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feat: add solutions to lc problems: No.2960~2963 (doocs#2083)
* No.2960.Count Tested Devices After Test Operations * No.2961.Double Modular Exponentiation * No.2962.Count Subarrays Where Max Element Appears at Least K Times * No.2963.Count the Number of Good Partitions
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‎solution/2900-2999/2960.Count Tested Devices After Test Operations/README.md

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<ul>
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<li>如果 <code>batteryPercentages[i]</code> <strong>大于</strong> <code>0</code>：
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<ul>
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<li><strong>增加</strong> 已测试设备的计数。</li>
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<li>将下标在 <code>[i + 1, n - 1]</code> 的所有设备的电池百分比减少 <code>1</code>，确保它们的电池百分比<strong> 不会低于</strong> <code>0</code> ，即 <code>batteryPercentages[j] = max(0, batteryPercentages[j] - 1)</code>。</li>
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<li>移动到下一个设备。</li>
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</ul>
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</li>
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<li>否则，移动到下一个设备而不执行任何测试。</li>
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</ul>
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<p>返回一个整数，表示按顺序执行测试操作后 <strong>已测试设备</strong> 的数量。</p>
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：模拟**
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假设我们当前已测试的设备数量为 $ans$，如果当测试新的一台设备 $i$ 时，它的剩余电量为 $\max(0, batteryPercentages[i] - ans)$，如果剩余电量大于 $0$，则说明这台设备可以进行测试，此时我们需要将 $ans$ 增加 $1$。
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最后返回 $ans$ 即可。
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时间复杂度 $O(n)$，其中 $n$ 为数组长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def countTestedDevices(self, batteryPercentages: List[int]) -> int:
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ans = 0
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for x in batteryPercentages:
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x -= ans
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ans += x > 0
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int countTestedDevices(int[] batteryPercentages) {
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int ans = 0;
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for (int x : batteryPercentages) {
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x -= ans;
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if (x > 0) {
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++ans;
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int countTestedDevices(vector<int>& batteryPercentages) {
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int ans = 0;
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for (int x : batteryPercentages) {
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x -= ans;
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if (x > 0) {
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++ans;
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func countTestedDevices(batteryPercentages []int) (ans int) {
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for _, x := range batteryPercentages {
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x -= ans
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if x > 0 {
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ans++
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}
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}
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return
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}
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```
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### **TypeScript**
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```ts
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function countTestedDevices(batteryPercentages: number[]): number {
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let ans = 0;
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for (let x of batteryPercentages) {
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x -= ans;
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if (x > 0) {
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++ans;
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}
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}
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return ans;
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}
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```
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### **...**

‎solution/2900-2999/2960.Count Tested Devices After Test Operations/README_EN.md

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<ul>
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<li>If <code>batteryPercentages[i]</code> is <strong>greater</strong> than <code>0</code>:
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<ul>
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<li><strong>Increment</strong> the count of tested devices.</li>
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<li><strong>Decrease</strong> the battery percentage of all devices with indices <code>j</code> in the range <code>[i + 1, n - 1]</code> by <code>1</code>, ensuring their battery percentage <strong>never goes below</strong> <code>0</code>, i.e, <code>batteryPercentages[j] = max(0, batteryPercentages[j] - 1)</code>.</li>
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<li>Move to the next device.</li>
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</ul>
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</li>
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<li>Otherwise, move to the next device without performing any test.</li>
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</ul>
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<p>Return <em>an integer denoting the number of devices that will be tested after performing the test operations in order.</em></p>
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## Solutions
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**Solution 1: Simulation**
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Assume that the current number of devices we have tested is $ans$. When testing a new device $i$, its remaining battery is $\max(0, batteryPercentages[i] - ans)$. If the remaining battery is greater than $0$, it means this device can be tested, and we need to increase $ans$ by $1$.
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Finally, return $ans$.
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The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def countTestedDevices(self, batteryPercentages: List[int]) -> int:
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ans = 0
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for x in batteryPercentages:
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x -= ans
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ans += x > 0
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int countTestedDevices(int[] batteryPercentages) {
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int ans = 0;
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for (int x : batteryPercentages) {
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x -= ans;
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if (x > 0) {
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++ans;
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int countTestedDevices(vector<int>& batteryPercentages) {
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int ans = 0;
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for (int x : batteryPercentages) {
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x -= ans;
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if (x > 0) {
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++ans;
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func countTestedDevices(batteryPercentages []int) (ans int) {
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for _, x := range batteryPercentages {
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x -= ans
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if x > 0 {
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ans++
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}
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}
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return
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}
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```
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### **TypeScript**
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```ts
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function countTestedDevices(batteryPercentages: number[]): number {
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let ans = 0;
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for (let x of batteryPercentages) {
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x -= ans;
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if (x > 0) {
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++ans;
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}
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}
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return ans;
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}
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```
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### **...**

‎solution/2900-2999/2960.Count Tested Devices After Test Operations/Solution.cpp

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

‎solution/2900-2999/2960.Count Tested Devices After Test Operations/Solution.go

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func countTestedDevices(batteryPercentages []int) (ans int) {
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for _, x := range batteryPercentages {
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x -= ans
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if x > 0 {
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ans++
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}
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}
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return
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}

‎solution/2900-2999/2960.Count Tested Devices After Test Operations/Solution.java

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class Solution {
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public int countTestedDevices(int[] batteryPercentages) {
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int ans = 0;
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for (int x : batteryPercentages) {
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x -= ans;
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if (x > 0) {
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++ans;
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}
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}
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return ans;
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}
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}

‎solution/2900-2999/2960.Count Tested Devices After Test Operations/Solution.py

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class Solution:
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def countTestedDevices(self, batteryPercentages: List[int]) -> int:
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ans = 0
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for x in batteryPercentages:
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x -= ans
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ans += x > 0
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return ans

‎solution/2900-2999/2960.Count Tested Devices After Test Operations/Solution.ts

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function countTestedDevices(batteryPercentages: number[]): number {
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let ans = 0;
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for (let x of batteryPercentages) {
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x -= ans;
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if (x > 0) {
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++ans;
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}
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}
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return ans;
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}

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