How to print numbers in N x (N+1) rectangles? - c++

Question

the program should take an integer from user and print pattern like below.
if n=3, then
333
313
323
333
if n=4, then
4444
4114
4224
4334
4444
if n=5, then
55555
51115
52225
53335
54445
55555
and so on
Here is what is have tried.
#include<iostream>
using namespace std;
int main()
{
int pattern[10][10], i, j, n;
cout << "Enter dimension of square matrix: ";
cin >> n;
for (i = 0;i <= n;i++)
{
pattern[i][0] = n;
pattern[i][n - 1] = n;
}
for (j = 1;j < n - 2;j++)
{
pattern[0][j] = n;
pattern[n][j] = n;
}
for (i = 1;i < n - 1;i++)
{
for (j = 1;j < n - 2;j++)
{
pattern[i][j] = i;
}
}
for (i = 0;i <=n;i++)
{
for (j = 0;j < n;j++)
{
cout << pattern[i][j];
cout << "\t";
}
cout << "\n";
}
return 0;
}
I am getting the right pattern but in some places, there are some garbage value (or something else)

Answer 1

Quite simple to do in a clear, concise manner:
#include<iostream>
int main()
{
int n = 0;
std::cout << "Enter dimension of square matrix: ";
std::cin >> n;
for (int i = 0; i < n + 1; ++i) {
for (int j = 0; j < n; ++j) {
// if on the first or last column or row: print n, else: print i
std::cout << ((i == 0 || j == 0 || i == n || j == n-1) ? n : i) << '\t';
}
std::cout << '\n';
}
return 0;
}
Example output:
Enter dimension of square matrix: 5
5 5 5 5 5
5 1 1 1 5
5 2 2 2 5
5 3 3 3 5
5 4 4 4 5
5 5 5 5 5

Answer 2

You are on the right way! Some simple change will give the right answer .
#include<iostream>
using namespace std;
int main()
{
int pattern[10][10], i, j, n;
cout << "Enter dimension of square matrix: ";
cin >> n;
for (i = 0;i <= n;i++)
{
pattern[i][0] = n;
pattern[i][n - 1] = n;
}
for (j = 1;j <n ;j++)
{
pattern[0][j] = n;
pattern[n][j] = n;
}
for (i = 1;i < n ;i++)
{
for (j = 1;j < n -1;j++)
{
pattern[i][j] = i;
}
}
for (i = 0;i <=n;i++)
{
for (j = 0;j < n;j++)
{
cout << pattern[i][j];
cout << "\t";
}
cout << "\n";
}
return 0;
}

Answer 3

I think you are overcomplicating things - there is absolutely no need for the two-dimensional array and all the loops.
You could just use a basic helper function printer, that prints the version with the nested number i and execute that in a loop from 1 to the last number, for example 5.
#include <cstdio>
#include <iostream>
int main() {
unsigned user_input = 5; // change to ask for actual user input
// Helper function, i.e. for 3 it prints 53335
auto printer = [](unsigned i) {
for (unsigned n = 0; n < user_input; ++n) {
if (n == 0 || n + 1 == user_input) std::cout << user_input;
else std::cout << i;
}
std::cout << '\n';
};
printer(user_input); // print 55555
// print the rest 51115, 52225 ... up to 55555 again
for (unsigned i = 1; i <= user_input; ++i) printer(i);
}
If you think you need the printer function elsewhere too or don't want to use a lambda, you can of course just move it and give it a signature like that:
void printer(unsigned i, unsigned user_input) {
/* copy-paste the code from printer */
}

Answer 4

In fact there is no need to define an array to output the pattern.
All you need is two loops.
Here is a demonstration program.
#include <iostream>
int main()
{
while ( true )
{
const unsigned int MAX_VALUE = 10;
std::cout << "Enter a non-negative number less than "
<< MAX_VALUE << " (0 - exit): ";
unsigned int n;
if ( not ( std::cin >> n ) or ( n == 0 ) ) break;
if ( MAX_VALUE - 1 < n ) n = MAX_VALUE - 1;
std::cout << '\n';
for ( unsigned int i = 0; i < n + 1; i++ )
{
for ( unsigned int j = 0; j < n; j++ )
{
if ( j == 0 || j == n - 1 )
{
std::cout << n;
}
else if ( i == 0 || i == n )
{
std::cout << n;
}
else
{
std::cout << i;
}
}
std::cout << '\n';
}
std::cout << '\n';
}
}
Its output might look like
Enter a non-negative number less than 10 (0 - exit): 1
1
1
Enter a non-negative number less than 10 (0 - exit): 2
22
22
22
Enter a non-negative number less than 10 (0 - exit): 3
333
313
323
333
Enter a non-negative number less than 10 (0 - exit): 4
4444
4114
4224
4334
4444
Enter a non-negative number less than 10 (0 - exit): 5
55555
51115
52225
53335
54445
55555
Enter a non-negative number less than 10 (0 - exit): 6
666666
611116
622226
633336
644446
655556
666666
Enter a non-negative number less than 10 (0 - exit): 7
7777777
7111117
7222227
7333337
7444447
7555557
7666667
7777777
Enter a non-negative number less than 10 (0 - exit): 8
88888888
81111118
82222228
83333338
84444448
85555558
86666668
87777778
88888888
Enter a non-negative number less than 10 (0 - exit): 9
999999999
911111119
922222229
933333339
944444449
955555559
966666669
977777779
988888889
999999999
Enter a non-negative number less than 10 (0 - exit): 0
That is you can substitute all your loops that fill the array for if statements in one pair of loops that output the pattern.
The if statements can be substituted for the conditional operator the following way
for ( unsigned int i = 0; i < n + 1; i++ )
{
for ( unsigned int j = 0; j < n; j++ )
{
std::cout << ( i % n == 0 || j % ( n - 1 ) == 0 ? n : i );
}
std::cout << '\n';
}
The restriction to output the pattern only for numbers [1, 9] is artificial.
It would be more reasonable to set the range at least to [1, 255] where 255 is the maximum value that can be stored in an object of type unsigned char.
You can get the value through the expression
std::numeric_limits<unsigned char>::max()
To make an alignment for outputted columns you can use standard function std::setw that sets the width of an outputted field.
Taking all this into account the program can look the following way.
#include <iostream>
#include <iomanip>
#include <limits>
int main()
{
while ( true )
{
const unsigned int Base = 10;
const unsigned int MAX_VALUE = std::numeric_limits<unsigned char>::max();
std::cout << "Enter a non-negative number less than or equal to "
<< MAX_VALUE << " (0 - exit): ";
unsigned int n = 0;
if ( not ( std::cin >> n ) || ( n == 0 ) ) break;
if ( MAX_VALUE < n ) n = MAX_VALUE;
// Calculating the width of numbers plus one space between them.
int number_width = 1;
unsigned int tmp = n;
do { ++number_width; } while ( tmp /= Base );
std::cout << '\n';
for ( unsigned int i = 0; i < n + 1; i++ )
{
for ( unsigned int j = 0; j < n; j++ )
{
std::cout << std::setw( number_width )
<< ( i % n == 0 || j % ( n - 1 ) == 0 ? n : i );
}
std::cout << '\n';
}
std::cout << '\n';
}
}
The program output might look like
Enter a non-negative number less than 255 (0 - exit): 20
20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20
20 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 20
20 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 20
20 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 20
20 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 20
20 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 20
20 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 20
20 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 20
20 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 20
20 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 20
20 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 20
20 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 11 20
20 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 12 20
20 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 13 20
20 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 20
20 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 15 20
20 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 16 20
20 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 17 20
20 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 20
20 19 19 19 19 19 19 19 19 19 19 19 19 19 19 19 19 19 19 20
20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20
Enter a non-negative number less than 255 (0 - exit): 0
As for your code then for starters you should check whether the user entered a number that is greater than 9
And for example in this pair of loops
for (i = 1;i < n - 1;i++)
{
for (j = 1;j < n - 2;j++)
{
pattern[i][j] = i;
}
}
the conditions should look at least like
for (i = 1; i < n; i++)
{
for (j = 1; j < n - 1; j++ )
{
pattern[i][j] = i;
}
}
because the height of the pattern is equal to n + 1 and the width is equal to n. So if you want to exclude the last row and last column you have to use correspondingly the expressions n and n - 1 in the conditions of the loops.