(Analysis by Nick Wu)

It isn't immediately clear how to fill the larger bucket using the size X bucket and the size Y bucket. Instead of trying to cleverly figure out how many times to use each bucket, we can fix the number of times we use the bucket of size X, and then continually use the bucket of size Y until we would overflow the old bucket. We can loop over all possible numbers for usages of the size X bucket.

Here is my Java code.

import java.io.*;
import java.util.*;
public class pails {
public static void main(String[] args) throws IOException {
// initialize file I/O
BufferedReader br = new BufferedReader(new FileReader("pails.in"));
PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("pails.out")));
StringTokenizer st = new StringTokenizer(br.readLine());
// read X, Y, and M
int x = Integer.parseInt(st.nextToken());
int y = Integer.parseInt(st.nextToken());
int m = Integer.parseInt(st.nextToken());
int ans = 0;
// loop over how many times we can pour the X-size bucket
for(int xPour = 0; xPour*x <= m; xPour++) {
// loop over how many times we can then pour over the Y-size bucket
for(int yPour = 0; xPour*x + yPour*y <= m; yPour++) {
// determine if we have filled the bucket more than any previous time
if(xPour*x + yPour*y > ans) {
ans = xPour*x + yPour*y;
}
}
}
// print the answer
pw.println(ans);
// close output stream
pw.close();
}
}