PERF: use c-division in nancorr
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There's three divisions in this function that can safely be done at the C-level:

        meanx += 1 / nobs * dx
        meany += 1 / nobs * dy

because `nobs` is at least `1` when we get here (it starts at `0`, `nobs+=1` is called a few lines
above, and `nobs` can't decrease), and

        result[xi, yi] = result[yi, xi] = covxy / divisor

because `divisor` is checked to not be equal to `0` in the preceding line.

This makes a noticeable difference, example:
```
arr1 = np.random.randn(100, 100)
```
On main:
```
%%timeit
nancorr(arr1)

5.39 ms ± 21.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
```

On this branch:
```
%%timeit
nancorr(arr1)

3.86 ms ± 58.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
```

I didn't find much difference in nancorr_spearman, I think because that one calls
the Python function rank_1d within the loop, so that might be a bigger bottleneck there
nvm, looks like this changes results (failing tests) - not sure I see why, will look into it, but closing for now

EDIT ok might have fixed this now, just a matter of marking the numerator in the division as float

running the same timing test as above now gives:

main:
```
5.68 ms ± 123 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
```

here:
```
4.97 ms ± 60.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
``` thanks @MarcoGorelli 