模擬：記錄當前已測試設備數量

class Solution {
public:int countTestedDevices(vector<int> &batteryPercentages) {int res = 0;int s = 0;for (auto x: batteryPercentages) {if (x - s > 0) {res++;s++;}}return res;}
};

快速冪

class Solution {
public:int fpow(int x, int n, int mod) {// x^n%modint res = 1;for (int e = x; n; e = e * e % mod, n >>= 1)if (n & 1)res = res * e % mod;return res;}vector<int> getGoodIndices(vector<vector<int>> &variables, int target) {vector<int> res;for (int i = 0; i < variables.size(); i++)if (fpow(fpow(variables[i][0], variables[i][1], 10), variables[i][2], variables[i][3]) == target)res.push_back(i);return res;}
};

滑動窗口：枚舉滑窗的左邊界 $l$，找到滿足條件的最小右邊界 $r$ ，則左邊界為 $l$ 的滿足條件的子數組數目為 $nums.size()?r$

class Solution {
public:long long countSubarrays(vector<int> &nums, int k) {int mx = *max_element(nums.begin(), nums.end());int n = nums.size();long long res = 0;int cnt = 0;for (int l = 0, r = -1; l < n;) {while (cnt < k && r + 1 < n)if (nums[++r] == mx)cnt++;if (cnt < k)break;res += n - r;if (nums[l++] == mx)cnt--;}return res;}
};

計數：因為相同數字必須在一個子數組中，所以可以預先求出 $nums$ 按要求最多可以劃分成的子數組數目 $cnt$，原數組好分割方案數目等于 $cnt$ 個互不相同的數形成的數組的好分割方案數目，即 $2^{cnt?1}$

class Solution {
public:int numberOfGoodPartitions(vector<int> &nums) {int n = nums.size();unordered_map<int, int> r;//記錄每個數出現的最右小標for (int i = 0; i < n; i++)r[nums[i]] = i;int cnt = 0;//按要求最多可以劃分成的子數組數目for (int i = 0, right = r[nums[0]]; i < n;) {if (i != right) {right = max(right, r[nums[i++]]);} else {cnt++;if (i == n - 1)break;right = r[nums[++i]];}}int mod = 1e9 + 7;int res = 1;for (int i = 0; i < cnt - 1; i++)res = res * 2 % mod;return (res + mod) % mod;}
};