题目
解法：
Weekly Contest 289的题目，自己感觉有点变态，当时没做出来
这道题需要解决一下几个问题：
- 如何数trailing zero
- 如何枚举这些符合条件形状的所有可能性
第一个问题最优解是，所有的0都只可能是2*5得到，所以可以把每个位置数字分解，记录每个数字以2为因子和以5为因子的个数。可以通过一个形状能组成多少个2，5的pair来得到多少个0。由于要累计这些个数，所以使用prefixsum技巧
枚举这些形状可能性最好操作的方案是把每个位置作为形状的turning point，这样就可以每个位置穷举四种组合即可。可以验证即使当前位置在第一行或者第一列（可能会有方向无法组成有效的形状），最后的答案也不会有问题
class Solution:
def maxTrailingZeros(self, grid: List[List[int]]) -> int:
# helper function to get the number of factors of 2 and 5 in a number
def helper(num,f):
if num % f != 0:
return 0
return 1 + helper(num // f,f)
m,n = len(grid),len(grid[0])
# left represents the rowwise left to right prefix sum of factor of 2 and 5 numbers
left = [[[0,0] for _ in range(n)] for _ in range(m)]
# top represents the columnwise top to down prefix sum of factor of 2 and 5 numbers
top = [[[0,0] for _ in range(n)] for _ in range(m)]
# get the prefix sum matrix for rowwise
for i in range(m):
for j in range(n):
if j == 0:
left[i][j] = [helper(grid[i][j],2),helper(grid[i][j],5)]
else:
a = helper(grid[i][j],2)
b = helper(grid[i][j],5)
left[i][j][0] = left[i][j-1][0] + a
left[i][j][1] = left[i][j-1][1] + b
# get the prefix sum matrix for columnwise
for j in range(n):
for i in range(m):
if i == 0:
top[i][j] = [helper(grid[i][j],2),helper(grid[i][j],5)]
else:
a = helper(grid[i][j],2)
b = helper(grid[i][j],5)
top[i][j][0] = top[i-1][j][0] + a
top[i][j][1] = top[i-1][j][1] + b
# for every grid position, take it as the elbow/tuning point, and span 4 different posibilities
ans = 0
for i in range(m):
for j in range(n):
# columnwise factor 2/5 prefix sum at last row column j;
fac2_last_row,fac5_last_row = top[m-1][j]
# rowwise factor 2/5 prefix sum at last column row i
fac2_last_col,fac5_last_col = left[i][n-1]
# factor 2/5 count for the number in current position; this is needed because this place might be added/subtracted twice
fac2_curr,fac5_curr = helper(grid[i][j],2),helper(grid[i][j],5)
# columnwise factor 2/5 prefix sum at current position
fac2_curr_row, fac5_curr_row = top[i][j]
# rowwise factor 2/5 prefix sum at current position
fac2_curr_col, fac5_curr_col= left[i][j]
'''
case1:
0
0
0 0 0
'''
case1 = min(fac2_curr_row+fac2_curr_col-fac2_curr,fac5_curr_row+fac5_curr_col-fac5_curr)
'''
case2:
0
0
0 0 0
'''
case2 = min(fac2_curr_row+fac2_last_col-fac2_curr_col,fac5_curr_row+fac5_last_col-fac5_curr_col)
'''
case3:
0 0 0
0
0
'''
case3 = min(fac2_curr_col+fac2_last_row-fac2_curr_row,fac5_curr_col+fac5_last_row-fac5_curr_row)
'''
case4:
0 0 0
0
0
'''
case4 = min(fac2_last_col-fac2_curr_col+fac2_last_row-fac2_curr_row+fac2_curr,fac5_last_col-fac5_curr_col+fac5_last_row-fac5_curr_row+fac5_curr)
ans = max(ans,case1,case2,case3,case4)
return ans
时间复杂度：O(mnk)，m为函数，n为列数，k可能不是一个常数，代表某个数字做2和5的因子分解的复杂度
空间复杂度：O(m*n)
参考答案链接：
https://leetcode.com/problems/maximum-trailing-zeros-in-a-cornered-path/discuss/1955607/Python3-Explanation-with-pictures-prefix-sum.
todo：
实现c++版本，可参考这里：https://leetcode.com/problems/maximum-trailing-zeros-in-a-cornered-path/discuss/1955515/Prefix-Sum-of-Factors-2-and-5
