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Lisa [10]
3 years ago
12

Write three statements to print the first three elements of array runTimes. Follow each statement with a newline. Ex: If runTime

= {800, 775, 790, 805, 808}, print:
800
775
790
Note: These activities may test code with different test values. This activity will perform two tests, the first with a 5-element array (int runTimes[5]), the second with a 4-element array (int runTimes[4]). See How to Use zyBooks.
Computers and Technology
1 answer:
Lady_Fox [76]3 years ago
5 0

Answer:

Following are the program in c language

#include <stdio.h> // header file

int main() // main function

{

  int runTimes[5]={800,775,790,805,808}; // declared the array

  for (int k = 0; k < 3; k++) // itearting the loop

  {

     printf("\n%d",runTimes[k]); // display array

  }

  return 0;

}

Output:

800

775

790

Explanation:

Following are the description of program

  • Declared a array "runTimes[5]" as the" int " type and store the five integer value in it .
  • After that iterating the for loop from the 0 index to the less then 3 index .
  • Inside the for loop we print the corresponding value that are stored in the particular index in the nextline .
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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
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
Write a loop that fills an array int values[10] with ten random numbers between 1 and 100. Write code for two nested loops that
zalisa [80]

Answer:

#include <iostream>

#include <cstdlib>

using namespace std;

int main() {

   int[] array = new int[10];

   int index = 0;

   while(index < array.size()){

           int number = (rand() % 100) + 1;

           for (int i = 0; i < 1; i++) {

               array[index] = number;

               cout<< "Position "<< index << "of the array = "<< number << endl;

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           }

     }

}

Explanation:

The while loop in the source code loops over a set of code ten times, The for loop only loops once to add the generated random number between 1 and 100 to the array of size 10. At the end of the for loop, the index location and the item of the array is printed out on the screen. The random number is generated from the 'rand()' function of the C++ standard library.

7 0
3 years ago
29
denis23 [38]

The correct answer is A it transfers control to the next loop in the program.


The most valid answer is that the break statement Exits the loop and continues executing the program, but assuming that there are other loops, the control of the program will go to them since the first loop is broken out of.


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never [62]

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