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You are given a state for a rectangular board game grid with chips in a binary matrix, where 1 is a cell with a chip and 0 is an empty cell. You are also given the coordinates for a cell in the form of row and column numbers (starting from 0). You should determine how many chips are close to this cell. Every cell interacts with its eight neighbours; those cells that are horizontally, vertically, or diagonally adjacent.

example For the given examples (see the schema) there is the same grid: ((1, 0, 0, 1, 0), (0, 1, 0, 0, 0), (0, 0, 1, 0, 1), (1, 0, 0, 0, 0), (0, 0, 1, 0, 0),)

For the first example coordinates of the cell is (1, 2) and as we can see from the schema this cell has 3 neighbour chips. For the second example coordinates is (0, 0) and this cell contains a chip, but we count only neighbours and the answer is 1.

Input: Three arguments. A grid as a tuple of tuples with integers (1/0), a row number and column number for a cell as integers. Output: How many neighbouring cells have chips as an integer.

给定一个二维元组。输出任意坐标周围的1的个数。

def count_neighbours(grid, row, col):    a=len(grid)    b=len(grid[0])    rangelistx=range(0,a)    rangelisty=range(0,b)    countrangex=[row-1,row,row+1]    countrangey=[col-1,col,col+1]    newrangex=[]    newrangey=[]    count=0    for i in countrangex:        if i in rangelistx:            newrangex.append(i)    for i in countrangey:        if i in rangelisty:            newrangey.append(i)    for i in newrangex:        for j in newrangey:            if grid[i][j]==1:                count+=1    if grid[row][col]==1:        count-=1       return countif __name__ == '__main__':    #These "asserts" using only for self-checking and not necessary for auto-testing    assert count_neighbours(((1, 0, 0, 1, 0),                             (0, 1, 0, 0, 0),                             (0, 0, 1, 0, 1),                             (1, 0, 0, 0, 0),                             (0, 0, 1, 0, 0),), 1, 2) == 3, "1st example"    assert count_neighbours(((1, 0, 0, 1, 0),                             (0, 1, 0, 0, 0),                             (0, 0, 1, 0, 1),                             (1, 0, 0, 0, 0),                             (0, 0, 1, 0, 0),), 0, 0) == 1, "2nd example"    assert count_neighbours(((1, 1, 1),                             (1, 1, 1),                             (1, 1, 1),), 0, 2) == 3, "Dense corner"    assert count_neighbours(((0, 0, 0),                             (0, 1, 0),                             (0, 0, 0),), 1, 1) == 0, "Single"

元组中行x列y都有各自的取值范围。

行：在[row-1,row,row+1]范围里，如果元素在x的范围里，就newrange.append(i)加入到新的范围list里面。 对于列也一样。 然后对新的行列list遍历，每有一个1出现，count+1。 最后在判断grid[row][col]这个值本身是不是1，是的话就count-1。 return count finish!!!!

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