您的答案很接近，但您误解了查询2中的操作

在y % size中找到元素seq的值（其中size是seq的大小）并将其分配给

因此，使用索引您将获得一个将成为列表的序列，但是随后您需要在列表中的位置y % size的索引中找到该值，其中size = len(seq[index])

def dynamicArray(n,queries):
    lastAnswer = 0
    seq = []
    result = []
    for k in range(0,n):
        seq.append([])

    for i in queries:
        N_querytype = i[0] #it can be either 1 or 2
        x = i[1]
        y = i[2]
        index = (x ^ lastAnswer) % n
        print("#",index,y)
        if(N_querytype == 1):
            seq[index].append(y)
        elif(N_querytype == 2):
            size = len(seq[index])   #Calculate the size of the sequnce at this index
            lastAnswer = seq[index][y % size] #look up the value in this sequence at index y % size
            result.append(lastAnswer)
    return result