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Commit 45f96c0

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feat: add solutions to lc problem: No.3021 (doocs#2283)
No.3021.Alice and Bob Playing Flower Game
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12 files changed

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‎solution/3000-3099/3021.Alice and Bob Playing Flower Game/README.md‎

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@@ -54,29 +54,107 @@
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## 解法
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### 方法一
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### 方法一:数学
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根据题目描述,每一次行动,玩家都会选择顺时针或者逆时针方向,然后摘一朵鲜花。由于 Alice 先行动,因此当 $x + y$ 为奇数时,Alice 一定会赢得游戏。
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因此,鲜花数目 $x$ 和 $y$ 满足以下条件:
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1. $x + y$ 为奇数;
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2. 1ドル \le x \le n$;
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3. 1ドル \le y \le m$。
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若 $x$ 为奇数,$y$ 一定为偶数。此时 $x$ 的取值个数为 $\lceil \frac{n}{2} \rceil,ドル$y$ 的取值个数为 $\lfloor \frac{m}{2} \rfloor,ドル因此满足条件的数对个数为 $\lceil \frac{n}{2} \rceil \times \lfloor \frac{m}{2} \rfloor$。
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若 $x$ 为偶数,$y$ 一定为奇数。此时 $x$ 的取值个数为 $\lfloor \frac{n}{2} \rfloor,ドル$y$ 的取值个数为 $\lceil \frac{m}{2} \rceil,ドル因此满足条件的数对个数为 $\lfloor \frac{n}{2} \rfloor \times \lceil \frac{m}{2} \rceil$。
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因此,满足条件的数对个数为 $\lceil \frac{n}{2} \rceil \times \lfloor \frac{m}{2} \rfloor + \lfloor \frac{n}{2} \rfloor \times \lceil \frac{m}{2} \rceil,ドル即 $\lfloor \frac{n + 1}{2} \rfloor \times \lfloor \frac{m}{2} \rfloor + \lfloor \frac{n}{2} \rfloor \times \lfloor \frac{m + 1}{2} \rfloor$。
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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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```python
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class Solution:
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def flowerGame(self, n: int, m: int) -> int:
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count = (n + 1) // 2
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tol = (m + 1) // 2
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ecount = n // 2
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etol = m // 2
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return count * etol + ecount * tol
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a1 = (n + 1) // 2
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b1 = (m + 1) // 2
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a2 = n // 2
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b2 = m // 2
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return a1 * b2 + a2 * b1
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```
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```java
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class Solution {
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public long flowerGame(int n, int m) {
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long a1 = (n + 1) / 2;
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long b1 = (m + 1) / 2;
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long a2 = n / 2;
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long b2 = m / 2;
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return a1 * b2 + a2 * b1;
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}
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}
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```
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```cpp
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class Solution {
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public:
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long long flowerGame(int n, int m) {
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long long a1 = (n + 1) / 2;
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long long b1 = (m + 1) / 2;
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long long a2 = n / 2;
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long long b2 = m / 2;
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return a1 * b2 + a2 * b1;
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}
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};
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```
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```go
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func flowerGame(n int, m int) int64 {
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a1, b1 := (n+1)/2, (m+1)/2
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a2, b2 := n/2, m/2
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return int64(a1*b2 + a2*b1)
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}
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```
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```ts
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function flowerGame(n: number, m: number): number {
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const [a1, b1] = [(n + 1) >> 1, (m + 1) >> 1];
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const [a2, b2] = [n >> 1, m >> 1];
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return a1 * b2 + a2 * b1;
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}
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```
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<!-- tabs:end -->
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### 方法二:数学(优化)
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方法一得出的结果为 $\lfloor \frac{n + 1}{2} \rfloor \times \lfloor \frac{m}{2} \rfloor + \lfloor \frac{n}{2} \rfloor \times \lfloor \frac{m + 1}{2} \rfloor$。
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如果 $n$ 和 $m$ 都是奇数,那么结果为 $\frac{n + 1}{2} \times \frac{m - 1}{2} + \frac{n - 1}{2} \times \frac{m + 1}{2},ドル即 $\frac{n \times m - 1}{2}$。
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如果 $n$ 和 $m$ 都是偶数,那么结果为 $\frac{n}{2} \times \frac{m}{2} + \frac{n}{2} \times \frac{m}{2},ドル即 $\frac{n \times m}{2}$。
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如果 $n$ 是奇数,且 $m$ 是偶数,那么结果为 $\frac{n + 1}{2} \times \frac{m}{2} + \frac{n - 1}{2} \times \frac{m}{2},ドル即 $\frac{n \times m}{2}$。
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如果 $n$ 是偶数,且 $m$ 是奇数,那么结果为 $\frac{n}{2} \times \frac{m - 1}{2} + \frac{n}{2} \times \frac{m + 1}{2},ドル即 $\frac{n \times m}{2}$。
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上面四种情况可以合并为 $\lfloor \frac{n \times m}{2} \rfloor$。
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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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```python
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class Solution:
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def flowerGame(self, n: int, m: int) -> int:
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return (n * m) // 2
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```
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```java
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class Solution {
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public long flowerGame(int n, int m) {
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long count = (n + 1) / 2;
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long tol = (m + 1) / 2;
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long ecount = n / 2;
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long etol = m / 2;
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return (count * etol + ecount * tol);
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return ((long) n * m) / 2;
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}
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}
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```
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class Solution {
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public:
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long long flowerGame(int n, int m) {
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long long count = (n + 1) / 2;
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long long tol = (m + 1) / 2;
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long long ecount = n / 2;
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long long etol = m / 2;
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return (count * etol + ecount * tol);
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return ((long long) n * m) / 2;
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}
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};
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```
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```go
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func flowerGame(n int, m int) int64 {
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count := int64((n + 1) / 2)
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tol := int64((m + 1) / 2)
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ecount := int64(n / 2)
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etol := int64(m / 2)
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return count*etol + ecount*tol
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return int64((n * m) / 2)
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}
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```
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```ts
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function flowerGame(n: number, m: number): number {
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return Number(((BigInt(n) * BigInt(m)) / 2n) | 0n);
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}
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```
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‎solution/3000-3099/3021.Alice and Bob Playing Flower Game/README_EN.md‎

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## Solutions
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### Solution 1
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### Solution 1: Mathematics
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According to the problem description, in each move, the player will choose to move in a clockwise or counterclockwise direction and then pick a flower. Since Alice moves first, when $x + y$ is odd, Alice will definitely win the game.
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Therefore, the number of flowers $x$ and $y$ meet the following conditions:
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1. $x + y$ is odd;
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2. 1ドル \le x \le n$;
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3. 1ドル \le y \le m$.
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If $x$ is odd, $y$ must be even. At this time, the number of values of $x$ is $\lceil \frac{n}{2} \rceil,ドル the number of values of $y$ is $\lfloor \frac{m}{2} \rfloor,ドル so the number of pairs that meet the conditions is $\lceil \frac{n}{2} \rceil \times \lfloor \frac{m}{2} \rfloor$.
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If $x$ is even, $y$ must be odd. At this time, the number of values of $x$ is $\lfloor \frac{n}{2} \rfloor,ドル the number of values of $y$ is $\lceil \frac{m}{2} \rceil,ドル so the number of pairs that meet the conditions is $\lfloor \frac{n}{2} \rfloor \times \lceil \frac{m}{2} \rceil$.
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Therefore, the number of pairs that meet the conditions is $\lceil \frac{n}{2} \rceil \times \lfloor \frac{m}{2} \rfloor + \lfloor \frac{n}{2} \rfloor \times \lceil \frac{m}{2} \rceil,ドル which is $\lfloor \frac{n + 1}{2} \rfloor \times \lfloor \frac{m}{2} \rfloor + \lfloor \frac{n}{2} \rfloor \times \lfloor \frac{m + 1}{2} \rfloor$.
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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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```python
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class Solution:
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def flowerGame(self, n: int, m: int) -> int:
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count = (n + 1) // 2
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tol = (m + 1) // 2
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ecount = n // 2
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etol = m // 2
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return count * etol + ecount * tol
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a1 = (n + 1) // 2
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b1 = (m + 1) // 2
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a2 = n // 2
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b2 = m // 2
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return a1 * b2 + a2 * b1
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```
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```java
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class Solution {
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public long flowerGame(int n, int m) {
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long a1 = (n + 1) / 2;
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long b1 = (m + 1) / 2;
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long a2 = n / 2;
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long b2 = m / 2;
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return a1 * b2 + a2 * b1;
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}
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}
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```
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```cpp
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class Solution {
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public:
98+
long long flowerGame(int n, int m) {
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long long a1 = (n + 1) / 2;
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long long b1 = (m + 1) / 2;
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long long a2 = n / 2;
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long long b2 = m / 2;
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return a1 * b2 + a2 * b1;
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}
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};
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```
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```go
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func flowerGame(n int, m int) int64 {
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a1, b1 := (n+1)/2, (m+1)/2
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a2, b2 := n/2, m/2
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return int64(a1*b2 + a2*b1)
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}
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```
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```ts
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function flowerGame(n: number, m: number): number {
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const [a1, b1] = [(n + 1) >> 1, (m + 1) >> 1];
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const [a2, b2] = [n >> 1, m >> 1];
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return a1 * b2 + a2 * b1;
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}
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```
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<!-- tabs:end -->
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### Solution 2: Mathematics (Optimized)
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The result obtained from Solution 1 is $\lfloor \frac{n + 1}{2} \rfloor \times \lfloor \frac{m}{2} \rfloor + \lfloor \frac{n}{2} \rfloor \times \lfloor \frac{m + 1}{2} \rfloor$.
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If both $n$ and $m$ are odd, then the result is $\frac{n + 1}{2} \times \frac{m - 1}{2} + \frac{n - 1}{2} \times \frac{m + 1}{2},ドル which is $\frac{n \times m - 1}{2}$.
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If both $n$ and $m$ are even, then the result is $\frac{n}{2} \times \frac{m}{2} + \frac{n}{2} \times \frac{m}{2},ドル which is $\frac{n \times m}{2}$.
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If $n$ is odd and $m$ is even, then the result is $\frac{n + 1}{2} \times \frac{m}{2} + \frac{n - 1}{2} \times \frac{m}{2},ドル which is $\frac{n \times m}{2}$.
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If $n$ is even and $m$ is odd, then the result is $\frac{n}{2} \times \frac{m - 1}{2} + \frac{n}{2} \times \frac{m + 1}{2},ドル which is $\frac{n \times m}{2}$.
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The above four cases can be combined into $\lfloor \frac{n \times m}{2} \rfloor$.
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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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```python
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class Solution:
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def flowerGame(self, n: int, m: int) -> int:
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return (n * m) // 2
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```
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```java
69151
class Solution {
70152
public long flowerGame(int n, int m) {
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long count = (n + 1) / 2;
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long tol = (m + 1) / 2;
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long ecount = n / 2;
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long etol = m / 2;
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return (count * etol + ecount * tol);
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return ((long) n * m) / 2;
76154
}
77155
}
78156
```
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class Solution {
82160
public:
83161
long long flowerGame(int n, int m) {
84-
long long count = (n + 1) / 2;
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long long tol = (m + 1) / 2;
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long long ecount = n / 2;
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long long etol = m / 2;
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return (count * etol + ecount * tol);
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return ((long long) n * m) / 2;
89163
}
90164
};
91165
```
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```go
94168
func flowerGame(n int, m int) int64 {
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count := int64((n + 1) / 2)
96-
tol := int64((m + 1) / 2)
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ecount := int64(n / 2)
98-
etol := int64(m / 2)
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return count*etol + ecount*tol
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return int64((n * m) / 2)
170+
}
171+
```
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```ts
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function flowerGame(n: number, m: number): number {
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return Number(((BigInt(n) * BigInt(m)) / 2n) | 0n);
100176
}
101177
```
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class Solution {
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public:
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long long flowerGame(int n, int m) {
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long long count = (n + 1) / 2;
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long long tol = (m + 1) / 2;
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long long ecount = n / 2;
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long long etol = m / 2;
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return (count * etol + ecount * tol);
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long long a1 = (n + 1) / 2;
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long long b1 = (m + 1) / 2;
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long long a2 = n / 2;
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long long b2 = m / 2;
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return a1 * b2 + a2 * b1;
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}
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};
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};
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@@ -1,7 +1,5 @@
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func flowerGame(n int, m int) int64 {
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count := int64((n + 1) / 2)
3-
tol := int64((m + 1) / 2)
4-
ecount := int64(n / 2)
5-
etol := int64(m / 2)
6-
return count*etol + ecount*tol
7-
}
2+
a1, b1 := (n+1)/2, (m+1)/2
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a2, b2 := n/2, m/2
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return int64(a1*b2 + a2*b1)
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}
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class Solution {
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public long flowerGame(int n, int m) {
3-
long count = (n + 1) / 2;
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long tol = (m + 1) / 2;
5-
long ecount = n / 2;
6-
long etol = m / 2;
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return (count * etol + ecount * tol);
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long a1 = (n + 1) / 2;
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long b1 = (m + 1) / 2;
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long a2 = n / 2;
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long b2 = m / 2;
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return a1 * b2 + a2 * b1;
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}
9-
}
9+
}
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@@ -1,7 +1,7 @@
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class Solution:
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def flowerGame(self, n: int, m: int) -> int:
3-
count = (n + 1) // 2
4-
tol = (m + 1) // 2
5-
ecount = n // 2
6-
etol = m // 2
7-
return count * etol + ecount * tol
3+
a1 = (n + 1) // 2
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b1 = (m + 1) // 2
5+
a2 = n // 2
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b2 = m // 2
7+
return a1 * b2 + a2 * b1
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,5 @@
1+
function flowerGame(n: number, m: number): number {
2+
const [a1, b1] = [(n + 1) >> 1, (m + 1) >> 1];
3+
const [a2, b2] = [n >> 1, m >> 1];
4+
return a1 * b2 + a2 * b1;
5+
}
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,6 @@
1+
class Solution {
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public:
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long long flowerGame(int n, int m) {
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return ((long long) n * m) / 2;
5+
}
6+
};
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@@ -0,0 +1,3 @@
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func flowerGame(n int, m int) int64 {
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return int64((n * m) / 2)
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}
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@@ -0,0 +1,5 @@
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class Solution {
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public long flowerGame(int n, int m) {
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return ((long) n * m) / 2;
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}
5+
}

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