Still takes more space than some O(N^2) solution i've found for some reasons. I am not really familiar with cpp compiler optimizations, so that might be linked !

#include <iostream>

int main(int argc, char* argv[]) {
    int size;
    if (argc != 2) {
        scanf("%d",&size);
    } else {
        size = std::atoi(argv[1]);
    }
    int halfSquare = (size * size) / 2;
    int half = size / 2;
    int previousY0 = halfSquare - (half) - size;
    int previous;
    for (int y = 0; y < size; y++) {
        previous = previousY0 + (size + 1 - y);
        printf("\n%d ", previous);
        for (int x = 1; x < size; x++) {
            previous = previous - (size - 1 - std::max(std::abs(x  - y), 1) + (x >= y));
            printf("%d ", previous);
        }
        previousY0 = previousY0 + (size + 1 - y);
    }
    return 0;
}

You are very stupid!!!
0<ans<9

test 27:  x==kx &&y==ky

good luck...

For this test case (n=10, k=20) answer is equal to 10.