CODEFORCES 1181A

CODEFORCES

1181A - Chunga Changa

Link to Ques - CodeForces 1181A

Approach

• It's easy to calculate how much coconuts we will buy: k=⌊x+yz⌋$k=\lfloor \frac{x+y}{z}\rfloor$ (suppose that all money transferred to a single person, this way the number of bought coconuts would be clearly maximal)
• If k=⌊xz⌋+⌊yz⌋$k=\lfloor \frac{x}{z}\rfloor +\lfloor \frac{y}{z}\rfloor$, then the answer is ⟨k,0⟩$⟨k,0⟩$. The remaining case is a bit harder.
• Let's notice, that there is no need to transfer ≥z$\ge z$ chizhiks, since the one transferring money could have used z$z$ chizhiks to buy one more coconut herself.
• Also it's optimal to transfer coins such that the remainder modulo z$z$ of the receiving part will turn to be exactly zero (we could have simply transfer less for the same effect).
• So the answer is ⟨k,min(z−(xmodz),z−(ymodz))⟩$⟨k,min(z-(xmodz),z-(ymodz))⟩$.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

import java.util.Scanner; public class CF1181A { public static void main(String[] args) { Scanner sc = new Scanner(System.in); long x= sc.nextLong(); long y= sc.nextLong(); long z= sc.nextLong(); long total=(x+y)/z; if((x/z+y/z)<((x+y)/z)) { long change = Math.min(z-(x%z),z-(y%z)); System.out.println(total+" "+change); } else { System.out.println(total+" 0"); } } }