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.
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.
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.
6 18 5 6 1 10 12 2
16 0 1 2 3 5 6 7 8 10 11 12 13 15 16 17 18
3 50 25 25 50
3 0 25 50
设dp[i][j] = 1表示和为i的元素集合中，部分元素之和可以为j. 状态转移方程dp[i+a][j] = dp[i+a][j+a] = 1;
因为已知元素可以组成i(a还没用)，那么i+a必定也能被组成。
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#define maxn 10005
#define INF 100000000
using namespace std;
int dp[505][505];
vector<int> ans;
int main(){
// freopen("in.txt", "r", stdin);
int n, k, a;
scanf("%d%d", &n, &k);
dp[0][0] = 1;
for(int i = 0; i < n; i++){
scanf("%d", &a);
for(int j = k-a; j >= 0; j--)
for(int h = j; h >= 0; h--){
if(dp[j][h]){
dp[j+a][h] = dp[j+a][h+a] = 1;
}
}
}
for(int i = 0; i <= k; i++)
if(dp[k][i])
ans.push_back(i);
printf("%d\n", ans.size());
printf("%d", 0);
for(int i = 1; i < ans.size(); i++)
printf(" %d", ans[i]);
puts("");
return 0;
}