xdamcnt2.c

 /*
       n Damen Problem loesen;
       DAMCOUNT;
       entworfen am 31.03.1985;
       geschrieben am 02.04.1985;
       revidiert am 18.04.1985 Do vorm.

       in C++ umgewandelt am 17.04.1994

     2   4    6     8      10         12           14
   1 0 0 2 10 4 40 92 352 724 2680 14200 73712 365596
*/


#include <stdio.h>
#include <stdlib.h>


#define abs(x)   ((x >= 0) ? x : -x)

/* Check if k-th queen is attacked by any other prior queen.
   Return nonzero if configuration is OK, zero otherwise.
*/
//inline	--> not faster, klm, 24-Apr-2020
int configOkay (int k, int a[]) {
	int z = a[k];

	for (int j=1; j<k; ++j) {
		int l = z - a[j];
		if (l == 0  ||  abs(l) == k - j) return 0;
	}
	return 1;
}



long solve (int N, int a[]) {  // return number of positions
	long cnt = 0;
	int k = a[1] = 1;
	int N2 = N;  //(N + 1) / 2;

	loop:
		if (configOkay(k,a)) {
			if (k < N)  { a[++k] = 1;  goto loop; }
			else ++cnt;
		}
		do
			if (a[k] < N)  { a[k] += 1;  goto loop; }
		while (--k > 1);
		a[1] += 1;
		if (a[1] > N2) return cnt;
		k = 2 ,  a[2] = 1;
	goto loop;
}



int main (int argc, char *argv[]) {
	enum {NMAX=100};
	int a[NMAX];

	puts(" n-queens problem.\n"
	"   2   4    6     8      10         12           14\n"
	" 1 0 0 2 10 4 40 92 352 724 2680 14200 73712 365596\n"
	);

	int start = 1;
	int end = 0;
	if (argc > 1)  end = atoi(argv[1]);
	if (end <= 0  ||  end >= NMAX)  end = 10;
	if (argc > 2) { start = end;  end = atoi(argv[2]); }
	if (end <= 0  ||  end >= NMAX)   end = 10;

	for (int n=start; n<=end; ++n)
		printf (" D(%2d) = %ld\n", n, solve(n,a));

	return 0;
}