⌈√n⌉ ( =ceil(sqrt(n)) )

After a few days a lot of those handshakes are going to be towards people that have already shook hands with someone, and some countries will take a long time to reach. If the handshake begins in a western country, most of Europe, the Middle East, the USA, Canada, Australia, New Zealand, Japan, and South Korea will have shook hands after a year, but places like Nigeria will probably be untouched. After 10 years, assuming every is still shaking the hand of two different people per day, the handshake will probably have reached most of Latin America, Africa, India, China, Siberia, and just about any well-populated settlement in every country. However, there will likely still be many untouched remote villages, and the handshake will probably only have reached about 99% of the Earth's population. Any longer timescale will have to take into account macroeconomic predictions for every country i.e. Subsaharan countries industrializing and becoming popular targets for incoming western handshaking businessmen in 2045 A.D. Also, handshakes are not going to be exchanged between the people of this island and the rest of the world any time soon, so we can simplify your question to "how long will it take everyone on that remote island to shake hands?". The time frame for that, however, is so long that you'll also have to specify whether or not your original question includes people who shook hands on Earth but later left for the Martian/ Moon colonies. This is also ignoring the cultural meanings of a handshake which may ensure that handshakes are directed towards already-handshaked people rather than new people who don't want to receive a handshake.

Best case scenario: 2. You start by chance on the floor one floor below, drop one, and it doesn't work. You move one floor up, and it does.
Worst case scenario: 7. to guarantee you find it in as few drops as possible no matter what floor your choose and what floor it's on: An idealized algorithm would attempt to rule out as many floors as possible per drop. The best option for this is to start from the middle at 50 (ruling out 1-49 or 50-100), then going halfway between your current floor and a boundary. The boundaries for the first round are 1 if the ball broke and 100 if it did not. On every other round the boundary is in the same direction you moved before (if the result is unchanged) or the floor of the previous round (if the result changed). Repeat until you have found the two adjacent floors for which the upper breaks the sphere and the lower does not. For example, if the lowest breaking floor is 32 the algorithm will find "50, 25, 37, 31, 34, 32". This algorithm requires five, six, or seven spheres depending on how it rounds, as 100*(1/2)^x drops below 1 for x=7. That is to say the algorithm is always guaranteed, regardless of rounding method, to find an answer after 7 spheres.

I'm not quite familiar with this particular syntax, could you explain clearly how this works?

From my limited experience with javascript, it seems to be telling the program to subtract from n 2x from all integer x's between 0 and n, exclusive, and then round up. That is to say, n - SUM(x=0,n-1,2x)=n-2[(n-1)^2+(n-1)]/2=n-(n^2-2n+1+n-1)=n-n^2+n=2n-n^2, then ceiling(2n-n^2). What am I missing?

EDIT: Nevermind, I read "return x" as "return n". When the problem reads "return x" you have to find the smallest integer x such that sum(k=0,x,2k)>n for all n. This sum can be simplified, leading to x^2+x>n. Since this has to work at infinite n, we can discard the x^1 and get x^2>n or x>sqrt(n). Therefore x is the first integer after sqrt(n).

Yes, it's a variation of binary search. For n floors, you can then always find the answer with ⌊log₂n⌋+1 spheres in this case (remember that we don't even know if it will break from the top floor).

In the case when the number or floors is infinite, there is no of course no minimum number of squares, but it's still likely that you wont need many. If the lowest floor from which it breaks is n, then you'll need 2⌊log₂n⌋+2 spheres, I believe. You would then start at 1st floor and double until it breaks, then do binary search.

Perfect reasoning and the right way to prove what it does! To clarify, though, x is incremented after n -= 2*x. This is equivalent: for (x = 0; n > x; n -= 2*x++);

The algorithm can easily be adapted to round the other way.

(Original conditions had eggs instead these spheres, because I can't imagine what kind of eggs it could be)

I think I might have the first part right:
1: 1c
2: 1c
3: 2c
4: 5c
5: 10c
6: 20c
But if I were to go on, I wouldn't use all 1000 coins. So I don't know.