poj3628 Bookshelf 2 (0/backpack) __integer

To give N and b, and then give n number, with this n number of some, to find a and, this and is the minimum value of >=b, the output of the minimum and B difference.

Analysis: This problem is very simple, is very obvious 01 knapsack problem, here n items, the weight of each item is C[i], value is w[i] and c[i]==w[i],, capacity for all C[i] and sum. Just start from scratch in f[] and find a minimal >=b is the solution to the problem

Description

Farmer John recently bought another bookshelf for the Cow library, but the shelf was getting filled up quite quickly, and N Ow the "only available" is in the top.

FJ has N cows (1≤n≤20) each with some height of Hi (1≤hi≤1,000,000-these are, very tall). The bookshelf has a height of B (1≤b≤s, where S is the sum of the heights of all cows).

To reach is the top of the bookshelf, one or more of the cows can stand on top of each of the "a stack, so" their total The is the sum of each of the their individual heights. This total height must is no less than the height of the bookshelf in order for the "cows to reach".

Since a taller stack of cows than necessary can be dangerous, your job are to find this set of cows that produces a stack of The smallest height possible such the stack can reach the bookshelf. Your program should print the minimal ' excess ' height between the optimal stack of cows and the bookshelf.

Input

* Line 1:two space-separated integers:n and B
* Lines 2..n+1:line i+1 contains a single Integer:hi

Output

* Line 1: A single integer representing the (non-negative) difference between the total height of the optimal set of cows And the height of the shelf.

Sample Input

5
3
1
3
5
6

Sample Output

1

Code:

#include <iostream>
using namespace std;
int f[1000050],c[10000];
int main ()
{
    int i,j,n,v,sum;
    while (scanf ("%d%d", &n,&v)!=eof)
    {
       memset (f,0,sizeof (f));
       Memset (C,0,sizeof (c));
       sum=0;
       for (i=1;i<=n;i++)
       {
         scanf ("%d", &c[i]);
         Sum+=c[i];
       }
       for (i=1;i<=n;i++)
       {for
         (j=sum;j>=c[i];j--)
           F[j]=max (F[j],f[j-c[i]]+c[i]);
       for (i=1;i<=sum;i++)
       {
         if (f[i]>=v)
         {
           printf ("%d\n", f[i]-v);
           break;
         }
    }} return 0;
}