How do you mark answers brainliest? it looks like they changed it and now idk how to

Licemer1 [7]

Answer:

People have been ringing the death knell for email on and off for a few years now. But should we be listening?

The latest peal came from French IT company Atos, which declared that it would phase out internal emails by 2013. And for reasons we’ll all recongize: too much time spent dealing with too many emails, of which too few are useful and too many are spam. CEO Thierry Breton said his staff would instead use good old face-to-face communication, as well as instant messaging (IM) and social media tools.

3 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

An indication of delayed fuel ignition could be

Sophie [7]

Answer:

The ignition delay in a diesel engine is defined as the time interval between the start of injection and the start of combustion. This delay period consists of (a) physical delay, wherein atomisation, vaporization and mixing of air fuel occur and (b) of chemical delay attributed to pre-combustion reactions. Physical and chemical delays occur simultaneously.

Hope This Helps! Have A Nice Day!!

6 0

3 years ago

Rewrite the following condition to avoid a possible arithmetic exception:

dangina [55]

Answer:

if (x > 7 && Math.sqrt(x) < 3)

Explanation:

The previous condition checks if the square root of x is less than 3, but this would raise an error if x is a value equal to or less than 0.

So, the condition checks for the value of x before evaluating the square root after the AND operation to prevent an arithmetic exception.

4 0

3 years ago

Suppose during a TCP connection between A and B, B acting as a TCP receiver, sometimes discards segments received out of sequenc

Shalnov [3]

Answer:

Yes

Explanation:

We known that TCP is the connection $\text{oriented}$ protocol. So the TCP expects for the target host to acknowledge or to check that the $\text{communication}$ session has been established or not.

Also the End stations that is running reliable protocols will be working together in order to verify transmission of data so as to ensure the accuracy and the integrity of the data. Hence it is reliable and so the data can be sent A will still be delivered reliably and it is in-sequence at B.

4 0

3 years ago