vivo 2020春招编程题(绘图解锁+数字乘积+手机产量）python实现

1. 绘图解锁 ，

手机解锁图案问题，至少链接[m,n]个键 解锁图案中所有键不同 若当前连线经过某一点，则该点必须选中 求合理的连线方式的种类

参考：https://blog.csdn.net/weixin_43647540/article/details/106385826

class Solution: def __init__(self): self.count = 0 def search(self, a, i, j, c, m, n): if c >= m: self.count += 1 if c >= n: return # 一个点走向周围包含24种情况,构成5 * 5 矩阵 for x in range(-i, 3-i): for y in range(-j, 3-j): # 将要访问的点位于棋盘内，且未被访问 if a[i + x][j + y] == 0: # 两点连线经过中间点时 需确认中间点已经被访问 if x % 2 or y % 2 or (~(x % 2) and ~(y % 2) and a[i + int(x / 2)][j + int(y / 2)] == 1): a[i + x][j + y] = 1 self.search(a, i + x, j + y, c + 1, m, n) a[i + x][j + y] = 0 def solution(self, m, n): a = [] for i in range(3): a_i = [] for j in range(3): a_i.append(0) a.append(a_i) if m > n or n <= 0: return 0 for a1 in range(3): for a2 in range(3): a[a1][a2] = 1 self.search(a, a1, a2, 1, m, n) a[a1][a2] = 0 return self.count if __name__ == '__main__': count = 0 s = Solution() print(s.solution(1, 2))

2. 数字乘积

计算m的数位之积等于n的数 例如： n = 36 m = 49

def solution(n): s = [] # while n > 1: i = 9 while i > 1: if n % i == 0: n = n / i s.append(i) i = 9 else: i -= 1 sum = 0 for i in range(len(s)): sum = sum +s[i] * 10 ** i if n != 1: return -1 else: return sum

3. 手机产量

统计手机产量 第一天能生成一台 2，3天生产2台， 4，5，6天生产3台 思路： 计算日期序列和 def solution(n): # write code here t = [] i = 1 sum = 0 while n > 0: if n > i: n = n-i t.append(i) i += 1 else: t.append(n) n = 0 print(t) for index, value in enumerate(t): sum = sum + (index+1) * value return sum