codeforces 360 E

   E - The Values You Can Make

Description

Pari wants to buy an expensive chocolate from Arya. She has n coins,

the value of the i-th coin is c_i. The price of the chocolate is k, so Pari

will take a subset of her coins with sum equal to k and give it to Arya.

Looking at her coins, a question came to her mind: after giving the

coins to Arya, what values does Arya can make with them? She is

jealous and she doesn't want Arya to make a lot of values. So she

wants to know all the values x, such that Arya will be able to make

 xusing some subset of coins with the sum k.

Formally, Pari wants to know the values x such that there exists a

subset of coins with the sum k such that some subset of this subset

has the sum x, i.e. there is exists some way to pay for the chocolate,

such that Arya will be able to make the sum x using these coins.

Input

The first line contains two integers n and k (1  ≤  n, k  ≤  500) — the

number of coins and the price of the chocolate, respectively.

Next line will contain n integers c₁, c₂, ..., c_n (1 ≤ c_i ≤ 500) — the

values of Pari's coins.It's guaranteed that one can make value k

 using these coins.

Output

First line of the output must contain a single integer q— the number

of suitable values x. Then print q integers in ascending order — the

values that Arya can make for some subset of coins of Pari that pays

for the chocolate.

Sample Input

Input

6 18
5 6 1 10 12 2

Output

16
0 1 2 3 5 6 7 8 10 11 12 13 15 16 17 18 

Input

3 50
25 25 50

Output

3
0 25 50 

  题意：

   给你n个数和k，问n个数所有能构成k的子集合中所有的可能的和是多少？

分析：

dp[i][j]表示当前和是i能否构成j,如果dp[i][j]是可以构成的话，

那由于是子集合的关系dp[i+m][j]和dp[i+m][j+m]也可以构成。

dp:  dp[i][j]=dp[i+m][j]+dp[i+m][j+m];

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[505][505];
int a[505];
int main()
{
    int n,k,m;
    scanf("%d%d",&n,&k);
    memset(dp,0,sizeof(dp));
    dp[0][0]=1;
    for(int i=0;i<n;i++)
    {
        scanf("%d",&m);
        for(int a=k;a>=m;a--)
        {
                for(int b=0;b+m<=k;b++)
                 {
                     if(dp[a-m][b])
                     {
                         dp[a][b]=1;
                       dp[a][b+m]=1;
                     }
                 }
        }

    }
    int len=0;
   for(int b=0;b<=k;b++)
     if(dp[k][b])
   {
        a[len]=b;
        len++;
        }

    sort(a,a+len);
    printf("%d
",len);
    for(int i=0;i<len-1;i++)
        printf("%d ",a[i]);
    printf("%d
",a[len-1]);
    return 0;
}