human
Solve in Python:
Given a rows * columns matrix mat of ones and zeros, return how many submatrices have all ones.
 
Example 1:
Input: mat = [[1,0,1],
              [1,1,0],
              [1,1,0]]
Output: 13
Explanation:
There are 6 rectangles of side 1x1.
There are 2 rectangles of side 1x2.
There are 3 rectangles of side 2x1.
There is 1 rectangle of side 2x2. 
There is 1 rectangle of side 3x1.
Total number of rectangles = 6 + 2 + 3 + 1 + 1 = 13.

Example 2:
Input: mat = [[0,1,1,0],
              [0,1,1,1],
              [1,1,1,0]]
Output: 24
Explanation:
There are 8 rectangles of side 1x1.
There are 5 rectangles of side 1x2.
There are 2 rectangles of side 1x3. 
There are 4 rectangles of side 2x1.
There are 2 rectangles of side 2x2. 
There are 2 rectangles of side 3x1. 
There is 1 rectangle of side 3x2. 
Total number of rectangles = 8 + 5 + 2 + 4 + 2 + 2 + 1 = 24.

Example 3:
Input: mat = [[1,1,1,1,1,1]]
Output: 21

Example 4:
Input: mat = [[1,0,1],[0,1,0],[1,0,1]]
Output: 5

 
Constraints:

1 <= rows <= 150
1 <= columns <= 150
0 <= mat[i][j] <= 1
assistant
#Here is the Python program you requested.

class Solution:
    def numSubmat(self, mat: List[List[int]]) -> int:
        m = len(mat)
        n = len(mat[0])
        pre = [[0]*(n+1) for _ in range(m)]
        
        for row in range(m):
            for col in range(1, n+1):
                pre[row][col] = mat[row][col-1] + pre[row][col-1]
        
        ans = 0
        for col1 in range(n):
            for col2 in range(col1+1, n+1):
                local_ans = 0
                curr_stack = 0
                for row in range(m):
                    if pre[row][col2] - pre[row][col1] == col2-col1:
                        curr_stack += 1
                        local_ans += curr_stack
                    else:
                        curr_stack = 0
                ans += local_ans
        return ans