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Commit 57032bd

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feat: add solutions to lc problem: No.3100 (doocs#4746)
No.3100.Water Bottles II
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‎solution/3100-3199/3100.Water Bottles II/README.md‎

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### 方法一:模拟
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我们可以在一开始就喝掉所有的满水瓶,因此初始时我们喝到的水数量为 `numBottles`。然后我们不断地进行以下操作:
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我们可以在一开始就喝掉所有的满水瓶,因此初始时我们喝到的水数量为 $\textit{numBottles}$。然后我们不断地进行以下操作:
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- 如果当前有 `numExchange` 个空水瓶,我们就可以用它们换一瓶满水瓶,换完后,`numExchange` 的值增加 1。然后,我们喝掉这瓶水,喝到的水数量增加 1ドル,ドル空水瓶数量增加 1ドル$。
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- 如果当前没有 `numExchange` 个空水瓶,那么我们就不能再换水了,此时我们就可以停止操作。
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- 如果当前有 $\textit{numExchange}$ 个空水瓶,我们就可以用它们换一瓶满水瓶,换完后,$\textit{numExchange}$ 的值增加 1ドル$。然后,我们喝掉这瓶水,喝到的水数量增加 1ドル,ドル空水瓶数量增加 1ドル$。
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- 如果当前没有 $\textit{numExchange}$ 个空水瓶,那么我们就不能再换水了,此时我们就可以停止操作。
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我们不断地进行上述操作,直到我们无法再换水为止。最终我们喝到的水的数量就是答案。
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时间复杂度 $O(\sqrt{numBottles}),ドル空间复杂度 $O(1)$。
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时间复杂度 $O(\sqrt{n}),ドル其中 $n$ 是初始的满水瓶数量。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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}
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```
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#### C#
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```cs
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public class Solution {
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public int MaxBottlesDrunk(int numBottles, int numExchange) {
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int ans = numBottles;
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while (numBottles >= numExchange) {
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numBottles -= numExchange;
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++numExchange;
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++ans;
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++numBottles;
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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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#### PHP
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```php
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class Solution {
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/**
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* @param Integer $numBottles
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* @param Integer $numExchange
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* @return Integer
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*/
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function maxBottlesDrunk($numBottles, $numExchange) {
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$ans = $numBottles;
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while ($numBottles >= $numExchange) {
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$numBottles -= $numExchange;
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$numExchange++;
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$ans++;
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$numBottles++;
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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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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/3100-3199/3100.Water Bottles II/README_EN.md‎

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### Solution 1: Simulation
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We can drink all the full water bottles at the beginning, so the initial amount of water we drink is `numBottles`. Then we continuously perform the following operations:
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We can drink all the full water bottles at the beginning, so initially the amount of water we drink is $\textit{numBottles}$. Then, we repeatedly perform the following operations:
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- If we currently have `numExchange` empty water bottles, we can exchange them for a full water bottle, after which the value of `numExchange` increases by 1. Then, we drink this bottle of water, the amount of water we drink increases by 1ドル,ドル and the number of empty water bottles increases by 1ドル$.
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- If we currently do not have `numExchange` empty water bottles, then we can no longer exchange for water, at which point we can stop the operation.
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- If we currently have $\textit{numExchange}$ empty bottles, we can exchange them for one full bottle. After the exchange, the value of $\textit{numExchange}$ increases by 1ドル$. Then, we drink this bottle, increasing the total amount of water drunk by 1ドル,ドル and the number of empty bottles increases by 1ドル$.
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- If we do not have $\textit{numExchange}$ empty bottles, we cannot exchange for more water and should stop.
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We continuously perform the above operations until we can no longer exchange for water. The final amount of water we drink is the answer.
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We repeat the above process until we can no longer exchange bottles. The total amount of water drunk is the answer.
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The time complexity is $O(\sqrt{numBottles})$ and the space complexity is $O(1)$.
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The time complexity is $O(\sqrt{n})$, where $n$ is the initial number of full bottles. The space complexity is $O(1)$.
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<!-- tabs:start -->
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}
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```
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#### C#
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```cs
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public class Solution {
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public int MaxBottlesDrunk(int numBottles, int numExchange) {
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int ans = numBottles;
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while (numBottles >= numExchange) {
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numBottles -= numExchange;
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++numExchange;
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++ans;
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++numBottles;
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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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#### PHP
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```php
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class Solution {
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/**
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* @param Integer $numBottles
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* @param Integer $numExchange
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* @return Integer
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*/
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function maxBottlesDrunk($numBottles, $numExchange) {
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$ans = $numBottles;
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while ($numBottles >= $numExchange) {
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$numBottles -= $numExchange;
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$numExchange++;
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$ans++;
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$numBottles++;
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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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<!-- tabs:end -->
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<!-- solution:end -->
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public class Solution {
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public int MaxBottlesDrunk(int numBottles, int numExchange) {
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int ans = numBottles;
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while (numBottles >= numExchange) {
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numBottles -= numExchange;
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++numExchange;
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++ans;
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++numBottles;
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}
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return ans;
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}
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}
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class Solution {
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/**
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* @param Integer $numBottles
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* @param Integer $numExchange
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* @return Integer
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*/
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function maxBottlesDrunk($numBottles, $numExchange) {
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$ans = $numBottles;
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while ($numBottles >= $numExchange) {
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$numBottles -= $numExchange;
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$numExchange++;
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$ans++;
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$numBottles++;
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}
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return $ans;
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}
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}

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