# 请你实现一个类 SubrectangleQueries ，它的构造函数的参数是一个 rows x cols 的矩形（这里用整数矩阵表示），并支持以下两种操作
# ： 
# 
#  1. updateSubrectangle(int row1, int col1, int row2, int col2, int newValue) 
# 
#  
#  用 newValue 更新以 (row1,col1) 为左上角且以 (row2,col2) 为右下角的子矩形。 
#  
# 
#  2. getValue(int row, int col) 
# 
#  
#  返回矩形中坐标 (row,col) 的当前值。 
#  
# 
#  
# 
#  示例 1： 
# 
#  输入：
# ["SubrectangleQueries","getValue","updateSubrectangle","getValue","getValue","
# updateSubrectangle","getValue","getValue"]
# [[[[1,2,1],[4,3,4],[3,2,1],[1,1,1]]],[0,2],[0,0,3,2,5],[0,2],[3,1],[3,0,3,2,10
# ],[3,1],[0,2]]
# 输出：
# [null,1,null,5,5,null,10,5]
# 解释：
# SubrectangleQueries subrectangleQueries = new SubrectangleQueries([[1,2,1],[4,
# 3,4],[3,2,1],[1,1,1]]);  
# // 初始的 (4x3) 矩形如下：
# // 1 2 1
# // 4 3 4
# // 3 2 1
# // 1 1 1
# subrectangleQueries.getValue(0, 2); // 返回 1
# subrectangleQueries.updateSubrectangle(0, 0, 3, 2, 5);
# // 此次更新后矩形变为：
# // 5 5 5
# // 5 5 5
# // 5 5 5
# // 5 5 5 
# subrectangleQueries.getValue(0, 2); // 返回 5
# subrectangleQueries.getValue(3, 1); // 返回 5
# subrectangleQueries.updateSubrectangle(3, 0, 3, 2, 10);
# // 此次更新后矩形变为：
# // 5   5   5
# // 5   5   5
# // 5   5   5
# // 10  10  10 
# subrectangleQueries.getValue(3, 1); // 返回 10
# subrectangleQueries.getValue(0, 2); // 返回 5
#  
# 
#  示例 2： 
# 
#  输入：
# ["SubrectangleQueries","getValue","updateSubrectangle","getValue","getValue","
# updateSubrectangle","getValue"]
# [[[[1,1,1],[2,2,2],[3,3,3]]],[0,0],[0,0,2,2,100],[0,0],[2,2],[1,1,2,2,20],[2,2
# ]]
# 输出：
# [null,1,null,100,100,null,20]
# 解释：
# SubrectangleQueries subrectangleQueries = new SubrectangleQueries([[1,1,1],[2,
# 2,2],[3,3,3]]);
# subrectangleQueries.getValue(0, 0); // 返回 1
# subrectangleQueries.updateSubrectangle(0, 0, 2, 2, 100);
# subrectangleQueries.getValue(0, 0); // 返回 100
# subrectangleQueries.getValue(2, 2); // 返回 100
# subrectangleQueries.updateSubrectangle(1, 1, 2, 2, 20);
# subrectangleQueries.getValue(2, 2); // 返回 20
#  
# 
#  
# 
#  提示： 
# 
#  
#  最多有 500 次updateSubrectangle 和 getValue 操作。 
#  1 <= rows, cols <= 100 
#  rows == rectangle.length 
#  cols == rectangle[i].length 
#  0 <= row1 <= row2 < rows 
#  0 <= col1 <= col2 < cols 
#  1 <= newValue, rectangle[i][j] <= 10^9 
#  0 <= row < rows 
#  0 <= col < cols 
#  
#  Related Topics 数组


# leetcode submit region begin(Prohibit modification and deletion)
class SubrectangleQueries(object):

    def __init__(self, rectangle):
        """
        :type rectangle: List[List[int]]
        """
        self.rectangle = rectangle

    def updateSubrectangle(self, row1, col1, row2, col2, newValue):
        """
        :type row1: int
        :type col1: int
        :type row2: int
        :type col2: int
        :type newValue: int
        :rtype: None
        """
        for row in range(row1, row2 + 1):
            for col in range(col1, col2 + 1):
                self.rectangle[row][col] = newValue

    def getValue(self, row, col):
        """
        :type row: int
        :type col: int
        :rtype: int
        """
        return self.rectangle[row][col]

# Your SubrectangleQueries object will be instantiated and called as such:
# obj = SubrectangleQueries(rectangle)
# obj.updateSubrectangle(row1,col1,row2,col2,newValue)
# param_2 = obj.getValue(row,col)
# leetcode submit region end(Prohibit modification and deletion)
