3212力扣：统计X和Y频数相等的子矩阵数量

class Solution {
public:
    int numberOfSubmatrices(vector<vector<char>>& grid) {
        int n=grid.size();
        int m=grid[0].size();
        vector<vector<int>> prex(n+1,vector<int>(m+1,0));
        vector<vector<int>> prey(n+1,vector<int>(m+1,0));
        for(int i=1;i<n+1;i++){
            for(int j=1;j<m+1;j++){
                if(grid[i-1][j-1]=='X'){
                    prex[i][j]=prex[i-1][j]+prex[i][j-1]-prex[i-1][j-1]+1;}
                else{
                    prex[i][j]=prex[i-1][j]+prex[i][j-1]-prex[i-1][j-1]+0;
                }
                if(grid[i-1][j-1]=='Y'){
                    prey[i][j]=prey[i-1][j]+prey[i][j-1]-prey[i-1][j-1]+1;
                }else{
                    prey[i][j]=prey[i-1][j]+prey[i][j-1]-prey[i-1][j-1]+0;
                }
            }
        }
        int res=0;
        for(int i=1;i<n+1;i++){
            for(int j=1;j<m+1;j++){
                int countx=prex[i][j]-prex[0][j]-prex[i][0]+prex[0][0];
                int county=prey[i][j]-prey[0][j]-prey[i][0]+prey[0][0];
                if(countx==county && countx>0){res++;}
            }
        }
        return res;
    }
};

二维前缀和 前缀和数组定义时扩宽一层