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Problem solution · Java

CCC 2015 J5 - Pi-Day

CCC 2015 J5 - Pi-Day: a Java solution using dynamic programming. Learn the idea, check the complexity, and read the full code, with credit to CCCSolutions.

Technique
Dynamic programming
Source
CCCSolutions
Length
27 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Dynamic programming

For CCC 2015 J5 - Pi-Day, the implementation records answers for smaller states and reuses them to build the requested result without repeating work.

  1. Define precisely what one DP state represents.
  2. Establish the base cases before transitions are evaluated.
  3. Process states in dependency order and combine only already-known values.

Code notes

  • 27 lines of Java from the credited upstream file ccc15j5.java.
  • The implementation visibly relies on sequence storage, cached states.
  • No explicit loop blocks detected.

Complexity

Multiply the number of reachable states by the work performed for each transition, then include the stored state table in memory usage.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from CCCSolutions by CCCSolutions contributors · Milliken Mills High School and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeCCC 2015 J5 - Pi-Day · JavaJava
Use this to learn the idea, then write your own version.
//Ivan Li, Markville secondary schoolimport java.io.*;import java.util.*;public class ccc15j5{   public static int memo[][] = new int[251][251];   public static int solve(int n, int k){    if(k<0||n<1) return 0;    if(k==0) return 1;    if(memo[n][k]!=0) return memo[n][k];    return memo[n][k]=solve(n-1, k)+solve(n, k-n);  }    public static void main(String[] args) {    Scanner sc = new Scanner(System.in);     int k  =sc.nextInt();    int n = sc.nextInt();     System.out.println(solve(n, k-n));   } }

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