Codeforces 688E The Values You Can Make【Dp】

经典01背包问题求解

最新推荐文章于 2020-11-12 22:43:48 发布

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本文探讨了一个经典的01背包问题变种，通过给定的一组数值和一个目标值K，找出能构成目标值K的所有子集，并进一步确定这些子集中所有可能组合的数值种类及其具体数值。

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

题目大意：

给你N个数值，再给你一个值K.每个数值只能用一次.问你能够组成数值K的集合S.其任意组合出来的值的集合有多少个数，分别都是什么。

思路：

经典的01背包思维。

设定dp【i】【j】表示组合值目标为K的集合已经凑成数字i.是否能够通过组合出来数字i的一些数来组合出来j.

dp【i】【j】=1表示可以，0表示不可以。

那么初始化dp【0】【0】=1；

接下来两层for维护dp【】【】；经典的01背包思维。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int ans[505];
int a[505];
int dp[505][505];
int main()
{
    int n,k;
    while(~scanf("%d%d",&n,&k))
    {
        for(int i=0;i<n;i++)scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
        for(int z=0;z<n;z++)
        {
            for(int i=k;i>=0;i--)
            {
                for(int j=k;j>=0;j--)
                {
                    if(i>=a[z])
                    dp[i][j]=max(dp[i][j],dp[i-a[z]][j]);
                    if(i>=a[z]&&j>=a[z])
                    dp[i][j]=max(dp[i][j],dp[i-a[z]][j-a[z]]);
                }
            }
        }
        int cnt=0;
        for(int i=0;i<=k;i++)
        {
            if(dp[k][i]>0)
            {
                ans[cnt++]=i;
            }
        }
        printf("%d\n",cnt);
        for(int i=0;i<cnt;i++)
        {
            printf("%d ",ans[i]);
        }
        printf("\n");
    }
}