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mote1985 [20]
2 years ago
9

Wesley purchased a word-processing software program. He used it for a year, during which he got regular updates every two months

. After a year, he was not allowed to update the software. However, he could continue using it. Why did the updates stop?
Computers and Technology
1 answer:
natka813 [3]2 years ago
3 0

Group of answer choices.

A. The software was corrupt and resulted in a bug.

B. He purchased a license with maintenance for a year.

C. The organization discontinued the software.

D. He purchased an open-source license.

E. He purchase a perpetual non-maintenance license.

Answer:

B. He purchased a license with maintenance for a year.

Explanation:

A software can be defined as a set of executable instructions (codes) or collection of data that is used typically to instruct a computer how to perform a specific task and to solve a particular problem.

Basically, softwares are categorized into two (2) main categories and these are;

I. Open-source software.

II. Proprietary software.

A proprietary software is also known as a closed-source software and it can be defined as any software application or program that has its source code copyrighted and as such cannot be used, modified or distributed without authorization from the software developer. Thus, it is typically published as a commercial software that may be sold, licensed or leased by the software developer (vendor) to the end users with terms and conditions.

In this scenario, Wesley purchased a word-processing software program. He used it for a year, during which he got regular updates every two months. However, after a year, he was not allowed to update the software but he could continue using it.

This ultimately implies that, Wesley purchased a licensed software with maintenance for a year and as such he would stop receiving an update from the software developer after his subscription expired.

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a cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits such that the giv
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Using the knowledge in computational language in C++ it is possible to write a code that  cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits

<h3>Writting the code:</h3>

<em>#include <bits/stdc++.h></em>

<em>using namespace std;</em>

<em>// chracter to digit mapping, and the inverse</em>

<em>// (if you want better performance: use array instead of unordered_map)</em>

<em>unordered_map<char, int> c2i;</em>

<em>unordered_map<int, char> i2c;</em>

<em>int ans = 0;</em>

<em>// limit: length of result</em>

<em>int limit = 0;</em>

<em>// digit: index of digit in a word, widx: index of a word in word list, sum: summation of all word[digit]  </em>

<em>bool helper(vector<string>& words, string& result, int digit, int widx, int sum) { </em>

<em>    if (digit == limit) {</em>

<em>        ans += (sum == 0);</em>

<em>        return sum == 0;</em>

<em>    }</em>

<em>    // if summation at digit position complete, validate it with result[digit].</em>

<em>    if (widx == words.size()) {</em>

<em>        if (c2i.count(result[digit]) == 0 && i2c.count(sum%10) == 0) {</em>

<em>            if (sum%10 == 0 && digit+1 == limit) // Avoid leading zero in result</em>

<em>                return false;</em>

<em>            c2i[result[digit]] = sum % 10;</em>

<em>            i2c[sum%10] = result[digit];</em>

<em>            bool tmp = helper(words, result, digit+1, 0, sum/10);</em>

<em>            c2i.erase(result[digit]);</em>

<em>            i2c.erase(sum%10);</em>

<em>            ans += tmp;</em>

<em>            return tmp;</em>

<em>        } else if (c2i.count(result[digit]) && c2i[result[digit]] == sum % 10){</em>

<em>            if (digit + 1 == limit && 0 == c2i[result[digit]]) {</em>

<em>                return false;</em>

<em>            }</em>

<em>            return helper(words, result, digit+1, 0, sum/10);</em>

<em>        } else {</em>

<em>            return false;</em>

<em>        }</em>

<em>    }</em>

<em>    // if word[widx] length less than digit, ignore and go to next word</em>

<em>    if (digit >= words[widx].length()) {</em>

<em>        return helper(words, result, digit, widx+1, sum);</em>

<em>    }</em>

<em>    // if word[widx][digit] already mapped to a value</em>

<em>    if (c2i.count(words[widx][digit])) {</em>

<em>        if (digit+1 == words[widx].length() && words[widx].length() > 1 && c2i[words[widx][digit]] == 0) </em>

<em>            return false;</em>

<em>        return helper(words, result, digit, widx+1, sum+c2i[words[widx][digit]]);</em>

<em>    }</em>

<em>    // if word[widx][digit] not mapped to a value yet</em>

<em>    for (int i = 0; i < 10; i++) {</em>

<em>        if (digit+1 == words[widx].length() && i == 0 && words[widx].length() > 1) continue;</em>

<em>        if (i2c.count(i)) continue;</em>

<em>        c2i[words[widx][digit]] = i;</em>

<em>        i2c[i] = words[widx][digit];</em>

<em>        bool tmp = helper(words, result, digit, widx+1, sum+i);</em>

<em>        c2i.erase(words[widx][digit]);</em>

<em>        i2c.erase(i);</em>

<em>    }</em>

<em>    return false;</em>

<em>}</em>

<em>void isSolvable(vector<string>& words, string result) {</em>

<em>    limit = result.length();</em>

<em>    for (auto &w: words) </em>

<em>        if (w.length() > limit) </em>

<em>            return;</em>

<em>    for (auto&w:words) </em>

<em>        reverse(w.begin(), w.end());</em>

<em>    reverse(result.begin(), result.end());</em>

<em>    int aa = helper(words, result, 0, 0, 0);</em>

<em>}</em>

<em />

<em>int main()</em>

<em>{</em>

<em>    ans = 0;</em>

<em>    vector<string> words={"GREEN" , "BLUE"} ;</em>

<em>    string result = "BLACK";</em>

<em>    isSolvable(words, result);</em>

<em>    cout << ans << "\n";</em>

<em>    return 0;</em>

<em>}</em>

See more about C++ code at brainly.com/question/19705654

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2 years ago
Write a method called findNames that meets the following specs: It takes two arguments: a list of strings, allNames, and a strin
Anna11 [10]

Answer:

public class Solution {

   public static void main(String args[]) {

       String[] allNames = new String[]{"Bob Smith", "Elroy Jetson", "Christina Johnson", "Rachael Baker", "cHRis", "Chris Conly"};

       String searchString = "cHRis";

       

       findNames(allNames, searchString);

   }

   

   public static void findNames(String[] listOfName, String nameToFind){

       ArrayList<String> resultName = new ArrayList<String>();

       

       for(String name : listOfName){

           if (name.toLowerCase().contains(nameToFind.toLowerCase())){

               resultName.add(name);

           }

       }

       

       for(String result : resultName){

           System.out.println(result);

       }

   }

}

Explanation:

The class was created called Solution. The second line define the main function. Inside the main function; we initialized and assign an array called allNames to hold the list of all name. Then a String called searchString was also defined. The searchString is the string to search for in each element of allNames array. The two variables (allNames and searchString) are passed as argument to the findNames method when it is called.

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Inside the findNames method, we initialized and assigned an ArrayList called resultName. The resultName variable is to hold list of element found that contain the searchString.

The first for-loop goes through the elements in the allNames array and compare it with the searchString. If any element is found containing the searchString; it is added to the resultName variable.

The second for-loop goes through the elements of the resultName array and output it. The output is empty if no element was found added to the resultName variable.

During comparison, the both string were converted to lower case before the comparison because same lowercase character does not equal same uppercase character. For instance 'A' is not same as 'a'.

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