Microsoft编程测试（2019春专业版）

今天实验室的师兄做微软实习编程题，我也来做一做吧。如有错误，请见谅，望及时指正。

Question #1

Playing with Beads

There are N people in a group labeled form 1 to N. People are connected to each other via threads in the following manner:

The person with label number K is connected to all the persons with label J such that J exactly divides K. Beads can be passed through the threads. If a person P has a head, in how many ways can the bead be passed in the network of threads so that it returns to the same person within X moves or less.

Move: Passing the bead from one person to the orther.

Input Specification:

Input1: N, denoting the number of people.

Input2: P, label of the person having the bead.

Input3: X, maximum number of moves that can be made.

Output Specification:

Your function should return the total number of ways in which the bead will return to its initial position within X moves.

Example:

Input1: 3

Input2: 2

Input3: 2

Output: 1

分析一波：邻接矩阵n次方的几何意义。

# 构造n×n的零阵
def matrix(n):
    return [[0 for _ in range(n)] for _ in range(n)]


# 定义矩阵乘法
def dot(A, B):
    row_len = len(A)
    column_len = len(B[0])
    cross_len = len(B)
    res_mat = matrix(column_len)
    for i in range(row_len):
        for j in range(column_len):
            for k in range(cross_len):
                temp = A[i][k] * B[k][j]
                res_mat[i][j] += temp
    return res_mat


def maxCircless(input1, input2, input3):
    m = matrix(input1)
    # 构造邻接矩阵
    for i in range(input1):
        for j in range(input1):
            if (i+1) % (j+1) == 0:
                if i == j:
                    pass
                else:
                    m[i][j] = 1
                    m[j][i] = 1

    num = 0
    sumt = m
    # 利用邻接矩阵n次方的几何意义
    for x in range(input3 - 1):
        sumt = dot(sumt, m)
        num += sumt[input2][input2]
    return num


if __name__ == "__main__":
    print(maxCircless(3, 2, 4))
    print(maxCircless(3, 2, 2))

Question #2

Archer

Ralph is a brilliant archer who always wines the archery. This time, to challenge him, the jubges have set a new type of challenge.

There are N bottles kept in a row with each bootle given a specific number. Ralph has to target and hit the bottles in a very interesting way. In one shot, Ralph can remove 1 or more than 1 bottles such that the numbers present on these sequence of bottlers from a palindrome. Also, these bottles must be present one after the other. After taking this shot, the remaining bottles are shifted so that all the bottles are again in a row and Ralph shoots the bottles again.

Now, if Ralph gets 1 point for each shot, find the minimum number of points Ralph can score.

Input Specification:

Input1: The number of bottles N.

Input2: The array representing the numbers written on the bottles with A[i] representing the number on the i-th bottles.

Output Specification:

The minimum number of points Ralph can score.

Example1:

Input1: 2

Input2: {1, 2}

Output: 2

Example2:

Input1: 5

Input2: {1, 4, 3, 1, 5}

Output: 3

分析：占位符

def dns(p):
    if len(p) == 0:
        return 0
    max_i = max(x[0] for x in p)
    min_j = min(x[1] for x in p)
    num = []
    for x in p:
        if x[0] == max_i or x[1] == min_j:
            num.append(dns([t for t in p if t[0] < x[0] and t[1] > x[1]]) + 1)
    return max(k for k in num)


def f(A, n):
    # 在左下矩阵，记录不同位置数字相同时的坐标
    flag = [(str(i), str(j)) for i in range(n) for j in range(n) if i > j and A[i] == A[j]]
    # 按照行升序排序，列降序排序（左下角优先级最高）
    flag.sort(key=lambda item: (item[0], tuple(map(lambda x: -ord(x), item[1]))), reverse=True)
    # 贪心选择
    sum = dns([(int(x[0]), int(x[1])) for x in flag])
    return n - 2 * sum


if __name__ == "__main__":
    a = [1, 2]
    b = [1, 4, 3, 1, 5]
    print(f(a, 2))
    print(f(b, 5))

Question #3

占位符

Question #4

占位符