codeforces 688E E. The Values You Can Make(dp)

题目链接：

E. The Values You Can Make

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 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, ..., 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.

Examples

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：

n个数,现在取和为k的数,用这些数能组成哪些数;

思路：

dp[i][j]表示和为i的那些数能否组成j,dp[i][j]=1表示可以,0表示不行,

现在面临第c[x],当把c[x]加到已有的集合中当dp[i][j]==1时dp[i+c[x]][j]=dp[i+c[x]][j+c[x]]=1;
转移的时候枚举i要从大到小枚举,否则当前转移的c[x]较小的和会对当前c[x]较大的和产生影响;

AC代码：

//#include <bits/stdc++.h>

#include <vector>

#include <iostream>

#include <queue>

#include <cmath>

#include <map>

#include <cstring>

#include <algorithm>

#include <cstdio>

using namespace std;

#define Riep(n) for(int i=1;i<=n;i++)

#define Riop(n) for(int i=0;i<n;i++)

#define Rjep(n) for(int j=1;j<=n;j++)

#define Rjop(n) for(int j=0;j<n;j++)

#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {

    char CH; bool F=false;

    for(CH=getchar();CH<''||CH>'';F= CH=='-',CH=getchar());

    for(num=;CH>=''&&CH<='';num=num*+CH-'',CH=getchar());

    F && (num=-num);

}

int stk[], tp;

template<class T> inline void print(T p) {

    if(!p) { puts(""); return; }

    while(p) stk[++ tp] = p%, p/=;

    while(tp) putchar(stk[tp--] + '');

    putchar('\n');

}

const LL mod=1e9+;

const double PI=acos(-1.0);

const LL inf=1e18;

const int N=+;

const int maxn=;

const double eps=1e-;

int n,k;

int c[N],dp[][*N];

int main()

{

        read(n);read(k);

        for(int i=;i<=n;i++)

            read(c[i]);

       // freopen("out.txt","w",stdout);

        dp[][]=;

        for(int i=;i<=n;i++)

        {

            for(int j=k;j>=;j--)

            {

                if(j+c[i]<=k)

                {

                    for(int x=;x<=k;x++)

                    {

                        if(dp[j][x])dp[j+c[i]][x]=dp[j+c[i]][x+c[i]]=;

                    }

                }

            }

        }

        int ans=;

        for(int i=;i<=k;i++)

        {

            if(dp[k][i])ans++;

        }

        cout<<ans<<"\n";

        for(int i=;i<=k;i++)

        {

            if(dp[k][i])printf("%d ",i);

        }

        return ;

}

巴特西

codeforces 688E E. The Values You Can Make(dp)

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