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

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

题意

给n个数，从中挑若干个元素使其和为k，组成集合S，S的某一子集元素之和为x，问所有x可能的值

思路

dp，状态dp[i][j][p]为考虑到第i个数，当前所有数的和为j，组成和为p的子集是否可能。
如果不使用第i个数，dp[i][j][p]=dp[i-1][j][p]
如果使用第i个数但子集不取他，dp[i][j][p]=dp[i-1][j-ci][p]
如果使用第i个数且子集取他，dp[i][j][p]=dp[i-1][j-ci][p-ci]

代码

#include <cstdio>
#include <vector>
using namespace std;
bool dp[505][505][505];
vector<int> ans;
int main()
{
    int n,k,c,t=0;
    scanf("%d %d",&n,&k);
    dp[0][0][0]=1;
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&c);
        for (int j=0;j<=500;j++)
        {
            for (int p=0;p<=j&&p<=k;p++)
            {
                if (dp[i-1][j][p] || j>=c&&dp[i-1][j-c][p] || p>=c&&dp[i-1][j-c][p-c]) dp[i][j][p]=1;
            }
        }
    }
    for (int i=0;i<=k;i++)
    {
        if (dp[n][k][i])
        {
            ans.push_back(i);
        }
    }
    int l=ans.size();
    printf("%d\n",l);
    for (int i=0;i<l;i++) printf("%d ",ans[i]);
}