# 861.Score After Flipping Matrix

**861.Score After Flipping Matrix**

难度:Medium

> 有一个二维矩阵 A 其中每个元素的值为 0 或 1 。

移动是指选择任一行或列，并转换该行或列中的每一个值：将所有 0 都更改为 1，将所有 1 都更改为 0。

在做出任意次数的移动后，将该矩阵的每一行都按照二进制数来解释，矩阵的得分就是这些数字的总和。

返回尽可能高的分数。

```
示例：

输入：[[0,0,1,1],[1,0,1,0],[1,1,0,0]]
输出：39
解释：
转换为 [[1,1,1,1],[1,0,0,1],[1,1,1,1]]
0b1111 + 0b1001 + 0b1111 = 15 + 9 + 15 = 39
 

提示：

1 <= A.length <= 20
1 <= A[0].length <= 20
A[i][j] 是 0 或 1
```

先对每行来说，保证最高位为1，然后对每列来说，保证至少大于一半的数为1.最后求和即可。

```
class Solution {
public:
    int matrixScore(vector<vector<int>>& A) {
        for(auto &a:A)
        {
            if(a[0] == 0 )
            {
                for(auto &n:a)
                 n=1-n;
            }
        }
        int len=A.size();
        for(int i=0;i<A[0].size();i++)
        {
            int sum=0;
            for(int j=0;j<len;j++)
            sum+=A[j][i];

            if(sum<= len/2)
            {
                for(int j=0;j<len;j++)
                A[j][i] = 1-A[j][i];
            }
        }
        vector<int>tmp(A[0].size());
        for(int j=0;j<A[0].size();j++)
        {
            for(int i=0;i<A.size();i++)
            tmp[j] +=A[i][j];
        }
        
        int res=0;
        for(int i=0;i<tmp.size();i++)
            res= res*2+ tmp[i];

        return res;
    }
};
```

> 执行用时 :4 ms, 在所有 C++ 提交中击败了86.54%的用户\
> 内存消耗 :8.7 MB, 在所有 C++ 提交中击败了100.00%的用户


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