The Values You Can Make CodeForces

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.

Example
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

大致题意：给你n个数，让你选出一些数使它们的和为k，然后问这些数所能构成的数有哪些？

思路：假设dp[i][j]表示 选择的数的和为i时，是否能构成数j。1表示能，0表示不能。那么，当dp[i-x][j]=1时（x为和为i的子集中的一个数）dp[i][j]=dp[i][j+x]=1.

代码如下

#include <iostream> #include <cstring>#include <cstdio>#include <vector>#include<algorithm>using namespace std; #define LL long long vector<int> ans;int a[505];int dp[505][505];int main(){    int n,k;    scanf("%d%d",&n,&k);    for(int i=1;i<=n;i++)    scanf("%d",&a[i]);    sort(a+1,a+1+n);    dp[0][0]=1;    for(int i=1;i<=n;i++)    for(int j=k;j>=a[i];j--)    for(int l=j-a[i];l>=0;l--)    if(dp[j-a[i]][l])    dp[j][l]=dp[j][l+a[i]]=1;    for(int i=0;i<=k;i++)    if(dp[k][i]) ans.push_back(i);    printf("%d\n",ans.size());    for(int i=0;i<ans.size();i++)    printf("%d ",ans[i]);    return 0; }