### Rei's blog

By Rei, 12 years ago, translation,
Problem A.
This problem is solved by simple emulation. After 2n jumps flea returns to initial position, because 1 + 2 + 3 + ... + 2n = n(2n + 1) is divisible by n. Moreover, after that, jumps would be the same as in the beginning, because 2n is divisible by n. So it is just enough to emulate first 2n jumps.
In fact one may see that it is enough to emulate first n jumps and moreover answer is "YES" exactly when n is power of two. Last gives alternative solution. For example: printf("%s", n&(n-1) : "NO" ? "YES");

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 12 years ago, # |   0 Can you please explain how did you deduce that answer will be yes only when n is a power of 2??
•  12 years ago, # ^ | ← Rev. 5 →   +4 Let n be a power of two: n = 2m. Let us prove that first n - 1 jumps will end at different hassocks. Really, jump with number k ends at hassock . Suppose that jumps k1 and k2 ends at the same hassock. It means that . So we get that k12 + k1 - k22 - k2 = (k1 - k2)(k1 + k2 + 1) is divisible by 2m + 1. But k1 - k2 and k1 + k2 + 1 have different parity and both less than 2m + 1. A contradiction. So we derive that all hassocks will be visited after first n - 1 jumps.If n isn't a power of two, then n is divisible by some odd prime p. Let's look at hassocks not mod n, but mod p. We will see that not all residues (mod p) will be visited. Indeed after p jumps flea returns to initial residue mod p, because is divisible by p. And after that jumps would be the same as in the beginning (mod p), because p is divisible by p. Moreover, even after p - 1 jumps flea returns to initial residue mod p. So there are p - 1 different residues as maximum. Thus not all hassocks will be visited.
 12 years ago, # |   0 Ya please explain why n should be a power of 2 ?
 » 10 years ago, # |   0 How is it enough to emulate only first n jumps ???