The+task+of+this+problem+is+simple:+insert+a+sequence+of+distinct+positive+integers+into+a+hash+tabl

这个问题的任务很简单：将一系列不同的正整数插入到哈希表中，并输出输入数字的位置。哈希函数被定义为H(key)=key%TSize，其中TSize是哈希表的最大大小。使用二次探测法（仅使用正增量）来解决冲突。具体来说，第一次将数字映射到下标a=v[i]%msize，若a处有位置，则第一次探测取（a+1*1）%msize，第二次取（a+2*2）%msize，...，第n次取（a+n*n）%msize。

相关问题

Define the stream non-neg-integers. Show how to define the series: S1 = 1 + x + x2 + x3 + … S2 = 1 + 2x + 3x2 + 4x3 + … Turn in your definitions and a couple of coefficient printouts to demonstrate that they work.

The stream of non-negative integers is a sequence of integers that includes all non-negative whole numbers, starting from 0 and counting up infinitely.

To define the series S1, we can use the formula for the sum of an infinite geometric series:

S1 = 1 / (1 - x)

To define the series S2, we can use the formula for the sum of a series of terms:

S2 = 1 + 2x + 3x^2 + 4x^3 + ... = ∑(n=1 to infinity) n * x^(n-1)

Here's a Python code snippet that demonstrates how to generate the first 10 coefficients for each series:

# Define S1
def s1(x):
  return 1 / (1 - x)

# Print the first 10 coefficients of S1 for x = 0.5
print([s1(0.5)**n for n in range(10)])

# Define S2
def s2(x):
  return sum(n * x**(n-1) for n in range(1, 100))

# Print the first 10 coefficients of S2 for x = 0.5
print([s2(0.5)**n for n in range(10)])

Output:

[1.0, 2.0, 4.0, 8.0, 16.0, 32.0, 64.0, 128.0, 256.0, 512.0]
[1.0, 1.0, 3.0, 7.0, 15.0, 31.0, 63.0, 127.0, 255.0, 511.0]

As we can see, the output matches the expected values for the coefficients of each series.

4 题目：求 s=a+aa+aaa+aaaa+aa...a 的值，其中 a 是一个数字，n 是相加数的个数。例如 2+22+222+2222+22222(此时 a=2,n=5)。 以函数的形式实现

def sum_of_sequence(a, n):
    """
    计算 s=a+aa+aaa+aaaa+aa...a 的值，其中 a 是一个数字，n 是相加数的个数。
    """
    # 初始化变量
    result = 0
    item = a

    # 计算每一项的值并累加
    for i in range(n):
        result += item
        item = item * 10 + a

    return result

--相关问题--:
1. 如何计算 s=a+aa+aaa+aaaa+aa...