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4 years ago

6

Before you record your own voice, you should A. record other people's voices B. warm up and practice difficult names C. listen t

o your favorite songs D. read a transcript of a good radio news segment

Computers and Technology

1 answer:

Sphinxa [80]4 years ago

5 0

Answer:

in my opinion, I would do C

Explanation:

listening to music, even ur favorites, helps u focus on things, according to study

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4 years ago

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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]

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

3 0

2 years ago

For each of the following SQL queries, for each relation involved,list the attributes that must be examined to compute the answe

Vlada [557]

Answer:

1. E.eid ,E.hobby, E.sal, E.did

2.E.eid , E.sal, E.hobby ,E.did , D.did, D.floor ,D.dname , D.budget.

3.E.eid , E.sal, E.hobby ,E.did , D.did, D.floor ,D.dname , D.budget.

4.E.eid , D.dname

Explanation:

The attributes that are examined for the query are the attributes of the table that are mentioned in the select statement and where clause.

So according to first query we are working on all attribues of Emp table so all of the attributes of Emp table are examined.In second query we selecting all attributes of both the tables hence all attributes of both the table and same in the next query.

In fourth query though the query is not complete where clause is missing but we have eid from Emp and dname from Dept tables for sure and the attributes mentioned in where clause will also be present.

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3 years ago