题目：
Given a 2D integer matrix M representing the gray scale of an image, you need to design a smoother to make the gray scale of each cell becomes the average gray scale (rounding down) of all the 8 surrounding cells and itself. If a cell has less than 8 surrounding cells, then use as many as you can.
Example 1:
Input: [[1,1,1], [1,0,1], [1,1,1]] Output: [[0, 0, 0], [0, 0, 0], [0, 0, 0]] Explanation: For the point (0,0), (0,2), (2,0), (2,2): floor(3/4) = floor(0.75) = 0 For the point (0,1), (1,0), (1,2), (2,1): floor(5/6) = floor(0.83333333) = 0 For the point (1,1): floor(8/9) = floor(0.88888889) = 0
Note:
The value in the given matrix is in the range of [0, 255].The length and width of the given matrix are in the range of [1, 150]. 思路：本题思路很简单，咱们是用最简单易懂的思路：额外构建两个二维数组，一个数组保留原数组的值不变，在四周包裹上0（用来统计每次计算的分子），一个数组是原数组所有值替换为1，再在四周包裹上0（用来统计每次计算的分母）
代码：
class Solution { public: vector<vector<int>> imageSmoother(vector<vector<int>>& M) { vector<vector<int>> res = M; vector<int> a(M[0].size(),1); vector<vector<int>> b(M.size(),a); for(int i=0;i<M.size();i++) { M[i].insert(M[i].begin(),0); M[i].push_back(0); b[i].insert(b[i].begin(),0); b[i].push_back(0); } vector<int> temp(M[0].size(),0); M.insert(M.begin(),temp); M.push_back(temp); b.insert(b.begin(),temp); b.push_back(temp); for(int i =1;i<M.size()-1;i++) { for(int j=1;j<M[0].size()-1;j++) { res[i-1][j-1] = floor((M[i][j]+M[i][j-1]+M[i][j+1]+M[i-1][j]+M[i-1][j-1]+M[i-1][j+1]+M[i+1][j]+M[i+1][j-1]+M[i+1][j+1])/(b[i][j]+b[i][j-1]+b[i][j+1]+b[i-1][j]+b[i-1][j-1]+b[i-1][j+1]+b[i+1][j]+b[i+1][j-1]+b[i+1][j+1])); } } return res; } };
