Score After Flipping Matrix

We have a two dimensional matrix A where each value is 0 or 1.

A move consists of choosing any row or column, and toggling each value in that row or column: changing all 0s to 1s, and all 1s to 0s.

After making any number of moves, every row of this matrix is interpreted as a binary number, and the score of the matrix is the sum of these numbers.

Return the highest possible score.

Example 1:

Input: [[0,0,1,1],[1,0,1,0],[1,1,0,0]]
Output: 39

Explanation:
Toggled to [[1,1,1,1],[1,0,0,1],[1,1,1,1]].
0b1111 + 0b1001 + 0b1111 = 15 + 9 + 15 = 39

Note:

1 <= A.length <= 20
1 <= A[0].length <= 20
A[i][j] is 0 or 1.

题意：只有0和1的矩阵每次翻转某行或某列，最后每行代表一个数的二进制，所有数之和最大为多少。
解法：第一列必须都变成1，才能保证每行的数更大，对每一列，看1的个数，翻转和不翻转谁的1多。

class Solution {
public:
    int matrixScore(vector>& A) {
        int m = A.size(), n = A[0].size();
        int ans = (1 << (n - 1)) * m;
        for(int j = 1; j < n; ++j){
            int cur = 0;
            for(int i = 0; i < m; ++i){
                cur += (A[i][j] == A[i][0]);
            }
            ans += max(cur, m - cur) * (1 << (n - j - 1));
        }
        return ans;
    }
};