How to create matrix output pattern with array 2D in C++

Question

I have a little difficulty in programming in C ++ to make patterned output using 2D arrays, I make the output like a matrix like this: input 4 (for column 4 & row 4)

I I I I
0 0 I 0
0 I 0 0
I I I I

I've tried this my programmed. For the top and bottom rows I've managed to change the value to 1

   #include <iostream>
using namespace std;

int main(){
    int elemen[100][100], n, i, j, k;
    
    cout <<"input element = ";
    cin >>n;
    
    for(i = 0;i < n;i++){
        for(j = 0;j < n;j++){
            elemen[0][j] = 1;
            elemen[n-1][j] = 1;
            elemen[n-i][n-j] = 1;           
            cout <<elemen[i][j]<<" ";
        }   
        cout<<endl;
    }
    
}

but the output that occurs in the above program is like this:

I I I I
0 0 0 0
0 0 I I
I I I I

whereas logically it is correct, and I've tried it on non-input arrays.whereas logically it is correct, and I've tried it on non-input arrays. because if we input the array element = 5, then the loop will automatically reduce the value of 5 one by one. Is there anyone who can help, sorry if you do not understand, because I am currently still learning to hone my logic

Ted Lyngmo · Accepted Answer · 2020-11-07 23:33:44Z

1

Unless you are actually using elemen for anything else than printing this pattern, I suggest dropping it and print the pattern directly.

Example:

#include <cstddef>  // size_t
#include <iostream>
#include <string>

int main() {
    std::size_t n = 4;
    
    std::cout << std::string(n, 'I') << '\n';

    for(std::size_t i=1; i < n - 1; ++i) {
        std::cout
            << std::string(n - i - 1, '0') 
            << 'I'
            << std::string(i, '0')
            << '\n'
        ; 
    }

    std::cout << std::string(n, 'I') << '\n';
}

Output:

IIII
00I0
0I00
IIII

edited Nov 7, 2020 at 23:33

answered Nov 7, 2020 at 23:07

Ted Lyngmo

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Comments

wcochran · Accepted Answer · 2020-11-07 23:07:52Z

0

Try this

 for(i = 0;i < n;i++){
      for(j = 0;j < n;j++){
            if (i == 0 || i == n-1 || i == n-j-1)
                elemen[i][j] = 1;
            else
                elemen[i][j] = 0;
            ...
       }
       ...
  }

i = 0 gives you the top row
i = n-1 gives you the bottom row
i = n-j-1 gives you reverse diagonal

answered Nov 7, 2020 at 23:07

wcochran

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