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snow_tiger [21]
3 years ago
5

my homework guides Write the definition of a method printDottedLine, which has no parameters and doesn't return anything. The me

thod prints to standard output a single line (terminated by a newline) consisting of five periods.
Computers and Technology
1 answer:
Masteriza [31]3 years ago
8 0

Answer:

   public static void printDottedLine(){

       System.out.print(".....\n");

   }

Explanation:

This method returns nothing so its return type is void

It also accepts no parameters so the argument list is empty

When called it executes the  System.out.print(".....\n"); which prints out 5 dots

See a complete program below:

public class TestClock {

   public static void main(String[] args) {

   printDottedLine();

   }

   public static void printDottedLine(){

       System.out.print(".....\n");

   }

}

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The java program for the given scenario is as follows.

import java.util.*;

import java.lang.*;

public class Main

{

   //variables for bill and tip declared and initialized

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   //variables for total bill and share declared

   static double total, share1;

public static void main(String[] args) {

    double total_tip= (bill*tip);

    //total bill computed

    total = bill + total_tip;

    //share of each friend computed

    share1 = total/2;

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}

}

Explanation:

1. The variables to hold the bill and tip percent are declared as double and initialized with the given values.

static double bill=47.28, tip=0.15;

2. The variables to hold the values of total bill amount and total tip are declared as double.

3. All the variables are declared outside main() and at class level, hence declared as static.

4. Inside main(), the values of total tip, total bill and share of each person are computed as shown.

double total_tip= (bill*tip);

total = bill + total_tip;

share1 = total/2;

5. The share of each person is displayed to the user. The value is displayed with only two decimal places which is assured by %.2f format modifier. The number of decimal places required can be changed by changing the number, i.e. 2. This format is used with printf() and not with println() method.

System.out.printf("Each person needs to pay: $%.2f", share1);  

6. The program is not designed to take any user input.

7. The program can be tested for any value of bill and tip percent.

8. The whole code is put inside a class since java is a purely object-oriented language.

9. Only variables can be declared outside method, the logic is put inside a method in a purely object-oriented language.

10. As shown, the logic is put inside the main() method and only variables are declared outside the method.

11. Due to simplicity, the program consists of only one class.

12. The output is attached.

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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

3 0
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