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den301095 [7]
4 years ago
12

Hello, please help write code in C++. Primary U.S. interstate highways are numbered 1-99. Odd numbers (like the 5 or 95) go nort

h/south, and evens (like the 10 90) go east/west. Auxiliary highways are numbered 100-999, and service the primary highway indicated by the rightmost two-digits. Thus, the 405 services the 5, and the 290 services the 90. Given a highway number, indicate whether it is a primary or auxiliary highway. If auxiliary, indicate what primary highway it serves. Also indicate if the (primary) highway runs north/south or east/west. If the input is 90, the output is: The 90 is primary, going east/west. If the input is 290, the output is: The 290 is auxiliary, serving the 90, going east/west. If the input is not 1-999 such as 0, the output is: 0 is not a valid interstate highway number.
Computers and Technology
1 answer:
krok68 [10]4 years ago
3 0

Answer:

Please Find attached  c++ code file.

Explanation:

  1. Ask User to enter highway number.
  2. Check if highway is primary or auxiliary.
  3. Check if highway number is even / odd.
  4. Find Primary high number for an auxiliary highway

Check if highway is primary or auxiliary.

Use if condition to check if the entered number is in between 0 and 99.If it's true than the highway is primary otherwise it's an auxiliary highway.

Check if highway number is even / odd.

To check if a number is even or odd we use simple modulo formula. In mathematics modulo returns the reminder of a division operation on two numbers.If modulo of a number and 2 is equal to zero than that number is an even number other wise its an odd number.

(Any number ) mod 2 = 0

Find Primary high number for an auxiliary highway

As mentioned in the question last two digits of highway number is the number of primary high.

To Find last two digits of number apply modulo operation on highway number and 100. it will provide last two digits of number specifying our primary highway.

Download cpp
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Answer:

The program in Python is as follows:

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my_list =[]

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Explanation:

This imports the operator function from itemgetter

from operator import itemgetter

This gets the number of records, m

m = int(input("Number of records: "))

This prints the format of input

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This initializes the list of tuples

my_list =[]

This iterates through m

for i in range(m):

This gets the details of each person

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This appends the details to the tuple

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This sorts the tuple

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This prints the sorted tuple

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Explanation:

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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
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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

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