(Analysis by Nathan Pinsker)

This is a standard knapsack problem with a bit of a twist. After Bessie drinks water once, her fullness value will decrease, but she won't be able to drink water again. A standard knapsack algorithm won't quite work here because the maximum fullness Bessie can achieve might not be the result of a non-decreasing function.

Since Bessie can only drink water once, we can separate the problem into two cases: the case where Bessie can still drink, and the case where she has already drunk. In both cases, we can run standard knapsack. Furthermore, notice that if we first handle the case where Bessie can still drink, then this allows us to obtain all possible points at which Bessie can start eating again, right after drinking. Then, we can just run another standard knapsack algorithm on the second case, now that we have the list of possible starting locations. Taking the maximum obtainable value over both cases gives us the answer.

Here is Nick Wu's code that implements this idea. (He sets seen[0][X] to true if Bessie can attain X units of fullness without drinking water, and seen[1][X] to true if Bessie can attain X units of fullness after drinking water.)

import java.io.*;
import java.util.*;
public class feast {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new FileReader("feast.in"));
StringTokenizer st = new StringTokenizer(br.readLine());
int t = Integer.parseInt(st.nextToken());
int x = Integer.parseInt(st.nextToken());
int y = Integer.parseInt(st.nextToken());
boolean[][] seen = new boolean[2][t+1];
seen[0][0] = true;
for(int a = 0; a < seen.length; a++) {
for(int i = 0; i < seen[a].length; i++) {
if(!seen[a][i]) {
continue;
}
if(i+x <= t) {
seen[a][i+x] = true;
}
if(i+y <= t) {
seen[a][i+y] = true;
}
if(a+1 < seen.length) {
seen[a+1][i/2] = true;
}
}
}
int ret = t;
while(!seen[1][ret]) {
ret--;
}
PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("feast.out")));
pw.println(ret);
pw.close();
}
}