Problem:
0-1 Knapsack Problem

You have N items that you want to put them into a knapsack. Item i has value vi and weight wi.

You want to find a subset of items to put such that:

The total value of the items is as large as possible.

The items have combined weight at most W, that is capacity of the knapsack.

Find the maximum total value of items in the knapsack.

Input

N W
v1 w1
v2 w2
:
vN wN

The first line consists of the integers N and W. In the following lines, the value and weight of the i-th item are given.

Output

Print the maximum total values of the items in a line.

Constraints

1 ≤ N ≤ 100

1 ≤ vi ≤ 1000

1 ≤ wi ≤ 1000

1 ≤ W ≤ 10000

Sample Input 1

4 5
4 2
5 2
2 1
8 3

Sample Output 1

13

Sample Input 2

2 20
5 9
4 10

Sample Output 2

9

Solution in JAVA (judge verdict: Runtime Error)

import java.util.Scanner;
import java.lang.Integer;
import java.lang.Math;

class Item {
  private int value;
  private int weight;

  public Item(int value, int weight) {
    this.set(value, weight);
  }

  public void set(int value, int weight) {
    this.value = value;
    this.weight = weight;
  }

  public int getValue() {
    return this.value;
  }

  public int getWeight() {
    return this.weight;
  }

}

class Main {
  private static int[][] dp;

  private static int calc(int N, int W, Item[] item) {
    if (N <= 0) {
      return 1;
    }

    for (int i = 1; i < N + 1; i++) {
      for (int j = 0; j < W + 1; j++)  {
        int candidate1 = dp[i - 1][j];
        int candidate2 = 0;

        if ((j - item[i].getWeight() >= 0) && (item[i].getWeight() <= W)) {
          if (j >= item[i].getWeight() * 2) {
            candidate2 = (dp[i][item[i].getWeight() * 2 - 1]);
          } else {
            candidate2 = (dp[i][j - item[i].getWeight()] + item[i].getValue());
          }
        } else {
          candidate2 = Integer.MIN_VALUE;
        }

        dp[i][j] = Math.max(candidate1, candidate2);
        //System.out.println(((j - item[i].getWeight() >= 0) && (item[i].getWeight() <= W)) + " candidate1 " + candidate1 + " candidate2 " + candidate2);
        //System.out.println("dp[" + i + "]" + "[" + j + "], " + dp[i][j]);
      }
    }

    return dp[N][W];
  }

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int W = sc.nextInt();
    Item[] item = new Item[N + 1];
    item[0] = new Item(0, Integer.MIN_VALUE);
    for (int i = 1; i < N + 1; i++) {
      int value = sc.nextInt();
      int weight = sc.nextInt();
      item[i] = new Item(value, weight);
    }
    dp = new int[N + 1][W + 1];

    // init kind of coin(i) == 0;
    int i = 0;
    for (int j = 0; j < N + 1; j++) {
      if (j == 0) {
        dp[i][j] = 0;
      } else {
        dp[i][j] = 0;
      }
    }

    int result = calc(N, W, item);

    System.out.println(result);
  }
}