Problem:
Score : 100 points

Problem Statement

There are N items, numbered 1, 2, \ldots, N.
For each i (1 \leq i \leq N), Item i has a weight of w_i and a value of v_i.

Taro has decided to choose some of the N items and carry them home in a knapsack.
The capacity of the knapsack is W, which means that the sum of the weights of items taken must be at most W.

Find the maximum possible sum of the values of items that Taro takes home.

Constraints

All values in input are integers.

1 \leq N \leq 100

1 \leq W \leq 10^9

1 \leq w_i \leq W

1 \leq v_i \leq 10^3

Input

Input is given from Standard Input in the following format:

N W
w_1 v_1
w_2 v_2
:
w_N v_N

Output

Print the maximum possible sum of the values of items that Taro takes home.

Sample Input 1

3 8
3 30
4 50
5 60

Sample Output 1

90

Items 1 and 3 should be taken.
Then, the sum of the weights is 3 + 5 = 8, and the sum of the values is 30 + 60 = 90.

Sample Input 2

1 1000000000
1000000000 10

Sample Output 2

10

Sample Input 3

6 15
6 5
5 6
6 4
6 6
3 5
7 2

Sample Output 3

17

Items 2, 4 and 5 should be taken.
Then, the sum of the weights is 5 + 6 + 3 = 14, and the sum of the values is 6 + 6 + 5 = 17.

Solution in Java7 (OpenJDK 1.7.0) (judge verdict: Runtime Error)



import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int numOfItems = sc.nextInt();
        int knapSackWeight = sc.nextInt();
        int[] itemValue = new int[numOfItems+1];
        int[] itemWeight = new int[numOfItems+1];
        for(int i =1;i<=numOfItems;i++){
            itemWeight[i] = sc.nextInt();
            itemValue[i] = sc.nextInt();
        }
//        Arrays.sort(itemWeight);
        System.out.println(KnapSack(itemValue,itemWeight,numOfItems,knapSackWeight));
    }
    static int KnapSack(int val[], int wt[],
                        int n, int W) {
        // matrix to store final result  
        int mat[][] = new int[2][W + 1];

        // iterate through all items  
        int i = 0;
        while (i < n) // one by one traverse each element  
        {
            int j = 0; // traverse all weights j <= W  

            // if i is odd that mean till now we have odd  
            // number of elements so we store result in 1th  
            // indexed row  
            if (i % 2 != 0) {
                while (++j <= W) // check for each value  
                {
                    if (wt[i] <= j) // include element  
                    {
                        mat[1][j] = Math.max(val[i] + mat[0][j - wt[i]],
                                mat[0][j]);
                    } else // exclude element  
                    {
                        mat[1][j] = mat[0][j];
                    }
                }

            }

            // if i is even that means till now  
            // we have even number of elements 
            // so we store result in 0th indexed row  
            else {
                while (++j <= W) {
                    if (wt[i] <= j) {
                        mat[0][j] = Math.max(val[i] + mat[1][j - wt[i]],
                                mat[1][j]);
                    } else {
                        mat[0][j] = mat[1][j];
                    }
                }
            }
            i++;
        }

        // Return mat[0][W] if n is odd, else mat[1][W]  
        return (n % 2 != 0) ? mat[0][W] : mat[1][W];
    }
}