# 题目描述：

Given an m * n matrix M initialized with all 0’s and several update operations.

Operations are represented by a 2D array, and each operation is represented by an array with two positive integers a and b, which means M[i][j] should be added by one for all 0 <= i < a and 0 <= j < b.

You need to count and return the number of maximum integers in the matrix after performing all the operations.

Example 1:

Input:
m = 3, n = 3
operations = [[2,2],[3,3]]
Output: 4

Explanation:
Initially, M = [[0, 0, 0], [0, 0, 0], [0, 0, 0]]

After performing [2,2], M = [[1, 1, 0], [1, 1, 0], [0, 0, 0]]

After performing [3,3], M = [[2, 2, 1], [2, 2, 1], [1, 1, 1]]

So the maximum integer in M is 2, and there are four of it in M. So return 4.

Note:

The range of m and n is [1,40000].
The range of a is [1,m], and the range of b is [1,n].
The range of operations size won’t exceed 10,000.\

# 代码实现：

``````class Solution {
public:
int maxCount(int m, int n, vector<vector<int>>& ops) {
int numl=m,numr=n;
for(int i=0;i<ops.size();i++){
numl = numl<ops[i]?numl:ops[i];
numr = numr<ops[i]?numr:ops[i];
}

return numl*numr;
}
};
``````