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3 years ago

What is the 5 basic steps of computer programing?

1 answer:

7 0

1. Learning Languages such as C+, VB, etc.
2. Gather data, proficient websites that will come to your need.
3. Install certain softwares to see it in action.
4. Do lots and lots of studying and research.
5. Ask for help, the Internet is here.

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Which line of code will only allow a symbol to be stored in a variable?

nataly862011 [7]

<h2>Answer:</h2><h2></h2><h2>phone = int(input("What is your phone number?"))</h2><h2></h2><h2>Hope this helps, have a great day, stay safe, and positive!!</h2>

8 0

3 years ago

Read 2 more answers

most dialog boxes in windows programs requiring navigation follow a similar procedure. true or false.

vfiekz [6]

I believe that is true

Hope this helps!! :)

8 0

3 years ago

By what decade were books readily available to the public across the United States and Europe? A. 1690 B. 1750 C. 1860 D. 1820

sladkih [1.3K]

Answer: 1820

Explanation:

I believe the answer is the 1820, as the introduction of steam powered printing presses and steam powered paper mills significantly lowered the cost of books to decrease and significantly increased their circulation/availability.

hope this helps

-lvr

6 0

3 years ago

What is the name of html command?

Makovka662 [10]

Here are some.

<html></html> This is the root element tag. ...
<head></head> ...
<title></title> ...
<body></body> ...
<h1></h1> ...
...
<a></a> ...
<img></img>

<h2>hope it helps.</h2><h2>stay safe healthy and happy....</h2>

3 0

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

#SPJ1

3 0

2 years ago