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Stella [2.4K]
3 years ago
11

How do i build a supercomputer.?

Computers and Technology
1 answer:
stira [4]3 years ago
3 0
You will need one head node, at least a dozen identical compute nodes, an Ethernet switch, a power distribution unit, and a rack. Determine the electrical demand, cooling and space required. Also decide on what IP address you want for your private networks, what to name the nodes, what software packages you want installed, and what technology you want to provide the parallel computing capabilities
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To provide for unobtrusive validation, you can install the ____________________ package for unobtrusive validation.
NARA [144]

Answer:

AspNet.ScriptManager.jQuery?

Explanation:

Unobtrusive validation means we can perform a simple client-side validation without writing a lot of validation code by adding suitable attributes and also by including the suitable script files.

One of the benefits of using a unobtrusive validation is that it help to reduce the amount of the Java script that is generated. We can install the AspNet.ScriptManager.jQuery? for the unobtrusive validation.

When the unobtrusive validation is used, validation of the client is being performed by using a JavaScript library.

4 0
3 years ago
The best time to visit a college is: A. during spring break. B. on a weekend. C. when the college is in session. D. during winte
GaryK [48]
When the college is in session
5 0
3 years ago
Read 2 more answers
Flesh out the body of the print_seconds function so that it prints the total amount of seconds given the hours, minutes, and sec
fiasKO [112]

Answer:

Step by step explanation along with code and output is provided below

Explanation:

#include<iostream>

using namespace std;

// print_seconds function that takes three input arguments hours, mints, and seconds. There are 60*60=3600 seconds in one hour and 60 seconds in a minute. Total seconds will be addition of these three  

void print_seconds(int hours, int mints, int seconds)

{

   int total_seconds= hours*3600 + mints*60 + seconds;

   cout<<"Total seconds are: "<<total_seconds<<endl;

}

// test code

// user inputs hours, minutes and seconds and can also leave any of them by  entering 0 that will not effect the program. Then function print_seconds is called to calculate and print the total seconds.

int main()

{

   int h,m,s;

   cout<<"enter hours if any or enter 0"<<endl;

   cin>>h;

   cout<<"enter mints if any or enter 0"<<endl;

   cin>>m;

   cout<<"enter seconds if any or enter 0"<<endl;

   cin>>s;

   print_seconds(h,m,s);

  return 0;

}

Output:

enter hours if any or enter 0

2

enter mints if any or enter 0

25

enter seconds if any or enter 0

10

Total seconds are: 8710

8 0
3 years ago
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
When two files are linked together, the __________ file receives the data from another workbook.?
Ket [755]
When two files are linked together, the <u>d</u><span><u>estination </u></span>file receives the data from another workbook.

If i'm not wrong.

3 0
3 years ago
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