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Commit 352832e

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feat: add solutions to lc problem: No.3560 (doocs#4443)
No.3560.Find Minimum Log Transportation Cost
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‎solution/3500-3599/3560.Find Minimum Log Transportation Cost/README.md‎

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<!-- solution:start -->
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### 方法一
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### 方法一:数学
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如果两根木材的长度都不超过卡车的最大载重 $k,ドル则不需要切割,直接返回 0ドル$。
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否则,说明只有一个木材的长度超过了 $k,ドル我们需要将其切割成两段。设较长的木材长度为 $x,ドル则切割成本为 $k \times (x - k)$。
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时间复杂度 $O(1),ドル空间复杂度 $O(1)$。
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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 minCuttingCost(self, n: int, m: int, k: int) -> int:
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x = max(n, m)
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return 0 if x <= k else k * (x - k)
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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:
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long long minCuttingCost(int n, int m, int k) {
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int x = max(n, m);
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return x <= k ? 0 : 1LL * k * (x - k);
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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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long long minCuttingCost(int n, int m, int k) {
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int x = max(n, m);
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return x <= k ? 0 : 1LL * k * (x - k);
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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 minCuttingCost(n int, m int, k int) int64 {
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x := max(n, m)
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if x <= k {
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return 0
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}
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return int64(k * (x - k))
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}
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```
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#### TypeScript
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```ts
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function minCuttingCost(n: number, m: number, k: number): number {
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const x = Math.max(n, m);
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return x <= k ? 0 : k * (x - k);
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}
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```
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<!-- tabs:end -->

‎solution/3500-3599/3560.Find Minimum Log Transportation Cost/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Mathematics
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If the lengths of both logs do not exceed the truck's maximum load $k,ドル then no cutting is needed, and we simply return 0ドル$.
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Otherwise, it means that only one log has a length greater than $k,ドル and we need to cut it into two pieces. Let the longer log have length $x,ドル then the cutting cost is $k \times (x - k)$.
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The time complexity is $O(1),ドル and 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 minCuttingCost(self, n: int, m: int, k: int) -> int:
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x = max(n, m)
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return 0 if x <= k else k * (x - k)
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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 long minCuttingCost(int n, int m, int k) {
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int x = Math.max(n, m);
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return x <= k ? 0 : 1L * k * (x - k);
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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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long long minCuttingCost(int n, int m, int k) {
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int x = max(n, m);
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return x <= k ? 0 : 1LL * k * (x - k);
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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 minCuttingCost(n int, m int, k int) int64 {
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x := max(n, m)
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if x <= k {
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return 0
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}
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return int64(k * (x - k))
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}
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```
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#### TypeScript
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```ts
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function minCuttingCost(n: number, m: number, k: number): number {
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const x = Math.max(n, m);
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return x <= k ? 0 : k * (x - k);
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}
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```
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<!-- tabs:end -->
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class Solution {
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public:
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long long minCuttingCost(int n, int m, int k) {
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int x = max(n, m);
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return x <= k ? 0 : 1LL * k * (x - k);
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}
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};
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func minCuttingCost(n int, m int, k int) int64 {
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x := max(n, m)
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if x <= k {
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return 0
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}
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return int64(k * (x - k))
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}
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class Solution {
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public long minCuttingCost(int n, int m, int k) {
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int x = Math.max(n, m);
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return x <= k ? 0 : 1L * k * (x - k);
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}
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}
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class Solution:
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def minCuttingCost(self, n: int, m: int, k: int) -> int:
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x = max(n, m)
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return 0 if x <= k else k * (x - k)
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function minCuttingCost(n: number, m: number, k: number): number {
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const x = Math.max(n, m);
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return x <= k ? 0 : k * (x - k);
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}

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