; char temp; for (i = 0; i < count; i++) { for (j = 0; j < 100; j++) { // labels are used with goto statements, not loops in C temp = getc(filePointer); if ((temp == ' ' || temp == '\n' || temp == EOF) && (temp != '\'') ) { allWords[i][j] = '\0'; break; } else { allWords[i][j] = temp; } printf ("%c", allWords[i][j]); } printf ("\n"); } // close file fclose (filePointer); /* Step 4: Use a simple selection sort algorithm to sort 2D char array */ // PStep 1: Compare two char arrays, and if // (a) c1 > c2, return 2 // (b) c1 == c2, return 1 // (c) c1 < c2, return 0 qsort(allWords, count, sizeof(char[][]), pstrcmp); /* int k = 0, l = 0, m = 0; char currentMax, comparedElement; int max; // the largest element in the current 2D array int elementToSort = 0; // elementToSort determines the element to swap with starting from the left // Outer a iterates through number of swaps needed for (k = 0; k < count - 1; k++) { // times of swaps max = k; // max element set to k // Inner b iterates through successive elements to fish out the largest element for (m = k + 1; m < count - k; m++) { currentMax = allWords[k][l]; comparedElement = allWords[m][l]; // Inner c iterates through successive chars to set the max vars to the largest for (l = 0; (currentMax != '\0' || comparedElement != '\0'); l++) { if (currentMax > comparedElement) break; else if (currentMax < comparedElement) { max = m; currentMax = allWords[m][l]; break; } else if (currentMax == comparedElement) continue; } } // After max (count and string) is determined, perform swap with temp variable char swapTemp[1][20]; int y = 0; do { swapTemp[0][y] = allWords[elementToSort][y]; allWords[elementToSort][y] = allWords[max][y]; allWords[max][y] = swapTemp[0][y]; } while (swapTemp[0][y++] != '\0'); elementToSort++; } */ int a, b; for (a = 0; a < count; a++) { for (b = 0; (temp = allWords[a][b]) != '\0'; b++) { printf ("%c", temp); } printf ("\n"); } // Copy rows to different array and print results /* char arrayCopy [count][20]; int ac, ad; char tempa; for (ac = 0; ac < count; ac++) { for (ad = 0; (tempa = allWords[ac][ad]) != '\0'; ad++) { arrayCopy[ac][ad] = tempa; printf("%c", arrayCopy[ac][ad]); } printf("\n"); } */ /* Step 5: Create two additional arrays: (a) One in which each element contains unique words from char array (b) One which holds the count for the corresponding word in the other array */ /* Step 6: Sort the count array in decreasing order, and print the corresponding array element as well as word count in the console */ return 0; } // Perform housekeeping tasks like freeing up memory and closing file I'm really stuck on the selection sort algorithm. I'm currently using 2D arrays to represent strings, and that worked out fine, but when it came to sorting, using three level nested loops didn't seem to work, I tried to use qsort instead, but I don't fully understand that function as well. Constructive feedback and criticism greatly welcome (...and needed)!

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  • File processing-Haskell

    - by Martinas Maria
    How can I implement in haskell the following: I receive an input file from the command line. This input file contains words separated with tabs,new lines and spaces.I have two replace this elements(tabs,new lines and spaces) with comma(,) .Observation:more newlines,tabs,spaces will be replaced with a single comma.The result has to be write in a new file(output.txt). Please help me with this.My haskell skills are very scarse. This is what I have so far: processFile::String->String processFile [] =[] processFile input =input process :: String -> IO String process fileName = do text <- readFile fileName return (processFile text) main :: IO () main = do n <- process "input.txt" print n In processFile function I should process the text from the input file. I'm stuck..Please help.

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  • Haskell - function (that returns a list) on each element in a list

    - by Ben
    The assignment is to create a multiples function and I essentially want todo the following code: map (\t -> scanl (\x y -> x+y) t (repeat t)) listofnumbers The problem is that the scanl function returns a list of results rather than the one which the map function requires. So is there a function that will allow the return of lists?

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  • Why does DataInputStream not support integers?

    - by Jason
    I need to read in a list of numbers from a file, none of which are larger than 32767. Originally I was going to use the Scanner class to pull in the data, then I read about DataInputStream. This would work well for me, except that according to the API, it supports all primitive variables EXCEPT ints! Listed are longs, shorts, bytes, chars, booleans, ect, but no ints. I have no need for double precision from the incoming data. Is this a deliberate or unintentional oversight?

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  • dynamic programming: speeding up this function

    - by aristotaly
    i have this program //h is our N static int g=0; int fun(int h){ if(h<=0){ g++; return g; } return g+fun(h-1)+fun(h-4); } is it possible to speed it up using dynamic programming i fugured out this function runs in O(2^n) it means that i suppose to reduce this time but the trouble is that i don get the idea of dinamic programming even a leading hint or a useful link to a resource will do it is a work assingment i do not ask for the solution :) just asking for the right direction

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  • Balancing heuristics (for timetable problem)

    - by genesiss
    I'm writing a genetic algorithm for generating timetables. At the moment I'm using these two heuristics: Number of holes between lectures in one day (related) (less holes - bigger score) Each hour has some value, so for each timetable I sum values for hours when lectures are on. (lectures at more appropriate hours - bigger score) I want to balance these two heuristics, so the algorithm wouldn't favor neither one. What would be the best way to achieve this?

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  • Difference between Logarithmic and Uniform cost criteria

    - by Marthin
    I'v got some problem to understand the difference between Logarithmic(Lcc) and Uniform(Ucc) cost criteria and also how to use it in calculations. Could someone please explain the difference between the two and perhaps show how to calculate the complexity for a problem like A+B*C (Yes this is part of an assignment =) ) Thx for any help! /Marthin

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  • Java Clock Assignment

    - by Mike S
    For my assignment we are suppose to make a clock. We need variables of hours, minutes, and seconds and methods like setHours/getHours, setMinutes/getMinutes, setSeconds/getSeconds. Now the parts of the assignment that I am having trouble on is that we need a addClock() method to make the sum of two clock objects and a tickDown() method which decrements the clock object and a tick() method that increments a Clock object by one second. Lastly, the part where I am really confused on is, I need to write a main() method in the Clock class to test the functionality of your objects with a separate Tester class with a main() method. Here is what I have so far... public class Clock { private int hr; //store hours private int min; //store minutes private int sec; //store seconds //Default constructor public Clock () { setClock (0, 0, 0); } public Clock (int hours, int minutes, int seconds) { setTimes (hours, minute, seconds); } public void setClock (int hours, int minutes, int seconds) { if(0 <= hours && hours < 24) { hr = hours; } else { hr = 0; } if(0 <= minutes && minutes < 60) { min = minutes; } else { min = 0; } if(0 <= seconds && seconds < 60) { sec = seconds; } else { sec = 0; } } public int getHours ( ) { return hr; } public int getMinutes ( ) { return min; } public int getSeconds ( ) { return sec; } //Method to increment the time by one second //Postcondition: The time is incremented by one second //If the before-increment time is 23:59:59, the time //is reset to 00:00:00 public void tickSeconds ( ) { sec++; if(sec > 59) { sec = 0; tickMinutes ( ); //increment minutes } } public void tickMinutes() { min++; If (min > 59) { min = 0; tickHours(); //increment hours } } public void tickHours() { hr++; If (hr > 23) hr = 0; } }

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  • Quickest way to write to file in java

    - by user1097772
    I'm writing an application which compares directory structure. First I wrote an application which writes gets info about files - one line about each file or directory. My soulution is: calling method toFile Static PrintWriter pw = new PrintWriter(new BufferedWriter( new FileWriter("DirStructure.dlis")), true); String line; // info about file or directory public void toFile(String line) { pw.println(line); } and of course pw.close(), at the end. My question is, can I do it quicker? What is the quickest way? Edit: quickest way = quickest writing in the file

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  • graph algorithms

    - by davit-datuashvili
    now one ask please help me to write a few graph algorithms for example http://en.wikipedia.org/wiki/Dijkstra%27s_algorithm there is given such graph my problem is that i want implement graph algorithms on arrays can anybody help me to imlement ddijkstra algorithm on array i want to see one example because it is difficulty for me to understand this pseudocodes which is in internet i mean classes edges and so on please help me

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  • Display numbers from 1 to 100 without loops or conditions

    - by Harsha
    Is there a way to print numbers from 1 to 100 without using any loops or conditions like "if"? We can easily do using recursion but that again has an if condition. Is there a way to do without using "if" as well? Also no repetitive print statements,or a single print statement containing all the numbers from 1 to 100. A solution in Java is preferable.

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  • Infinite loop when adding a row to a list in a class in python3

    - by Margaret
    I have a script which contains two classes. (I'm obviously deleting a lot of stuff that I don't believe is relevant to the error I'm dealing with.) The eventual task is to create a decision tree, as I mentioned in this question. Unfortunately, I'm getting an infinite loop, and I'm having difficulty identifying why. I've identified the line of code that's going haywire, but I would have thought the iterator and the list I'm adding to would be different objects. Is there some side effect of list's .append functionality that I'm not aware of? Or am I making some other blindingly obvious mistake? class Dataset: individuals = [] #Becomes a list of dictionaries, in which each dictionary is a row from the CSV with the headers as keys def field_set(self): #Returns a list of the fields in individuals[] that can be used to split the data (i.e. have more than one value amongst the individuals def classified(self, predicted_value): #Returns True if all the individuals have the same value for predicted_value def fields_exhausted(self, predicted_value): #Returns True if all the individuals are identical except for predicted_value def lowest_entropy_value(self, predicted_value): #Returns the field that will reduce <a href="http://en.wikipedia.org/wiki/Entropy_%28information_theory%29">entropy</a> the most def __init__(self, individuals=[]): and class Node: ds = Dataset() #The data that is associated with this Node links = [] #List of Nodes, the offspring Nodes of this node level = 0 #Tree depth of this Node split_value = '' #Field used to split out this Node from the parent node node_value = '' #Value used to split out this Node from the parent Node def split_dataset(self, split_value): fields = [] #List of options for split_value amongst the individuals datasets = {} #Dictionary of Datasets, each one with a value from fields[] as its key for field in self.ds.field_set()[split_value]: #Populates the keys of fields[] fields.append(field) datasets[field] = Dataset() for i in self.ds.individuals: #Adds individuals to the datasets.dataset that matches their result for split_value datasets[i[split_value]].individuals.append(i) #<---Causes an infinite loop on the second hit for field in fields: #Creates subnodes from each of the datasets.Dataset options self.add_subnode(datasets[field],split_value,field) def add_subnode(self, dataset, split_value='', node_value=''): def __init__(self, level, dataset=Dataset()): My initialisation code is currently: if __name__ == '__main__': filename = (sys.argv[1]) #Takes in a CSV file predicted_value = "# class" #Identifies the field from the CSV file that should be predicted base_dataset = parse_csv(filename) #Turns the CSV file into a list of lists parsed_dataset = individual_list(base_dataset) #Turns the list of lists into a list of dictionaries root = Node(0, Dataset(parsed_dataset)) #Creates a root node, passing it the full dataset root.split_dataset(root.ds.lowest_entropy_value(predicted_value)) #Performs the first split, creating multiple subnodes n = root.links[0] n.split_dataset(n.ds.lowest_entropy_value(predicted_value)) #Attempts to split the first subnode.

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  • SQL Query with computed column

    - by plotnick
    help me please with a query. Assume that we have a table with columns: Transaction StartTime EndTime Now, I need a query with computed column of (value = EndTime-Startime). Actually I need to group Users(Transaction has a FK for Users) and sort them by average time spent for transaction.

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  • Finding the average of two number using classes and methods

    - by Have alook
    I want to use methods inside class. Q: find the average of two number using classes and methods. import java.util.*; class aaa { int a,b,sum,avrg; void average() { System.out.println("The average is ="+avrg); avrg=(sum/2); } } class ave { public static void main(String args[]){ aaa n=new aaa(); Scanner m=new Scanner(System.in); System.out.println("write two number"); n.a=m.nextInt(); n.b=m.nextInt(); n.average(); } }

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  • how to pass an array into an function and in the function count how many numbers are in a range?

    - by user320950
    #include <iostream> #include <fstream> using namespace std; int calculate_total(int exam1[], int exam2[], int exam3[]); // function that calcualates grades to see how many 90,80,70,60 int exam1[100];// array that can hold 100 numbers for 1st column int exam2[100];// array that can hold 100 numbers for 2nd column int exam3[100];// array that can hold 100 numbers for 3rd column // here i am passing an array into the function calcualate_total int calculate_total(exam1[],exam2[],exam3[]) { int above90=0, above80=0, above70=0, above60=0; if((num<=90) && (num >=100)) { above90++; { if((num<=80) && (num >=89)) { above80++; { if((num<=70) && (num >=79)) { above70++; { if((num<=60) && (num >=69)) { above60++; } } } } } } } }

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  • Why isn't this file reading/writing program working?

    - by user320950
    This program is supposed to read files and write them. I took the file open checks out because they kept causing errors. The problem is that the files open like they are supposed to and the names are correct but nothing is on any of the text screens. Do you know what is wrong? #include<iostream> #include<fstream> #include<cstdlib> #include<iomanip> using namespace std; int main() { ifstream in_stream; // reads itemlist.txt ofstream out_stream1; // writes in items.txt ifstream in_stream2; // reads pricelist.txt ofstream out_stream3;// writes in plist.txt ifstream in_stream4;// read recipt.txt ofstream out_stream5;// write display.txt float price=' ',curr_total=0.0; int wrong=0; int itemnum=' '; char next; in_stream.open("ITEMLIST.txt", ios::in); // list of avaliable items out_stream1.open("listWititems.txt", ios::out); // list of avaliable items in_stream2.open("PRICELIST.txt", ios::in); out_stream3.open("listWitdollars.txt", ios::out); in_stream4.open("display.txt", ios::in); out_stream5.open("showitems.txt", ios::out); in_stream.close(); // closing files. out_stream1.close(); in_stream2.close(); out_stream3.close(); in_stream4.close(); out_stream5.close(); system("pause"); in_stream.setf(ios::fixed); while(in_stream.eof()) { in_stream >> itemnum; cin.clear(); cin >> next; } out_stream1.setf(ios::fixed); while (out_stream1.eof()) { out_stream1 << itemnum; cin.clear(); cin >> next; } in_stream2.setf(ios::fixed); in_stream2.setf(ios::showpoint); in_stream2.precision(2); while((price== (price*1.00)) && (itemnum == (itemnum*1))) { while (in_stream2 >> itemnum >> price) // gets itemnum and price { while (in_stream2.eof()) // reads file to end of file { in_stream2 >> itemnum; in_stream2 >> price; price++; curr_total= price++; in_stream2 >> curr_total; cin.clear(); // allows more reading cin >> next; } } } out_stream3.setf(ios::fixed); out_stream3.setf(ios::showpoint); out_stream3.precision(2); while((price== (price*1.00)) && (itemnum == (itemnum*1))) { while (out_stream3 << itemnum << price) { while (out_stream3.eof()) // reads file to end of file { out_stream3 << itemnum; out_stream3 << price; price++; curr_total= price++; out_stream3 << curr_total; cin.clear(); // allows more reading cin >> next; } return itemnum, price; } } in_stream4.setf(ios::fixed); in_stream4.setf(ios::showpoint); in_stream4.precision(2); while ( in_stream4.eof()) { in_stream4 >> itemnum >> price >> curr_total; cin.clear(); cin >> next; } out_stream5.setf(ios::fixed); out_stream5.setf(ios::showpoint); out_stream5.precision(2); out_stream5 <<setw(5)<< " itemnum " <<setw(5)<<" price "<<setw(5)<<" curr_total " <<endl; // sends items and prices to receipt.txt out_stream5 << setw(5) << itemnum << setw(5) <<price << setw(5)<< curr_total; // sends items and prices to receipt.txt out_stream5 << " You have a total of " << wrong++ << " errors " << endl; }

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  • VB.net Network Graph code/algorithm

    - by Jens
    For a school project we need to visualise a computer network graph. The number of computers with specific properties are read from an XML file, and then a graph should be created. Ad random computers are added and removed. Is there any open source project or algorithm that could help us visualising this in VB.net? Or would you suggest us to switch to java. Update: We eventually switched java and used the Jung libraries because this was easier for us to understand and implement.

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  • Incorrect logic flow? function that gets coordinates for a sudoku game

    - by igor
    This function of mine keeps on failing an autograder, I am trying to figure out if there is a problem with its logic flow? Any thoughts? Basically, if the row is wrong, "invalid row" should be printed, and clearInput(); called, and return false. When y is wrong, "invalid column" printed, and clearInput(); called and return false. When both are wrong, only "invalid row" is to be printed (and still clearInput and return false. Obviously when row and y are correct, print no error and return true. My function gets through most of the test cases, but fails towards the end, I'm a little lost as to why. bool getCoords(int & x, int & y) { char row; bool noError=true; cin>>row>>y; row=toupper(row); if(row>='A' && row<='I' && isalpha(row) && y>=1 && y<=9) { x=row-'A'; y=y-1; return true; } else if(!(row>='A' && row<='I')) { cout<<"Invalid row"<<endl; noError=false; clearInput(); return false; } else { if(noError) { cout<<"Invalid column"<<endl; } clearInput(); return false; } }

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  • mv() while reading

    - by K'
    on Linux ext3 filesystem, what happens if mv() is called on the same file (file descriptor) while reading the file? It is actually an exam question and I can only say something like: CPU traps OS for interrupt handling etc, etc. I would appreciate if OS guys out there can help me out, please :D

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