What is meant by the term text?

I don't understand how to write code for this in Java using nested for loops only. The official question is: write a program tha

Bingel [31]

Answer:

public class Triangle

{

public static void main( String[] args )

{

show( 5 );

}

public static void show( int n )

{

int i,j,k;

for (i = 0; i < n - 1; i++ )

{

for (j = 0; j < i; j++ )

{

System.out.print( " " );

}

for (k = n - i; k > 0; k-- )

{

System.out.print( "* " );

}

System.out.println();

}

for (i = 0; i < n; i++ )

{

for (j = n - i; j > 1; j-- )

{

System.out.print( " " );

}

for (k = 0; k < i + 1; k++ )

{

System.out.print( "* " );

}

System.out.println();

}

4 0

2 years ago

Who's better Kapkan or Tachanka?

navik [9.2K]

If you want fun operator, get chanka, if you want good operator, get kapkan it also depends if depends you are a stationary man who likes big guns or a mobile man who likes traps

4 0

3 years ago

1. Who was able to complete the puzzle the fastest in Trial 1?

KATRIN_1 [288]

Answer:

You...answered all of the questions.

Explanation:

All of them have a response!

8 0

2 years ago

Mike has never used a slide software before, but he needs to create a presentation by the end of the week. what recourse would b

ankoles [38]

This is an incomplete question. The complete question is given below:

Mike has never used slide presentation software before but he needs to create a presentation by the end of the week what resource would be most helpful to mike

a. The 350-page printed manual from the slide presentation software publisher

b. A free tutorial the slide presentation software publisher has posted on the company website

c. A trouble-shooting website created by a third party

d. The 350-page online manual from the slide presentation software publisher

Answer:

b - A free tutorial the slide presentation software publisher has posted on the company website

Explanation:

As Mike has a short time and no prior experience with a slide software, then in this scenario, the best, simplest and fastest way to learn and create a presentation a free tutorial which the slide presentation software publisher has posted on the company website as this is the same company that has created this particular software so he can be rest-assured that the resource he is relying on is authentic and up-to-date with information on latest features.

Moreover, it's efficient and quick way to learn from a free tutorial rather than from 350-page printed or online manual especially for a beginner.

Besides, his purpose is to create the presentation using the software and not trouble-shooting so trouble-shooting website created by a third party is not useful for him and it also might not be authentic or updated as well.

6 0

2 years ago

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

1 year ago