How to write 2D array with diagonally numbers? [closed] - c++

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How to write in C++ 2D array with diagonally numbers for
n - size of array (width and height)
x - how many the same number in a row
c - how many numbers must be used
example for
n = 5
x = 2
c = 2
output is:
0 0 1 1 0
0 1 1 0 0
1 1 0 0 1
1 0 0 1 1
0 0 1 1 0
My current code:
#include <iostream>
#include <string>
using namespace std;
int main()
{
int n=0, x=0, c=0;
int temp_x=0,temp_c=-1;
cin >> n >> x >> c;
c--;
for(int i=0; i<n;i++){
for(int j=0; j<n;j++){
cout << ++temp_c;
temp_x++;
if(temp_x>x){
temp_x=0;
if(temp_c=c){
temp_c=-1;
}
}
}
cout << endl;
}
}
I will be grateful for your help. :)
But my code return incorrectly number :(

Answer 1

Are you trying to do this?
int main()
{
int n=0, x=0, c=0;
int temp_x=0,temp_c=0;
cin >> n >> x >> c;
c--;
for(int i=0; i<n;i++){
for(int j=0; j<n;j++){
if(temp_x<x)
{
temp_x++;
cout << temp_c << " ";
continue;
}
temp_c++;
temp_x=0;
if(temp_c>c)
{
temp_c=0;
}
cout << temp_c << " ";
temp_x++;
}
cout << endl;
}
}
Output:
5 2 2
0 0 1 1 0
0 1 1 0 0
1 1 0 0 1
1 0 0 1 1
0 0 1 1 0
5 2 3
0 0 1 1 2
2 0 0 1 1
2 2 0 0 1
1 2 2 0 0
1 1 2 2 0
5 3 2
0 0 0 1 1
1 0 0 0 1
1 1 0 0 0
1 1 1 0 0
0 1 1 1 0

Answer 2

I'd like to propose another algorithm:
Run It Online !
#include <iostream>
#include <vector>
#include <numeric> // iota
using std::cout;
using std::endl;
void fill(const size_t n ///< size of array (width and height)
, const size_t x ///< how many the same number in a row
, const size_t c) ///< how many numbers must be used
{
// generate the sequence of possible numbers
std::vector<int> numbers(c);
std::iota(numbers.begin(), numbers.end(), 0);
//std::vector<int> all(n * n); // for storing the output, if needed
for (size_t i = 0, // element index
k = 0, // "number" index
elements = n * n; // the square matrix can also be viewed as a n*n-long, 1D array
i < elements;
k = (k + 1) % c) // next number (and the modulus is for circling back to index 0)
{
// print the number "x" times
for (size_t j = 0; j < x && i < elements; ++j, ++i)
{
// break the line every "n" prints
if ((i % n) == 0)
{
cout << endl;
}
//all[i] = numbers[k];
cout << numbers[k] << " ";
}
}
cout << endl;
}
int main()
{
fill(5, 2, 2);
}
Output for fill(5, 2, 2)
0 0 1 1 0
0 1 1 0 0
1 1 0 0 1
1 0 0 1 1
0 0 1 1 0