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2 years ago

What would be the total width of the div in the code below?

Computers and Technology

2 answers:

Artyom0805 [142]2 years ago

4 0

Answer:

The answer to this question is given below in the explanation section.

Explanation:

The correct answer to this question is d- 346px.

The complete code of this example is given below:

<div>

style="margin:20px; border:solid 3px #888888;">

</div>

while the image width is 300px.

It is noted that the image width is 300px and the margin is 20 px.

Margin:20px means that image margin from four side is 20 px each.

So the width of dive from both side increase to 40 px and the image has 300 px, then total width becomes 340px. Now the border also has 3px of four sides of the image, then the width of the border becomes 6px.

so the total width of the div tag is 346 px.

In short, we can calculate the width of tag as below:

total width of div= width of image+ width of margin+ width of border

total width of dive= 300px+40px+6px

total width of div=346px.

adell [148]2 years ago

4 0

Answer:

Explanation:

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Phantasy [73]

Answer:

c. your friend can hash all possible options and discover your secret.

Explanation:

SHA-256 is a set of hash functions that was designed by the NSA. SHA-2 is considered an upgrade on the set that was its predecessor, SHA-1. A hash is a mathematical function that condenses data in a process of one-way encryption. SHA-256 creates hash algoritms that are considered irreversible and unique. However, one of the properties of hashing algorithms is determinism, which means that any computer in the world would be able to compute a particular hash and get the same answer.

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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]

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<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;

}

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sergey [27]

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