展开F(i,c), F(i,c+ci), F(i,c+2ci), 可以看到一个sliding window，但是入队的时候需要dp[i-1][j]-mci, m是当前ci的倍数。拿到最大值再把mci加回来即可。
一个单调队列是不够的，需要ci个，分别是F(i,c),F(i,c+1),F(i,c+2),....,F(i,c+ci-1)

private static int knapsackDPSlidingWindow(int N, int C, int[] ss, int[] vs, int[] qs) {
    int[][] dp = new int[N+1][C+1];
    for (int i=1; i<=N; i++) {
        int s = ss[i-1], v = vs[i-1], q = qs[i-1];
        List<Deque<int[]>> dqs = new ArrayList<>();
        List<Integer> ws = new ArrayList<>(); // window start index
        List<Integer> we = new ArrayList<>(); // window end index
        for (int j=0, k=0; j<=C; j++, k=(k+1)%s) {
            if (k == dqs.size()) {
                dqs.add(new LinkedList<>());
                ws.add(j);
                we.add(j);
            }
            Deque<int[]> dq = dqs.get(k);
            if ((we.get(k)-ws.get(k))/s == q) { // slide window already max out
                if (dq.size() > 0 && dq.peekFirst()[0] == ws.get(k)) { // if window start already the max, will eject it
                    dq.removeFirst();
                }
                ws.set(k, ws.get(k)+s); // increment the window start
            }
            int m = (j-k)/s;
            // try to add the current index
            while (!dq.isEmpty() && dq.peekLast()[1] <= dp[i-1][j]-m*v) {
                dq.removeLast();
            }
            dq.addLast(new int[]{j, dp[i-1][j]-m*v});
            we.set(k, j);
            dp[i][j] = dq.peekFirst()[1]+m*v;
        }
    }
    return dp[N][C];
}