Which of the following is not a web browser? A. safari B. explorer C. chrome D. google
2 answers:
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Answer:
Given code output is "786".
Explanation:
Code:
list1 = [ 'cyber', 786 , 2.23, 'square', 70.2 ]#defining a list list1 that holds value in parameter
print (list1[1] )#use a print method that prints list index value
In the given code a list "list1" is declared that holds value in it and in the next step a print method is declared that prints the first index value of the list.
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
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Answer:
The longest codeword that could possibly be for "n" symbol is n-1 bits.
Explanation:
From the given information:
Suppose we are to consider a set of frequencies , for which f is a symbol for the length n. Therefore, the longest codeword that could possibly be for "n" symbol is n-1 bits.
However, during the encoding for "n" in conjunction with n-2, then the possibilities for n are;
We can conclude that the longest codeword that could possibly be for "n" symbol is n-1 bits.