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167 changes: 167 additions & 0 deletions
solution/2900-2999/2927.Distribute Candies Among Children III/README.md
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| # [2927. Distribute Candies Among Children III](https://leetcode.cn/problems/distribute-candies-among-children-iii) | ||
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| [English Version](/solution/2900-2999/2927.Distribute%20Candies%20Among%20Children%20III/README_EN.md) | ||
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| ## 题目描述 | ||
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| <!-- 这里写题目描述 --> | ||
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| <p>You are given two positive integers <code>n</code> and <code>limit</code>.</p> | ||
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| <p>Return <em>the <strong>total number</strong> of ways to distribute </em><code>n</code> <em>candies among </em><code>3</code><em> children such that no child gets more than </em><code>limit</code><em> candies.</em></p> | ||
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| <p> </p> | ||
| <p><strong class="example">Example 1:</strong></p> | ||
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| <pre> | ||
| <strong>Input:</strong> n = 5, limit = 2 | ||
| <strong>Output:</strong> 3 | ||
| <strong>Explanation:</strong> There are 3 ways to distribute 5 candies such that no child gets more than 2 candies: (1, 2, 2), (2, 1, 2) and (2, 2, 1). | ||
| </pre> | ||
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| <p><strong class="example">Example 2:</strong></p> | ||
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| <pre> | ||
| <strong>Input:</strong> n = 3, limit = 3 | ||
| <strong>Output:</strong> 10 | ||
| <strong>Explanation:</strong> There are 10 ways to distribute 3 candies such that no child gets more than 3 candies: (0, 0, 3), (0, 1, 2), (0, 2, 1), (0, 3, 0), (1, 0, 2), (1, 1, 1), (1, 2, 0), (2, 0, 1), (2, 1, 0) and (3, 0, 0). | ||
| </pre> | ||
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| <p> </p> | ||
| <p><strong>Constraints:</strong></p> | ||
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| <ul> | ||
| <li><code>1 <= n <= 10<sup>8</sup></code></li> | ||
| <li><code>1 <= limit <= 10<sup>8</sup></code></li> | ||
| </ul> | ||
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| ## 解法 | ||
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| <!-- 这里可写通用的实现逻辑 --> | ||
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| **方法一:组合数学 + 容斥原理** | ||
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| 根据题目描述,我们需要将 $n$ 个糖果分给 3ドル$ 个小孩,每个小孩分到的糖果数在 $[0, limit]$ 之间。 | ||
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| 这实际上等价于把 $n$ 个球放入 3ドル$ 个盒子中。由于盒子可以为空,我们可以再增加 3ドル$ 个虚拟球,然后再利用隔板法,即一共有 $n + 3$ 个球,我们在其中 $n + 3 - 1$ 个位置插入 2ドル$ 个隔板,从而将实际的 $n$ 个球分成 3ドル$ 组,并且允许盒子为空,因此初始方案数为 $C_{n + 2}^2$。 | ||
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| 我们需要在这些方案中,排除掉存在盒子分到的小球数超过 $limit$ 的方案。考虑其中有一个盒子分到的小球数超过 $limit,ドル那么剩下的球(包括虚拟球)最多有 $n + 3 - (limit + 1) = n - limit + 2$ 个,位置数为 $n - limit + 1,ドル因此方案数为 $C_{n - limit + 1}^2$。由于存在 3ドル$ 个盒子,因此这样的方案数为 3ドル \times C_{n - limit + 1}^2$。这样子算,我们会多排除掉同时存在两个盒子分到的小球数超过 $limit$ 的方案,因此我们需要再加上这样的方案数,即 3ドル \times C_{n - 2 \times limit}^2$。 | ||
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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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| <!-- 这里可写当前语言的特殊实现逻辑 --> | ||
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| ```python | ||
| class Solution: | ||
| def distributeCandies(self, n: int, limit: int) -> int: | ||
| if n > 3 * limit: | ||
| return 0 | ||
| ans = comb(n + 2, 2) | ||
| if n > limit: | ||
| ans -= 3 * comb(n - limit + 1, 2) | ||
| if n - 2 >= 2 * limit: | ||
| ans += 3 * comb(n - 2 * limit, 2) | ||
| return ans | ||
| ``` | ||
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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 | ||
| class Solution { | ||
| public long distributeCandies(int n, int limit) { | ||
| if (n > 3 * limit) { | ||
| return 0; | ||
| } | ||
| long ans = comb2(n + 2); | ||
| if (n > limit) { | ||
| ans -= 3 * comb2(n - limit + 1); | ||
| } | ||
| if (n - 2 >= 2 * limit) { | ||
| ans += 3 * comb2(n - 2 * limit); | ||
| } | ||
| return ans; | ||
| } | ||
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| private long comb2(int n) { | ||
| return 1L * n * (n - 1) / 2; | ||
| } | ||
| } | ||
| ``` | ||
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| ### **C++** | ||
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| ```cpp | ||
| class Solution { | ||
| public: | ||
| long long distributeCandies(int n, int limit) { | ||
| auto comb2 = [](int n) { | ||
| return 1LL * n * (n - 1) / 2; | ||
| }; | ||
| if (n > 3 * limit) { | ||
| return 0; | ||
| } | ||
| long long ans = comb2(n + 2); | ||
| if (n > limit) { | ||
| ans -= 3 * comb2(n - limit + 1); | ||
| } | ||
| if (n - 2 >= 2 * limit) { | ||
| ans += 3 * comb2(n - 2 * limit); | ||
| } | ||
| return ans; | ||
| } | ||
| }; | ||
| ``` | ||
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| ### **Go** | ||
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| ```go | ||
| func distributeCandies(n int, limit int) int64 { | ||
| comb2 := func(n int) int { | ||
| return n * (n - 1) / 2 | ||
| } | ||
| if n > 3*limit { | ||
| return 0 | ||
| } | ||
| ans := comb2(n+2) | ||
| if n > limit { | ||
| ans -= 3 * comb2(n-limit+1) | ||
| } | ||
| if n-2 >= 2*limit { | ||
| ans += 3 * comb2(n-2*limit) | ||
| } | ||
| return int64(ans) | ||
| } | ||
| ``` | ||
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| ### **TypeScript** | ||
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| ```ts | ||
| function distributeCandies(n: number, limit: number): number { | ||
| const comb2 = (n: number) => (n * (n - 1)) / 2; | ||
| if (n > 3 * limit) { | ||
| return 0; | ||
| } | ||
| let ans = comb2(n + 2); | ||
| if (n > limit) { | ||
| ans -= 3 * comb2(n - limit + 1); | ||
| } | ||
| if (n - 2 >= 2 * limit) { | ||
| ans += 3 * comb2(n - 2 * limit); | ||
| } | ||
| 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/2900-2999/2927.Distribute Candies Among Children III/README_EN.md
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| # [2927. Distribute Candies Among Children III](https://leetcode.com/problems/distribute-candies-among-children-iii) | ||
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| [中文文档](/solution/2900-2999/2927.Distribute%20Candies%20Among%20Children%20III/README.md) | ||
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| ## Description | ||
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| <p>You are given two positive integers <code>n</code> and <code>limit</code>.</p> | ||
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| <p>Return <em>the <strong>total number</strong> of ways to distribute </em><code>n</code> <em>candies among </em><code>3</code><em> children such that no child gets more than </em><code>limit</code><em> candies.</em></p> | ||
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| <p> </p> | ||
| <p><strong class="example">Example 1:</strong></p> | ||
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| <pre> | ||
| <strong>Input:</strong> n = 5, limit = 2 | ||
| <strong>Output:</strong> 3 | ||
| <strong>Explanation:</strong> There are 3 ways to distribute 5 candies such that no child gets more than 2 candies: (1, 2, 2), (2, 1, 2) and (2, 2, 1). | ||
| </pre> | ||
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| <p><strong class="example">Example 2:</strong></p> | ||
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| <pre> | ||
| <strong>Input:</strong> n = 3, limit = 3 | ||
| <strong>Output:</strong> 10 | ||
| <strong>Explanation:</strong> There are 10 ways to distribute 3 candies such that no child gets more than 3 candies: (0, 0, 3), (0, 1, 2), (0, 2, 1), (0, 3, 0), (1, 0, 2), (1, 1, 1), (1, 2, 0), (2, 0, 1), (2, 1, 0) and (3, 0, 0). | ||
| </pre> | ||
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| <p> </p> | ||
| <p><strong>Constraints:</strong></p> | ||
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| <ul> | ||
| <li><code>1 <= n <= 10<sup>8</sup></code></li> | ||
| <li><code>1 <= limit <= 10<sup>8</sup></code></li> | ||
| </ul> | ||
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| ## Solutions | ||
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| **Solution 1: Combinatorial Mathematics + Principle of Inclusion-Exclusion** | ||
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| According to the problem description, we need to distribute $n$ candies to 3ドル$ children, with each child receiving between $[0, limit]$ candies. | ||
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| This is equivalent to placing $n$ balls into 3ドル$ boxes. Since the boxes can be empty, we can add 3ドル$ virtual balls, and then use the method of inserting partitions, i.e., there are a total of $n + 3$ balls, and we insert 2ドル$ partitions among the $n + 3 - 1$ positions, thus dividing the actual $n$ balls into 3ドル$ groups, and allowing the boxes to be empty. Therefore, the initial number of schemes is $C_{n + 2}^2$. | ||
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| We need to exclude the schemes where the number of balls in a box exceeds $limit$. Consider that there is a box where the number of balls exceeds $limit,ドル then the remaining balls (including virtual balls) have at most $n + 3 - (limit + 1) = n - limit + 2,ドル and the number of positions is $n - limit + 1,ドル so the number of schemes is $C_{n - limit + 1}^2$. Since there are 3ドル$ boxes, the number of such schemes is 3ドル \times C_{n - limit + 1}^2$. In this way, we will exclude too many schemes where the number of balls in two boxes exceeds $limit$ at the same time, so we need to add the number of such schemes, i.e., 3ドル \times C_{n - 2 \times limit}^2$. | ||
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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 | ||
| class Solution: | ||
| def distributeCandies(self, n: int, limit: int) -> int: | ||
| if n > 3 * limit: | ||
| return 0 | ||
| ans = comb(n + 2, 2) | ||
| if n > limit: | ||
| ans -= 3 * comb(n - limit + 1, 2) | ||
| if n - 2 >= 2 * limit: | ||
| ans += 3 * comb(n - 2 * limit, 2) | ||
| return ans | ||
| ``` | ||
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| ### **Java** | ||
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| ```java | ||
| class Solution { | ||
| public long distributeCandies(int n, int limit) { | ||
| if (n > 3 * limit) { | ||
| return 0; | ||
| } | ||
| long ans = comb2(n + 2); | ||
| if (n > limit) { | ||
| ans -= 3 * comb2(n - limit + 1); | ||
| } | ||
| if (n - 2 >= 2 * limit) { | ||
| ans += 3 * comb2(n - 2 * limit); | ||
| } | ||
| return ans; | ||
| } | ||
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| private long comb2(int n) { | ||
| return 1L * n * (n - 1) / 2; | ||
| } | ||
| } | ||
| ``` | ||
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| ### **C++** | ||
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| ```cpp | ||
| class Solution { | ||
| public: | ||
| long long distributeCandies(int n, int limit) { | ||
| auto comb2 = [](int n) { | ||
| return 1LL * n * (n - 1) / 2; | ||
| }; | ||
| if (n > 3 * limit) { | ||
| return 0; | ||
| } | ||
| long long ans = comb2(n + 2); | ||
| if (n > limit) { | ||
| ans -= 3 * comb2(n - limit + 1); | ||
| } | ||
| if (n - 2 >= 2 * limit) { | ||
| ans += 3 * comb2(n - 2 * limit); | ||
| } | ||
| return ans; | ||
| } | ||
| }; | ||
| ``` | ||
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| ### **Go** | ||
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| ```go | ||
| func distributeCandies(n int, limit int) int64 { | ||
| comb2 := func(n int) int { | ||
| return n * (n - 1) / 2 | ||
| } | ||
| if n > 3*limit { | ||
| return 0 | ||
| } | ||
| ans := comb2(n+2) | ||
| if n > limit { | ||
| ans -= 3 * comb2(n-limit+1) | ||
| } | ||
| if n-2 >= 2*limit { | ||
| ans += 3 * comb2(n-2*limit) | ||
| } | ||
| return int64(ans) | ||
| } | ||
| ``` | ||
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| ### **TypeScript** | ||
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| ```ts | ||
| function distributeCandies(n: number, limit: number): number { | ||
| const comb2 = (n: number) => (n * (n - 1)) / 2; | ||
| if (n > 3 * limit) { | ||
| return 0; | ||
| } | ||
| let ans = comb2(n + 2); | ||
| if (n > limit) { | ||
| ans -= 3 * comb2(n - limit + 1); | ||
| } | ||
| if (n - 2 >= 2 * limit) { | ||
| ans += 3 * comb2(n - 2 * limit); | ||
| } | ||
| return ans; | ||
| } | ||
| ``` | ||
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| ### **...** | ||
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| ``` | ||
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| ``` | ||
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| <!-- tabs:end --> |
19 changes: 19 additions & 0 deletions
solution/2900-2999/2927.Distribute Candies Among Children III/Solution.cpp
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,19 @@ | ||
| class Solution { | ||
| public: | ||
| long long distributeCandies(int n, int limit) { | ||
| auto comb2 = [](int n) { | ||
| return 1LL * n * (n - 1) / 2; | ||
| }; | ||
| if (n > 3 * limit) { | ||
| return 0; | ||
| } | ||
| long long ans = comb2(n + 2); | ||
| if (n > limit) { | ||
| ans -= 3 * comb2(n - limit + 1); | ||
| } | ||
| if (n - 2 >= 2 * limit) { | ||
| ans += 3 * comb2(n - 2 * limit); | ||
| } | ||
| return ans; | ||
| } | ||
| }; |
16 changes: 16 additions & 0 deletions
solution/2900-2999/2927.Distribute Candies Among Children III/Solution.go
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,16 @@ | ||
| func distributeCandies(n int, limit int) int64 { | ||
| comb2 := func(n int) int { | ||
| return n * (n - 1) / 2 | ||
| } | ||
| if n > 3*limit { | ||
| return 0 | ||
| } | ||
| ans := comb2(n+2) | ||
| if n > limit { | ||
| ans -= 3 * comb2(n-limit+1) | ||
| } | ||
| if n-2 >= 2*limit { | ||
| ans += 3 * comb2(n-2*limit) | ||
| } | ||
| return int64(ans) | ||
| } |
19 changes: 19 additions & 0 deletions
solution/2900-2999/2927.Distribute Candies Among Children III/Solution.java
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,19 @@ | ||
| class Solution { | ||
| public long distributeCandies(int n, int limit) { | ||
| if (n > 3 * limit) { | ||
| return 0; | ||
| } | ||
| long ans = comb2(n + 2); | ||
| if (n > limit) { | ||
| ans -= 3 * comb2(n - limit + 1); | ||
| } | ||
| if (n - 2 >= 2 * limit) { | ||
| ans += 3 * comb2(n - 2 * limit); | ||
| } | ||
| return ans; | ||
| } | ||
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| private long comb2(int n) { | ||
| return 1L * n * (n - 1) / 2; | ||
| } | ||
| } |
10 changes: 10 additions & 0 deletions
solution/2900-2999/2927.Distribute Candies Among Children III/Solution.py
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,10 @@ | ||
| class Solution: | ||
| def distributeCandies(self, n: int, limit: int) -> int: | ||
| if n > 3 * limit: | ||
| return 0 | ||
| ans = comb(n + 2, 2) | ||
| if n > limit: | ||
| ans -= 3 * comb(n - limit + 1, 2) | ||
| if n - 2 >= 2 * limit: | ||
| ans += 3 * comb(n - 2 * limit, 2) | ||
| return ans |
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