I am a bit confused regarding the time complexity here.

X <= 1000, and number of elements is also at-max 1000.

In the worst case, assume 1 insertion and then 1 subset query alternating one after another.

500 subset queries and for each query you'll need O(N*|X|) operations. So the total operation turns out to be around 500*500*1000 ≈ 2.5e8 operations? Maybe brute forcing using knapsack won't give TLE.

I think its very common to add in knapsack in O(size), similarly we can delete from knapsack also in O(size).

    vector<int> dp(1001,0);
    dp[0] = 1;
    const int mod = 1e9 + 7;
    int Q;
    cin >> Q;
    while(Q--) {
        int tt,x;
        cin >> tt >> x;
        if(tt == 0) {
            for(int i=1000;i>=x;--i)
                dp[i] = (dp[i] + dp[i-x]) % mod;
        }
        else if(tt == 1) {
            for(int i=x;i<=1000;++i)
                dp[i] = (dp[i] - dp[i-x] + mod) % mod;
        }
        else {
            cout << dp[x] << "\n";
        }
    }

Note that the addition part is quite easy, and the subtraction also works the same way(it is clear that the order doesn't matter when we add in the knapsack, intutively you can assume that the last element added was the one to be deleted and try to think how would you undo it).

Hope it helps.

We could solve this in $$$ Q * X * sqrt(Q)$$$ with square root decomposition, i already explained this on our group to someone. Hope at that time it was not live.

So basically you can decompose array into square root buckets or now when you remove the element, you can recalculate the current knapsack bucket, as you know the all the element in square root bucket. this will handle for the update part. Complexity of update is $$$ X * sqrt(Q)$$$ , as there will be only sqrt elements

Now for the query part you know the knapsack states, so you can iterate on square root buckets to get number of ways to get sum = X, complexity will be same $$$ X * sqrt(Q)$$$, for finding the ways part to get sum = X, on first two buckets its just $$$ \sum _{i = 0} ^{ i = X} ways1[i] * ways2[X-i] $$$, similary you can keep combining those buckets and after this remaining atmost square root element, can be added in this in same way as we did in update.

Another way is kind of FFT and segtree, i am not sure about this approach, but we can build convolution tree, someway, i will think this in more detail and write the answer.

I think knapsack is the way to go. Because the third type of queries only ask for subset sums of $$$X \lt =10^3$$$, we only need the knapsack DP array to be of length 1001. Say we have calculated the DP array for some numbers, then it's easy to calculate a new DP array in $$$O(maxX)$$$, extending the set of numbers by adding number X (query type add). Remove queries make it harder. We can use offline deletion from datastructure only supporting insertion and a stack of DP arrays to answer all the queries offline, and support delete. This only costs an extra log factor, and would give a runtime of $$$O(maxX*Q*log(Q))$$$

Edit: Oh, removing is just as easy as adding, because instead of adding, we know subtract all DP transitions. My approach could still be useful for queries of the form: Is a subset sum of X possible?

What is goc33?

You didn't write the correct constraints. The actual constraints were 7*1e3 for both 'X' and the queries. That is surely going to make a difference right?

Would it be possible for you to share the first question of your test as well?? It would really help!!

Its addition and deletion on knapsack, I saw this idea here IZho017 bootfall