How to make square of numbers in opposite way? - c++

Question

I have made a code that makes this kind of output:
1 2 3 4 5 6
7 8 9 10 11 12
13 14 15 16 17 18
19 20 21 22 23 24
25 26 27 28 29 30
31 32 33 34 35 36
37 38 39 40 41 42
43 44 45 46 47 48
#include <iostream>
using namespace std;
int main() {
// dimensions:
int x=6,y=8;
int sum=0;
int numery[8][6]={};
for (i = 0; i < y; i++) {
for (j = 0; j < x; j++) {
numery[i][j]=++sum;
if (numery[i][j]<=9) cout << " ";
cout << numery[i][j] << " ";
}
cout << endl;
}
return 0;
}
But I don't know how to edit it to get this kind of output:
6 5 4 3 2 1
7 8 9 10 11 12
18 17 16 15 14 13
19 20 21 22 23 24
30 29 28 27 26 25
31 32 33 34 35 36
37 38 39 40 41 42
43 44 45 46 47 48
I can think of making if statement for each i%2==1 that it should go backwards, but I don't know how to make program do such thing. Otherwise it should go normally. So if it find an even row it should go like 7 8 9 10 11 12, whereas if it's i%2==1 it should go like 6 5 4 3 2 1 and so on.
Some suggestions?

Answer 1

One solution would be the following (I assume that you do not really need to store the numbers). Notice that setw() can be used to print the numbers with fixed width:
#include <iostream>
#include <iomanip>
using namespace std;
int main() {
// dimensions:
int x=6,y=8;
for (int i = 0; i < y; i++)
{
for (int j = 0; j < x; j++)
{
cout << setw(3) << ((i%2) ? i*x+j+1 : i*x+x-j) << " ";
}
cout << endl;
}
return 0;
}

Answer 2

First get rid of the array. It is unnecessary.
#include <iostream>
using namespace std;
int number(int i,int j,int width) { return j+i*width +1 ;}
int main() {
// dimensions:
int x=6,y=8;
for (int i = 0; i < y; i++) {
for (int j = 0; j < x; j++) {
auto n = number(i,j,x);
if (n <= 9) cout << " ";
cout << n << " ";
}
cout << endl;
}
return 0;
}
This has same output as your code.
Next consider what happens to j when you go reverse. You should think that through, the answer is: You replace j with width-j-1.
int number_reverse(int i,int j, int width) { return number(i,width-j-1,width); }
Eventually you just need to call the right function in each iteration of the outer loop:
#include <iostream>
using namespace std;
int number(int i,int j,int width) { return j+i*width +1 ;}
int number_reverse(int i,int j, int width) { return number(i,width-j-1,width); }
int main() {
// dimensions:
int x=6,y=8;
for (int i = 0; i < y; i++) {
for (int j = 0; j < x; j++) {
auto n = (i%2) ? number_reverse(i,j,x) : number(i,j,x);
if (n <= 9) cout << " ";
cout << n << " ";
}
cout << endl;
}
return 0;
}

Answer 3

Here is a solution in case you DO need to store the numbers in an array.
#include <iostream>
using namespace std;
int main() {
// dimensions:
int x=6,y=8;
int sum=0;
int numery[y][x];
for (int i = 0; i < y; i++)
for (int j = 0; j < x; j++)
numery[i][j]=++sum;
for(int i = 0; i < y; i++)
{
if(i%2 == 0)
{
for(int j = x-1; j >= 0; --j)
{
if(numery[i][j] < 10)
{
cout << " ";
}
cout << numery[i][j] << " ";
}
cout << endl;
}
else
{
for(int j = 0; j < x; j++)
{
if(numery[i][j] < 10)
{
cout << " ";
}
cout << numery[i][j] << " ";
}
cout << endl;
}
}
return 0;
}
you assign the numbers to your array first. then you print the numbers from back-to-front or front-to-back based on if are on an even or odd row.

Answer 4

If you are willing to fill the array in non-linear order, you can fill every second line from right to left: array[i][x-1-j]=++sum;.
Another approach is to use one outer loop and two inner loops, noticing that all the rows are of form a*i + b*j + c, where b is either +1 or -1. This simplifies to array[i][j] = 6*i + 6 - j for even rows and to array[i][j]=6*i + 1 + j for the odd rows.
One more option is to have a single inner loop, which will increase or decrease by some amount d = i % 2 == 0 ? -1 : 1;
for (int j = 0; j<x; j++) { arr[i][j]=sum; sum += d;}
sum+=6-d;
After each row we need to compensate the variable sum so that it begins with the correct value for the next index i+1.