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3 years ago

Which technology concept uses computer resources from multiple locations to solve a common problem?

Computers and Technology

1 answer:

umka2103 [35]3 years ago

5 0

Answer:

Distributed memory systems

Distributed memory systems use multiple computers to solve a common problem, with computation distributed among the connected computers (nodes) and using message-passing to communicate between the nodes.

Explanation:

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3 years ago

Which type of portfolio might a young investor who is not afraid of risk choose?

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3 years ago

Write a program FindDuplicate.java that reads n integer arguments from the command line into an integer array of length n, where

kozerog [31]

Answer:

public class FindDuplicate{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int n = 5;
int arr[] = new int[n];
for(int i=0; i < arr.length; i++){
int inputNum = input.nextInt();
if(inputNum >=1 && inputNum <=n) {
arr[i] = inputNum;
}
}
for(int j =0; j < arr.length; j++){
for(int k = 0; k < arr.length; k++){
if(j == k){
continue;
}else{
if(arr[j] == arr[k]){
System.out.println("True");
return;
}
}
}
}
System.out.println("False");
}
}

Explanation:

Firstly, create a Scanner object to get user input (Line 4).

Next, create an array with n-size (Line 7) and then create a for-loop to get user repeatedly enter an integer and assign the input value to the array (Line 9 - 14).

Next, create a double layer for-loop to check the each element in the array against the other elements to see if there is any duplication detected and display "True" (Line 21 - 22). If duplication is found the program will display True and terminate the whole program using return (Line 23). The condition set in Line 18 is to ensure the comparison is not between the same element.

If all the elements in the array are unique the if block (Line 21 - 23) won't run and it will proceed to Line 28 to display message "False".

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2 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

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3 0

1 year ago

Which technology concept uses computer resources from multiple locations to solve a common problem?​

Which technology concept uses computer resources from multiple locations to solve a common problem?