$\large\color{orange}{比赛入口：}$$\color{orange}{LeetCode104}$ $\large\color{red}{双}$$\large\color{orange}{周赛}$

$\large\color{green}{6366. 老人的数目}$

字符串操作, 时间复杂度：$O(2n)$

class Solution {
public:
    int countSeniors(vector<string>& details) {
        int cnt = 0;
        for(auto s : details)   cnt += (s.substr(11, 2)) > "60";
        return cnt;
    }
};

$\large\color{orange}{2679. 矩阵中的和}$

模拟, 时间复杂度：$O(nm^{2})$

class Solution {
public:
    static const int N = 510;
    bool vis[N][N];
    int matrixSum(vector<vector<int>>& nums) {
        memset(vis, 0, sizeof vis);
        int n = nums.size(), m = nums[0].size();
        int res = 0;
        for(int k = 0; k < m; k ++ ){
            int score = -1;
            for(int row = 0; row < n; row ++){
                int maxv = -1, index;
                for(int col = 0; col < m; col ++ ){
                    if(!vis[row][col] && maxv < nums[row][col])   maxv = nums[row][col], index = col;
                }    
                vis[row][index] = true;
                score = max(score, maxv);
            }
            res += score;
        }
        return res;
    }
};

$\large\color{orange}{2680. 最大或值}$

贪心 + 前缀和, 时间复杂度：$O(n)$

class Solution {
public:
    static const int N = 1E5 + 10;
    typedef long long ll;
    ll pre[N], last[N];
    long long maximumOr(vector<int>& nums, int k) {
        memset(pre, 0, sizeof pre);
        memset(last, 0, sizeof last);
        int n = nums.size();
        for(int i = 0; i < n; i ++ ){
            if(i)   pre[i] = nums[i] | pre[i - 1];
            else    pre[i] = nums[i];
        } 
        for(int i = n - 1; i >= 0; i -- ){
            if(i == n - 1)  last[i] = nums[i];
            else   last[i] = last[i + 1] | nums[i];
        }
        ll res = 0;
        for(int i = 0; i < n; i ++ ){
            ll x = nums[i];
            if(!i)  res = last[1] | (x << k);
            else    res = max(res, pre[i - 1] | (x << k) | last[i + 1]);
        }
        return res;
    }
};

$\large\color{red}{2681. 英雄的力量}$

贪心 + 前缀和, 时间复杂度：$O(n)$

class Solution {
public:
    typedef long long ll;
    static const ll MOD = 1E9 + 7;
    int sumOfPower(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        ll res = 0, S = 0;
        for(auto x : nums){
            ll u = x;
            res = (res + ((S * ((u * u) % MOD)) % MOD) + (((u * u) % MOD) * u) % MOD) % MOD;
            S = (S * 2 + u) % MOD;
        }
        return res;
    }
};