Problem:
Knapsack Problem

You have N kinds of 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.

You can select as many items as possible into a knapsack for each kind.

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 8
4 2
5 2
2 1
8 3

Sample Output 1

21

Sample Input 2

2 20
5 9
4 10

Sample Output 2

10

Sample Input 3

3 9
2 1
3 1
5 2

Sample Output 3

27

Solution in JAVA (judge verdict: Time Limit Exceeded)


import java.util.Scanner;

public class Main {

    public static class KnapsackComplete {

        int capacity;
        int n;
        int[] cost;
        int[] value;

        public KnapsackComplete(int n, int capacity, int[] cost, int[] value) {
            this.n = n;
            this.capacity = capacity;
            this.cost = cost;
            this.value = value;
        }

        int solve() {
            int[] rowCurr;
            int[] rowPrev;
            rowCurr = new int[capacity + 1];
            rowPrev = new int[capacity + 1];
            for (int numOfItems = 1; numOfItems <= n; numOfItems++) {
                int[] tmp = rowCurr;
                rowCurr = rowPrev;
                rowPrev = tmp;
                int costThis = cost[numOfItems - 1];
                for (int weight = 1; weight <= capacity; weight++) {
                    rowCurr[weight] = rowPrev[weight];
                    for (int num = 1; weight >= costThis * num; num++) {
                        rowCurr[weight] = Math.max(rowCurr[weight],
                                rowPrev[weight - costThis * num] + value[numOfItems - 1] * num);
                    }
                }
            }
            return rowCurr[capacity];
        }

    }

    static Scanner in = new Scanner(System.in);

    static KnapsackComplete readCase() {
        while (in.hasNext()) {
            int n = in.nextInt();
            int capacity = in.nextInt();
            int[] cost = new int[n];
            int[] value = new int[n];
            for (int i = 0; i < n; i++) {
                value[i] = in.nextInt();
                cost[i] = in.nextInt();
            }
            KnapsackComplete knapsack = new KnapsackComplete(n, capacity, cost, value);
            return knapsack;
        }
        return null;
    }

    public static void main(String[] args) {
        KnapsackComplete knapsack = null;
        while ((knapsack = readCase()) != null) {
            System.out.println(knapsack.solve());
        }
    }

}