To solve this problem, the coding isn't hard -- it's more about understanding the underlying structure of the cow herding process in the first place.

Suppose our three cows are at locations $a < b < c$. For the minimum number of moves, there are a few different cases to differentiate. The answer is clearly zero if the three cows are already consecutive, and one the cows are consecutive except for a single unit-sized gap (e.g., "3, 4, 6"). In all other cases, the answer is two. For example, if there is a gap between $a$ and $b$ and also between $b$ and $c$, then we can move $a$ to $b+1$ and then $c$ to $b-1$. Otherwise, we have two adjacent cows (say, $a$ and $b$) and a gap of size at least two with the third cow --- in which case we can move $a$ to $b+2$ and $c$ to $b+1$.

For the maximum number of moves, our main observation is the following: consider the gap between $a$ and $b$, and that gap between $b$ and $c$. After the first move, one of these gaps essentially "goes away", meaning there cannot be any cows landing inside the gap. On the other hand, we can strategically move cows so the other gap has every empty space used by a cow at some point in time --- the main idea here is to ensure we always have two adjacent cows at one endpoint, flipping back and forth between which side has the two adjacent cows. So the answer for the maximum is related to which of the $a \ldots b$ and $b \ldots c$ gaps is largest, since we can land cows in all the empty spaces in this gap, and none in the other gap.

Here is my code that expresses this idea:

#include <iostream> #include <fstream> using namespace std; int main(void) { int a, b, c; ifstream fin ("herding.in"); fin >> a >> b >> c; // Arrange in sorted order if (a > b) swap(a,b); if (b > c) swap(b,c); if (a > b) swap(a,b); ofstream fout ("herding.out"); if (c==a+2) fout << "0\n"; else if (b==a+2 || c==b+2) fout << "1\n"; else fout << "2\n"; fout << max(b-a, c-b) - 1 << "\n"; return 0; }