按喜好程度排序, 每次按取的件数和当前花费金额分类记录, 剩下普通零食按完全背包更新就可以了, 这是后来改的, 不一定完全正确, 欢迎讨论..
觉得不错的点个赞呗

#include<bits/stdc++.h>
using namespace std;
vector<int> f;
bool cmp(const int& a, const int& b) {
    return f[a] < f[b];
}

int main(){
    int m, n, v, t;
    while (cin >> v >> n) {
        vector<int> a(n);
        for (int i = 0; i < n; i++) {
            cin >> a[i];
        }
        cin >> m;
        f = vector<int>(n, n);
        for (int i = 0; i < m; i++) {
            cin >> t;
            f[t - 1] = i;
        }
        sort(a.begin(), a.end(), cmp);
        vector<vector<int> > dp(v / a[0] + 1, vector<int>(v + 1, 0));
        int res = 0;
        // 先挑选最喜欢的零食, 无约束
        for (int i = 0; i < v / a[0] + 1; i++) {
            dp[i][i * a[0]] = 1;
        }
        // 按喜欢程度依次挑选其他特别喜欢的零食, 
        for(int i = 1; i < m; i++) {
            vector<vector<int>> t = move(dp);
            int size = v / a[i] + 1;
            dp = vector<vector<int> >(size, vector<int>(v + 1, 0));
            for (int j = 1; j < t.size(); j++) {
                // 每次取的要比前一种要少
                for (int k = min(size - 1, j - 1); k >= 0; k--) {
                    for (int l = k * a[i]; l <= v; l++){
                        dp[k][l] = (dp[k][l] + t[j][l - k * a[i]]) % 10000007;
                    }
                }
            }
        }
        // 统计各花费的方案数
        vector<int> cur(v + 1, 0);
        for (int i = 0; i <= v; i++) {
            for (int j = 0; j < dp.size(); j++) {
                cur[i] = (cur[i] + dp[j][i]) % 10000007;
            }
        }
        // 按完全背包更新其余零食
        for (int i = m; i < n; i++) {
            for (int j = a[i]; j <= v; j++) {
                cur[j] = (cur[j] + cur[j - a[i]]) % 10000007;
            }
        }
        cout << cur[v] << endl;
    }
    return 0;
}