Codeforces Round #360 (Div. 2) E. The Values You Can Make dp

E. The Values You Can Make

题目连接：

http://www.codeforces.com/contest/688/problem/E

Description

Pari wants to buy an expensive chocolate from Arya. She has n coins, the value of the i-th coin is ci. 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 x using 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, ..., cn (1 ≤ ci ≤ 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

6 18
5 6 1 10 12 2

Sample Output

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

Hint

题意

所有能组成K的C的子集方案中，能拼出哪些面额

题解

n^3dp

dp[i][j][k]表示用到了第i个数，当前和为j，子集和为k可不可行

裸的会稍微卡一下空间，这个滚动数组优化，或者直接用bool就可以了。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 505;

int n,K;
bool dp[maxn][maxn][maxn];
int main(){
    scanf("%d%d",&n,&K);
    dp[0][0][0]=1;
    for(int i=1;i<=n;i++){
        int c;
        scanf("%d",&c);
        for(int j=0;j<=500;j++){
            for(int k=0;k<=j;k++){
                dp[i][j][k]|=dp[i-1][j][k];
                if(j>=c)dp[i][j][k]|=dp[i-1][j-c][k];
                if(k>=c)dp[i][j][k]|=dp[i-1][j-c][k-c];
            }
        }
    }
    vector<int> ans;
    for(int i=0;i<=500;i++)
        if(dp[n][K][i])ans.push_back(i);
    cout<<ans.size()<<endl;
    for(int i=0;i<ans.size();i++)
        cout<<ans[i]<<" ";
    cout<<endl;
}