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yarga [219]
3 years ago
14

2. Create a file with the follow integer/string content and save it as fun.txt. 6 fun. 3 hello 10 <> 2 25 4 wow! Write an

interactive program called PrintStrings that: • Generates file name based on the file path entered by the user • Then read the file fun.txt and produce the following output on the console. fun.fun.fun.fun.fun.fun. hellohellohello <><><><><><><><><><> 2525 wow!wow!wow!wow! Notice that there is one line of output for each integer/string pair. The first line has 6 occurrences of "fun.", the second line has 3 occurrences of "hello", the third line has 10 occurrences of "<>", the fourth line has 2 occurrences of "25" the fifth line has 4 occurrences of "wow!". Notice that there are no extra spaces included in the output. You are to exactly reproduce the format of this sample output. You may assume that the input values always come in pairs with an integer followed by a String (which could be numeric, such as "25" above).

Computers and Technology
1 answer:
Nesterboy [21]3 years ago
8 0

Answer:

see explaination

Explanation:

I made use of python program to solve this.

text file name with fun.txt.

6 fun. 3 hello 10 <> 2 25 4 wow!

Program code:

import re

file = open('fun.txt','r') #for reading file

strings="" #declaring empty string

for k in file:

strings=strings+k #all character in file is storing in strings variable for do operations

pattern = '\s' #for pattern \s is for space

result = re.split(pattern, strings) #split the string with space

for k in range(len(result)): #loop through the list of string

if(k%2) == 0: #checking for integer to time of string

p=int(result[k])

print(result[k+1] *p) #print times of the string(k+1)

Check attachment for output

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The OSI model is a useful tool in troubleshooting a network because it enables you to isolate a problem to a particular software
lutik1710 [3]

Answer:

1)  Layer 4, transport

2) Layer 2, datalink

3) Layer 5, session

4) Layer 7, application

Description of Problems are as below:

1) One of your servers has been exhibiting sluggish network performance. you use a network-monitoring program to try to evaluate the problem. You find considerable TCP retries occurring because the server is being overwhelmed by data, and packets are being discarded.

2) You check some statistics generated by a network-monitoring program and discovers that an abnormally high number of CRC errors were detected. (Hint think of the cause of CRC errors).

3) A user is trying to connect to another computer, but the logon attempt is continually rejected.

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Explanation:

OSI Model is a reference model to determine how applications communicate over a network. OSI consists of seven layers, and each layer performs a particular network function. The seven layers are:

   Layer 7 - Application.

   Layer 6 - Presentation.

   Layer 5 - Session.

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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
Some people use the term _____ to refer to the case that contains and protects the motherboard, internal hard drive, memory, and
Volgvan

Answer:

System unit.

Explanation:

Some people use the term system unit to refer to the case that contains and protects the motherboard, internal hard drive, memory, and other electronic components of the computer from damage.

A system unit also referred to as chassis or tower can be defined as a hardware case that is typically used as a protective case for the main component or primary devices such as power supply, random access memory (RAM), graphics card, CD-ROM drive, Harddisk drive, motherboard, internal cables, central processing unit (CPU) that makes up a computer system.

Basically, the main purpose of a system unit is to properly house the main components of a computer, serve as an electrical insulator and to prevent them from any form of damage.

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