How can you easily calculate the square root of an unsigned long long in C?

There is no need for long double; the square root can be calculated with double (if it is IEEE-754 64-bit binary). The rounding error in converting a 64-bit integer to double is nearly irrelevant in this problem.

The rounding error is at most one part in 2⁵³. This causes an error in the square root of at most one part in 2⁵⁴. The sqrt itself has a rounding error of less than one part in 2⁵³, due to rounding the mathematical result to the double format. The sum of these errors is tiny; the largest possible square root of a 64-bit integer (rounded to 53 bits) is 2³², so an error of three parts in 2⁵⁴ is less than .00000072.

For a uint64_t x, consider sqrt(x). We know this value is within .00000072 of the exact square root of x, but we do not know its direction. If we adjust it to sqrt(x) - 0x1p-20, then we know we have a value that is less than, but very close to, the square root of x.

Then this code calculates the square root of x, truncated to an integer, provided the operations conform to IEEE 754:

uint64_t y = sqrt(x) - 0x1p-20;
if (2*y < x - y*y)
    ++y;

(2*y < x - y*y is equivalent to (y+1)*(y+1) <= x except that it avoids wrapping the 64-bit integer if y+1 is 2³².)

Function sqrtl(), taking a long double, is part of C99.

Note that your compilation platform does not have to implement long double as 80-bit extended-precision. It is only required to be as wide as double, and Visual Studio implements is as a plain double. GCC and Clang do compile long double to 80-bit extended-precision on Intel processors.

Yes, the standard library has sqrtl() (since C99).