Here you go!

C (http://rosettacode.org/wiki/N-queens_problem#C)

Program 1 (compiled with gcc -O3 c1.c -o bin/c1)[0]

    real	0m0.003s
    user	0m0.000s
    sys	        0m0.002s

Program 2: (compiled with gcc -O3 c2.c -o bin/c2)[1]

Same as above, second code listing (faster while still readable,excluded unreadable fastest c version)

    real	0m0.001s
    user	0m0.000s
    sys	        0m0.001s

Haskell (http://rosettacode.org/wiki/N-queens_problem#Haskell)

Program 1 (modified to do 8 queens like the c versions, compiled with: ghc -O2 -threaded haskell2.hs -o bin/haskell2)[2]

    real	0m0.010s
    user	0m0.005s
    sys	        0m0.005s

Program 2 (modified to do 8 queens like the c version, compiled with: ghc -O2 -threaded haskell2.hs -o bin/haskell2)[3] (word of warning, this program is only 52% efficient according to ghc and has a lot of optimization that could be done)

    real	0m0.006s
    user	0m0.006s
    sys	        0m0.000s

Note that the fastest Haskell solution was also the shortest by quite a bit. Are you convinced yet?

EDIT: Found an ATS solution, but I'm not setup to compile it: http://www.ats-lang.org/SERVER/MYCODE/Patsoptaas_serve.php?m...

0: C, solution 1

    #include <stdio.h>
    #include <stdlib.h>
 
    int count = 0;
    void solve(int n, int col, int *hist)
    {
    	if (col == n) {
    		printf("\nNo. %d\n-----\n", ++count);
    		for (int i = 0; i < n; i++, putchar('\n'))
    			for (int j = 0; j < n; j++)
    				putchar(j == hist[i] ? 'Q' : ((i + j) & 1) ? ' ' : '.');
     
    		return;
    	}
     
    #	define attack(i, j) (hist[j] == i || abs(hist[j] - i) == col - j)
    	for (int i = 0, j = 0; i < n; i++) {
    		for (j = 0; j < col && !attack(i, j); j++);
    		if (j < col) continue;
     
    		hist[col] = i;
    		solve(n, col + 1, hist);
    	}
    }
     
    int main(int n, char **argv)
    {
    	if (n <= 1 || (n = atoi(argv[1])) <= 0) n = 8;
    	int hist[n];
    	solve(n, 0, hist);
    }

1: c solution 2

    #include <stdio.h>
    #include <stdlib.h>
    #include <stdint.h>
     
    typedef uint32_t uint;
    uint full, *qs, count = 0, nn;
     
    void solve(uint d, uint c, uint l, uint r)
    {
    	uint b, a, *s;
    	if (!d) {
    		count++;
    #if 0
    		printf("\nNo. %d\n===========\n", count);
    		for (a = 0; a < nn; a++, putchar('\n'))
    			for (b = 0; b < nn; b++, putchar(' '))
    				putchar(" -QQ"[((b == qs[a])<<1)|((a + b)&1)]);
    #endif
    		return;
    	}
     
    	a = (c | (l <<= 1) | (r >>= 1)) & full;
    	if (a != full)
    		for (*(s = qs + --d) = 0, b = 1; b <= full; (*s)++, b <<= 1)
    			if (!(b & a)) solve(d, b|c, b|l, b|r);
    }
     
    int main(int n, char **argv)
    {
    	if (n <= 1 || (nn = atoi(argv[1])) <= 0) nn = 8;
     
    	qs = calloc(nn, sizeof(int));
    	full = (1U << nn) - 1;
     
    	solve(nn, 0, 0, 0);
    	printf("\nSolutions: %d\n", count);
    	return 0;
    }

2: Haskell solution 1 (comments stripped out for comparison)

    import           Control.Monad
    import           Data.List
    
    queens :: Int -> [[Int]]
    queens n = map fst $ foldM oneMoreQueen ([],[1..n]) [1..n]  where
      oneMoreQueen (y,d) _ = [(x:y, delete x d) | x <- d, safe x]  where
        safe x = and [x /= c + n && x /= c - n | (n,c) <- zip [1..] y]
    printSolution y = do
         let n = length y
         mapM_ (\x -> putStrLn [if z == x then 'Q' else '.' | z <- [1..n]]) y
         putStrLn ""
    
    main = mapM_ printSolution $ queens 8

3: Haskell solution 2

    import           Control.Monad (foldM)
    import           Data.List     ((\\))
    
    main :: IO ()
    main = mapM_ print $ queens 8
    
    queens :: Int -> [[Int]]
    queens n = foldM f [] [1..n]
        where
          f qs _ = [q:qs | q <- [1..n] \\ qs, q `notDiag` qs]
          q `notDiag` qs = and [abs (q - qi) /= i | (qi,i) <- qs `zip` [1..]]