Clunker Motors Inc. is recalling all vehicles from model years 2001-2006. A bool variable named norecall has been declared . 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
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Answer:
Here is the JAVA program. Let me know if you need the program in some other programming language.
import java.util.Scanner; // used for taking input from user
public class Main{ //Main class
public static void main(String[] args) {//start of main() function body
Scanner scan = new Scanner(System.in); // creates a Scanner type object
String input; // stores input series
int sum = 0; //stores the sum of the series
System.out.println("Enter a series of numbers separated by commas: "); //prompts user to enter a series of numbers separated by comma
input = scan.nextLine(); //reads entire input series
String[] numbers = input.split("[, ]"); //breaks the input series based on delimiter i.e. comma and stores the split sub strings (numbers) into the numbers array
for (int i = 0; i < numbers.length; i++) { //loops through each element of numbers array until the length of the numbers reaches
sum += Integer.parseInt(numbers[i]); } // parses the each String element of numbers array as a signed decimal integer object and takes the sum of all the integer objects
System.out.println("Sum of all the numbers: " + sum); } } //displays the sum of all the numbers
Explanation:
The program prompts the user to enter a series separated by commas.
Then split() method is used to split or break the series of numbers which are in String form, into sub strings based on comma (,) . This means the series is split into separate numbers. These are stored in numbers[] array.
Next the for loop iterate through each sub string i.e. each number of the series, converts each String type decimal number to integer using Integer.parseInt and computes the sum of all these integers. The last print statement displays the sum of all numbers in series.
If the input is 7,9,10,2,18,6
Then the output is: 7+9+10+2+18+6
sum = 52
The program and its output is attached.
