6909. 最长奇偶子数组 - 力扣（LeetCode）

思路：模拟

class Solution {
public:int longestAlternatingSubarray(vector<int>& nums, int threshold) {int res = 0;int n = nums.size();for(int i = 0; i < n; i ++ ){if(nums[i] % 2 == 0 && nums[i] <= threshold){for(int j = i + 1; j < n; j ++ ){if(nums[j] % 2 == nums[j - 1] % 2 || nums[j] > threshold) break;res = max(res, j - i + 1);}}}if(res == 0){for(int i = 0; i < n; i ++ ) if(nums[i] % 2 == 0 && nums[i] <= threshold) return 1;}return res;}
};

6916. 和等于目标值的质数对 - 力扣（LeetCode）

思路：线性筛质数预处理出n以内的质数，然后双指针模拟

class Solution {
public:vector<vector<int>> findPrimePairs(int n) {vector<vector<int>> res;int primes[1000010], cnt = 0;bool st[1000010];memset(primes, -1, sizeof primes);memset(st, false, sizeof st);for(int i = 2; i <= n; i ++ ){if(!st[i]) primes[cnt ++ ] = i;for(int j = 0; primes[j] <= n / i; j ++ ){st[primes[j] * i] = true;if(i % primes[j] == 0) break;}}int l = 0, r = cnt - 1;while (l <= r){if(primes[l] + primes[r] > n) r -- ;else if(primes[l] + primes[r] < n) l ++ ;else{res.push_back({primes[l], primes[r]});l ++ , r -- ;}}return res;}
};

6911. 不间断子数组 - 力扣（LeetCode）

思路：用multiset来维护滑动窗口，比赛中没想到，自己暴力+剪枝过不了最后几个

class Solution {
public:long long continuousSubarrays(vector<int>& nums) {int n = nums.size();long long res = 0;multiset<int> s;for(int i = 0, j = 0; i < n; i ++ ){s.insert(nums[i]);while (j <= i && *s.rbegin() - *s.begin() > 2){s.erase(s.find(nums[j]));j ++ ;}res += (long long)i - j + 1;}return res;}
};

2763. 所有子数组中不平衡数字之和 - 力扣（LeetCode）

思路：枚举，计算不平衡贡献值

class Solution {
public:int sumImbalanceNumbers(vector<int>& nums) {int res = 0, n = nums.size();bool st[n + 2];for(int i = 0; i < n; i ++ ){int cnt = 0;memset(st, false, sizeof st);st[nums[i]] = true;for(int j = i + 1; j < n; j ++ ){int x = nums[j];if(!st[x]){cnt += 1 - st[x - 1] - st[x + 1];st[x] = true;}res += cnt;}}return res;}
};