The best ways to navigate through your computer? a)Start button b)My Computer c)Windows explorer d)All of the above

. Consider the problem of finding the largest element in a list of n elements. What will be the basic operation of an algorithm

natali 33 [55]

Answer:

Comparison.

Explanation:

When we have to find a largest in a list of n elements.First we have to iterate over the list so we can access all the elements of the list in one go.Then to find the largest element in the list we have to initialize a variable outside the loop with the minimum value possible and in the loop compare each element with this value,if the element is greater than the variable assign the element to the variable.Then the loop will find the largest element and it will be the variable.

4 0

3 years ago

Creating parts that are interchangeable in manufacturing terms is key to the ____ line

denis-greek [22]

Answer:

Assembly line.

Explanation:

The interchangeable parts was a game-changing concept for the manufacturing industry during the Industrial Revolution.

It was first introduced by Eli Whitney, also the inventor of the Cotton Gin, and later was perfected by Henry Ford, who was the first to create a continuous moving assembly line. The Interchangeable parts are identical pieces created from a master model and are so similar to each other, that they can fit into any line of production of the same kind.

Thanks to these advances, the manufacturing process across all industries could be now faster, more cost-efficient, and profitable.

7 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>

#include <bits/stdc++.h>

using namespace std;

// chracter to digit mapping, and the inverse

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

unordered_map<char, int> c2i;

unordered_map<int, char> i2c;

int ans = 0;

// limit: length of result

int limit = 0;

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

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

if (digit == limit) {

ans += (sum == 0);

return sum == 0;

}

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

if (widx == words.size()) {

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

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

return false;

c2i[result[digit]] = sum % 10;

i2c[sum%10] = result[digit];

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

c2i.erase(result[digit]);

i2c.erase(sum%10);

ans += tmp;

return tmp;

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

if (digit + 1 == limit && 0 == c2i[result[digit]]) {

return false;

}

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

} else {

return false;

}

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

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

return helper(words, result, digit, widx+1, sum);

}

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

if (c2i.count(words[widx][digit])) {

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

return false;

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

}

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

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

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

if (i2c.count(i)) continue;

c2i[words[widx][digit]] = i;

i2c[i] = words[widx][digit];

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

c2i.erase(words[widx][digit]);

i2c.erase(i);

}

return false;

}

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

limit = result.length();

for (auto &w: words)

if (w.length() > limit)

return;

for (auto&w:words)

reverse(w.begin(), w.end());

reverse(result.begin(), result.end());

int aa = helper(words, result, 0, 0, 0);

}

int main()

{

ans = 0;

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

string result = "BLACK";

isSolvable(words, result);

cout << ans << "\n";

return 0;

}

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

#SPJ1

3 0

2 years ago

g You and your friend both send documents to the printer, but the printer can only print one document at a time. What does it do

Sedbober [7]

Answer:

It places the document in a buffer

Explanation:

Since the printer can only print a document at a time, the other document is going to be placed in a buffer.

The print buffer can be described as a location in memory that has the function of holding data that is going to be sent to a computers printer. A print job such as in this scenario may remain in the buffer because another document is to be printed first.

Sometimes a document could be put in the buffer because the computer is waiting for the printer to respond. This is just another reason why documents are placed in a buffer.

5 0

3 years ago

What did czarnowski and triantafyllou learn from observing boat propulsion systems?

noname [10]

Czarnowski and Triantafyllou learned that boat propellers are not very efficient, except penguin propulsion systems.

7 0

3 years ago

The best ways to navigate through your computer? a)Start button b)My Computer c)Windows explorer d)All of the above​

The best ways to navigate through your computer? a)Start button b)My Computer c)Windows explorer d)All of the above