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Ann [662]
3 years ago
9

Visit on social networking site and look for the site's privacy policy.The link is typically found at the bottom of the page and

sometimes labelled only as "Privacy" write a summary on how the website handles both your private and public information.
pahelp po​
Computers and Technology
1 answer:
Mekhanik [1.2K]3 years ago
4 0

Answer:

Explanation:

Your profile. Most social networks allow users to create detailed online profiles and connect with other users in some way. This may involve users sharing information with other users, such as one’s gender, age, familial information, interests, educational background and employment.

Your status. Most social networks also allow users to post status updates in order to communicate with other users quickly. Though there may be privacy settings to restrict access to status updates, these networks are frequently designed to broadcast information quickly and publicly.

Your location.  Many social networks are designed to broadcast your real-time location, either as public information or as an update viewable to authorized contacts. This might allow users to “check in” to a local event or business, or share one’s location with contacts within their network.

Shared content. Many social networks encourage users to share content, such as music, photographs, videos and links to other webpages.

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a cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits such that the giv
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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>

<em>#include <bits/stdc++.h></em>

<em>using namespace std;</em>

<em>// chracter to digit mapping, and the inverse</em>

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

<em>unordered_map<char, int> c2i;</em>

<em>unordered_map<int, char> i2c;</em>

<em>int ans = 0;</em>

<em>// limit: length of result</em>

<em>int limit = 0;</em>

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

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

<em>    if (digit == limit) {</em>

<em>        ans += (sum == 0);</em>

<em>        return sum == 0;</em>

<em>    }</em>

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

<em>    if (widx == words.size()) {</em>

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

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

<em>                return false;</em>

<em>            c2i[result[digit]] = sum % 10;</em>

<em>            i2c[sum%10] = result[digit];</em>

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

<em>            c2i.erase(result[digit]);</em>

<em>            i2c.erase(sum%10);</em>

<em>            ans += tmp;</em>

<em>            return tmp;</em>

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

<em>            if (digit + 1 == limit && 0 == c2i[result[digit]]) {</em>

<em>                return false;</em>

<em>            }</em>

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

<em>        } else {</em>

<em>            return false;</em>

<em>        }</em>

<em>    }</em>

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

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

<em>        return helper(words, result, digit, widx+1, sum);</em>

<em>    }</em>

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

<em>    if (c2i.count(words[widx][digit])) {</em>

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

<em>            return false;</em>

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

<em>    }</em>

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

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

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

<em>        if (i2c.count(i)) continue;</em>

<em>        c2i[words[widx][digit]] = i;</em>

<em>        i2c[i] = words[widx][digit];</em>

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

<em>        c2i.erase(words[widx][digit]);</em>

<em>        i2c.erase(i);</em>

<em>    }</em>

<em>    return false;</em>

<em>}</em>

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

<em>    limit = result.length();</em>

<em>    for (auto &w: words) </em>

<em>        if (w.length() > limit) </em>

<em>            return;</em>

<em>    for (auto&w:words) </em>

<em>        reverse(w.begin(), w.end());</em>

<em>    reverse(result.begin(), result.end());</em>

<em>    int aa = helper(words, result, 0, 0, 0);</em>

<em>}</em>

<em />

<em>int main()</em>

<em>{</em>

<em>    ans = 0;</em>

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

<em>    string result = "BLACK";</em>

<em>    isSolvable(words, result);</em>

<em>    cout << ans << "\n";</em>

<em>    return 0;</em>

<em>}</em>

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

#SPJ1

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Answer:

def LetterGame():

   vowels = ["a", "e", "i", "o", "u", "A", "E", "I", "O", "U"]

   count1 = count2 = 0

   while True:

       choice = input("Enter a letter / digit to stop: ")

       if choice.isdigit():

           break

       elif choice.isalpha():

           count1 += 1

           if choice in vowels:

               count2 += 1

   print("You entered " + str(count1) + " letters, " + str(count2) + " of which weere vowels.")

   print("The percentage of vowels was " + str(count2 / count1 * 100) + "%")

LetterGame()

Explanation:

Create a function called LetterGame

Inside the function:

Create a list of vowels

Initialize count1, counts the total letters, and count2, counts the vowels

Initialize a while loop that iterates until the specified condition is met in the loop

Get the input from the user. If it is a digit, stop the loop. If it is an alphabet, increase the count1 by 1, and also check if it is a vowel. If it is a vowel, increment the count2 by 1

When the loop is done, print the required information

Finally, call the function

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