Can someone write this in java? Also, does anyone know how to do Edhesive assignments?

Write a method called printRangeOfNumbers that accepts a minimum, maximum numbers as parameters and prints each number from mini

timama [110]

Answer:

The method in python is as follows:

class myClass:

def printRange(min,max):

for i in range(min, max+1):

print("{"+str(i)+"} ", end = '')

Explanation:

This line declares the class

class myClass:

This line defines the method

def printRange(min,max):

This line iterates from min to max

for i in range(min, max+1):

This line prints the output in its required format

print("{"+str(i)+"} ", end = '')

5 0

3 years ago

Question 9. which describes an str?

Nadya [2.5K]

STR stands for Short Term repeat. It is a region of a DNA molecule that contains short segments of three to seven repeating base pairs
Hundreds of different types of STR's are found in the human genomes.

3 0

3 years ago

Which software is used to play, create, and modify audio and video files?

igor_vitrenko [27]

Video Editing software is used to play, create, and modify audio and video files.

7 0

3 years ago

Read 2 more answers

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>

#include <bits/stdc++.h>

using namespace std;

// chracter to digit mapping, and the inverse

// (if you want better performance: use array instead of unordered_map)

unordered_map<char, int> c2i;

unordered_map<int, char> i2c;

int ans = 0;

// limit: length of result

int limit = 0;

// digit: index of digit in a word, widx: index of a word in word list, sum: summation of all word[digit]

bool helper(vector<string>& words, string& result, int digit, int widx, int sum) {

if (digit == limit) {

ans += (sum == 0);

return sum == 0;

}

// if summation at digit position complete, validate it with result[digit].

if (widx == words.size()) {

if (c2i.count(result[digit]) == 0 && i2c.count(sum%10) == 0) {

if (sum%10 == 0 && digit+1 == limit) // Avoid leading zero in result

return false;

c2i[result[digit]] = sum % 10;

i2c[sum%10] = result[digit];

bool tmp = helper(words, result, digit+1, 0, sum/10);

c2i.erase(result[digit]);

i2c.erase(sum%10);

ans += tmp;

return tmp;

} else if (c2i.count(result[digit]) && c2i[result[digit]] == sum % 10){

if (digit + 1 == limit && 0 == c2i[result[digit]]) {

return false;

}

return helper(words, result, digit+1, 0, sum/10);

} else {

return false;

}

// if word[widx] length less than digit, ignore and go to next word

if (digit >= words[widx].length()) {

return helper(words, result, digit, widx+1, sum);

}

// if word[widx][digit] already mapped to a value

if (c2i.count(words[widx][digit])) {

if (digit+1 == words[widx].length() && words[widx].length() > 1 && c2i[words[widx][digit]] == 0)

return false;

return helper(words, result, digit, widx+1, sum+c2i[words[widx][digit]]);

}

// if word[widx][digit] not mapped to a value yet

for (int i = 0; i < 10; i++) {

if (digit+1 == words[widx].length() && i == 0 && words[widx].length() > 1) continue;

if (i2c.count(i)) continue;

c2i[words[widx][digit]] = i;

i2c[i] = words[widx][digit];

bool tmp = helper(words, result, digit, widx+1, sum+i);

c2i.erase(words[widx][digit]);

i2c.erase(i);

}

return false;

}

void isSolvable(vector<string>& words, string result) {

limit = result.length();

for (auto &w: words)

if (w.length() > limit)

return;

for (auto&w:words)

reverse(w.begin(), w.end());

reverse(result.begin(), result.end());

int aa = helper(words, result, 0, 0, 0);

}

int main()

{

ans = 0;

vector<string> words={"GREEN" , "BLUE"} ;

string result = "BLACK";

isSolvable(words, result);

cout << ans << "\n";

return 0;

}

See more about C++ code at brainly.com/question/19705654

#SPJ1

3 0

2 years ago

A developer needs to create a visualforce page that displays case data. The page will be used by both support reps and support m

shutvik [7]

Answer:

B. Use a new support manager permission sets

Explanation:

According to the requirements, field level security is required so, 1st options is not suitable because it may reduce the maintenance cost but increase the risk of security.

By creating a separate page for each of the two, it will leads to increase in the maintenance cost of the system. So Option C is also not suitable.

Option B is more Suitable as compared to others, that Create a new support manager permission set, with the help of this, both of Support rep and Support manager can visualize their required information with the help of support manager permission. This solution will not compromise the security and not increase the maintenance cost.

7 0

3 years ago