Question

Given numRows and numColumns, print a list of all seats in a theater. Rows are numbered, columns lettered, as in 1A or 3E. Print a space after each seat, including after the last. Ex: numRows = 2 and numColumns = 3 prints:
1A 1B 1C 2A 2B 2C
#include
using namespace std;

int main() {
int numRows;
int numColumns;
int currentRow;
int currentColumn;
char currentColumnLetter;

cin >> numRows;
cin >> numColumns;

/* Your solution goes here */

cout << endl;

return 0;
}

Answer 1

Answer:

First for loop is for printing the rows.

Then we set currColumnLetter to 'A' for first time

Second for loop is for printing the columnletter with each row

Explanation:

Answer 2

Answer:

char rowList[26] = {'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};

for(char row = 'A';row<rowList[numRows];row++) {

for(int i = 1;i<=numColumns;i++)  {

 cout<<row<<i<<" ";

}

}

Explanation:

Replace /* Your solution goes here */ with the above code

and replace #include with #include<iostream>

The line 1 of the above code segment declares a list of all alphabets to represent the rows.

Line 2 and 3 make use of a loop to iterate through rows and column (depending on the input values)

Line 4 prints the resulting row and column values

Take for instance:

numRows = 3

numColumns = 4

What the compiler does is that;

it iterates from A to C --- for rows and

it iterates from 1 to 4 --- for columns

For the first row, it takes alphabet A and digit 1 to 4 to print

A1 A2 A3 A4

Same is applicable to B and C and even if the row is more than 3 as used in this instance