顯然，n<=20, m<=4 的資料範圍一眼爆搜。
直接搜尋一下不用哪4個砝碼，再做一遍01揹包即可。
可能是本人太菜雞，01揹包部分調了半天QAQ……

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
using namespace std;
const int maxn = 2002;
int ans,n,target,sumv, w[30];
bool f[maxn], mark[30];
inline int solve()
{
    memset
 
(f,0,sizeof(f));
    int fin = 0, tot = 0;
    for(int i = 1;i <= n;++i)
    {
        if(mark[i])continue;
        tot += w[i];
        for(int j = sumv;j >= w[i];--j)if(f[j-w[i]])f[j] = 1;
        f[w[i]] = 1;
    }
    for(int i = 1;i <= tot; ++i)if(f[i]) ++ fin;
    return fin;
}
void dfs(int
 
 nums,int cur)
{
    if(nums == target)
    {
        ans = max(ans,solve());
        return;
    }
    for(int i  = cur + 1;i <= n;++i)
    {
        mark[i] = 1;
        dfs(nums + 1, i);
        mark[i] = 0;
    }
}
int main()
{
   // freopen("input.txt","r",stdin);
    scanf("%d%d",&n,&target);
    for
 
(int i = 1;i <= n;++i)
    {
        scanf("%d",&w[i]);
        sumv += w[i];
    }
    if(target == 0){printf("%d",solve());return 0;}
    for(int i = 1;i <= n;++i)
    {
        mark[i] = 1;
        dfs(1,i);
        mark[i] = 0;
    }
    printf("%d",ans);
    return 0;
}