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ryzh [129]
3 years ago
10

Clarissa is a computer programming consultant. When a client states that they have an extremely complex programming task, which

of these programming approaches is Clarissa most likely to recommend?
A. converting from binary to decimal
B. creating algorithms
C. a top-down approach
D. a bottom-up approach

I'm looking for an answer my last question that was answered a lot was just people taking my points. I lost almost 300 points literally just because of that so I will report you if you don't put an answer.
Thank you
Computers and Technology
2 answers:
ANEK [815]3 years ago
7 0
The answer indeed is D it makes more sense into that consideration
Ann [662]3 years ago
5 0

Answer:i think its D it makes more sense also can i get brainliest

Explanation:

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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
Leokris [45]

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

#SPJ1

3 0
2 years ago
Intranet giving unlimited information <br><br>True<br><br>False​
leva [86]

Answer:

true

Explanation:

An intranet is a computer network for sharing information, collaboration tools, operational systems, and other computing services within an organization, usually to the exclusion of access by outsiders. ... Intranets can also be used to facilitate working in groups or via teleconferences.

7 0
2 years ago
Which group on the Home Ribbon allows you to add shapes to a PowerPoint slide?
skad [1K]
It is under the drawings group.  
8 0
3 years ago
Read 2 more answers
Alright whoever can answer these the quickest will be marked the brainliest
Levart [38]

The number of pixels displayed on the screen is known as screen resolution.

The coding applied to a text document to change it into an HTML is HTTP

Scripting language for webpages is Javascript.

Hypertext markup language is HTML

The address, or the item that specifies where on the Internet a site can be found is the URL.

--

Hope this helped :)


4 0
3 years ago
Which version of Windows 10 is better?
IrinaVladis [17]

Answer:

Windows 10 has diverse editions and your purpose of use.

Explanation:

Windows 10 Home edition is consumer-focused desktop edition and good for home use alone.

Windows 10 Mobile edition is designed to meet and deliver the best user experience on smaller, mobile, touch-centric devices like smartphones and small tablets.

Windows 10 Pro is also a desktop edition for PCs, tablets. It has upon both it, the familiar and innovative features of Windows 10 Home, it has many extra features designed to meet the various needs of small businesses.

Windows 10 Enterprise builds on Windows 10 Pro, adding advanced features designed to meet the demands of medium and large sized organizations.

This  also supports the broadest a very wide range of options for operating system deployment but on a LIVE server or a Test server and equally have a comprehensive device and app management added.

8 0
2 years ago
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