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Commit f5f8de8

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feat: add solutions to lc problems: No.2928,2929 (doocs#1956)
* No.2928.Distribute Candies Among Children I * No.2929.Distribute Candies Among Children II
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‎solution/2900-2999/2928.Distribute Candies Among Children I/README.md‎

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@@ -41,34 +41,123 @@
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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
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class Solution:
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def distributeCandies(self, n: int, limit: int) -> int:
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if n > 3 * limit:
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return 0
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ans = comb(n + 2, 2)
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if n > limit:
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ans -= 3 * comb(n - limit + 1, 2)
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if n - 2 >= 2 * limit:
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ans += 3 * comb(n - 2 * limit, 2)
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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 distributeCandies(int n, int limit) {
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if (n > 3 * limit) {
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return 0;
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}
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long ans = comb2(n + 2);
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if (n > limit) {
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ans -= 3 * comb2(n - limit + 1);
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}
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if (n - 2 >= 2 * limit) {
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ans += 3 * comb2(n - 2 * limit);
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}
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return (int) ans;
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}
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private long comb2(int n) {
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return 1L * n * (n - 1) / 2;
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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 distributeCandies(int n, int limit) {
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auto comb2 = [](int n) {
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return 1LL * n * (n - 1) / 2;
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};
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if (n > 3 * limit) {
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return 0;
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}
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long long ans = comb2(n + 2);
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if (n > limit) {
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ans -= 3 * comb2(n - limit + 1);
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}
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if (n - 2 >= 2 * limit) {
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ans += 3 * comb2(n - 2 * limit);
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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 distributeCandies(n int, limit int) int {
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comb2 := func(n int) int {
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return n * (n - 1) / 2
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}
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if n > 3*limit {
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return 0
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}
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ans := comb2(n + 2)
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if n > limit {
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ans -= 3 * comb2(n-limit+1)
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}
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if n-2 >= 2*limit {
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ans += 3 * comb2(n-2*limit)
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}
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return ans
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}
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```
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### **TypeScript**
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```ts
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function distributeCandies(n: number, limit: number): number {
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const comb2 = (n: number) => (n * (n - 1)) / 2;
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if (n > 3 * limit) {
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return 0;
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}
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let ans = comb2(n + 2);
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if (n > limit) {
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ans -= 3 * comb2(n - limit + 1);
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}
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if (n - 2 >= 2 * limit) {
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ans += 3 * comb2(n - 2 * limit);
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}
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return ans;
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}
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```
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### **...**

‎solution/2900-2999/2928.Distribute Candies Among Children I/README_EN.md‎

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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
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class Solution:
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def distributeCandies(self, n: int, limit: int) -> int:
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if n > 3 * limit:
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return 0
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ans = comb(n + 2, 2)
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if n > limit:
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ans -= 3 * comb(n - limit + 1, 2)
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if n - 2 >= 2 * limit:
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ans += 3 * comb(n - 2 * limit, 2)
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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 distributeCandies(int n, int limit) {
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if (n > 3 * limit) {
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return 0;
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}
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long ans = comb2(n + 2);
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if (n > limit) {
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ans -= 3 * comb2(n - limit + 1);
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}
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if (n - 2 >= 2 * limit) {
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ans += 3 * comb2(n - 2 * limit);
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}
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return (int) ans;
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}
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private long comb2(int n) {
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return 1L * n * (n - 1) / 2;
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}
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}
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```
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### **C++**
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```cpp
55-
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class Solution {
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public:
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int distributeCandies(int n, int limit) {
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auto comb2 = [](int n) {
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return 1LL * n * (n - 1) / 2;
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};
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if (n > 3 * limit) {
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return 0;
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}
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long long ans = comb2(n + 2);
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if (n > limit) {
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ans -= 3 * comb2(n - limit + 1);
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}
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if (n - 2 >= 2 * limit) {
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ans += 3 * comb2(n - 2 * limit);
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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 distributeCandies(n int, limit int) int {
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comb2 := func(n int) int {
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return n * (n - 1) / 2
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}
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if n > 3*limit {
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return 0
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}
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ans := comb2(n + 2)
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if n > limit {
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ans -= 3 * comb2(n-limit+1)
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}
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if n-2 >= 2*limit {
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ans += 3 * comb2(n-2*limit)
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}
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return ans
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}
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```
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### **TypeScript**
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```ts
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function distributeCandies(n: number, limit: number): number {
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const comb2 = (n: number) => (n * (n - 1)) / 2;
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if (n > 3 * limit) {
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return 0;
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}
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let ans = comb2(n + 2);
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if (n > limit) {
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ans -= 3 * comb2(n - limit + 1);
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}
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if (n - 2 >= 2 * limit) {
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ans += 3 * comb2(n - 2 * limit);
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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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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,19 @@
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class Solution {
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public:
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int distributeCandies(int n, int limit) {
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auto comb2 = [](int n) {
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return 1LL * n * (n - 1) / 2;
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};
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if (n > 3 * limit) {
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return 0;
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}
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long long ans = comb2(n + 2);
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if (n > limit) {
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ans -= 3 * comb2(n - limit + 1);
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}
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if (n - 2 >= 2 * limit) {
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ans += 3 * comb2(n - 2 * limit);
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}
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return ans;
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}
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};
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Original file line numberDiff line numberDiff line change
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func distributeCandies(n int, limit int) int {
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comb2 := func(n int) int {
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return n * (n - 1) / 2
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}
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if n > 3*limit {
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return 0
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}
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ans := comb2(n + 2)
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if n > limit {
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ans -= 3 * comb2(n-limit+1)
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}
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if n-2 >= 2*limit {
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ans += 3 * comb2(n-2*limit)
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}
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return ans
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}
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,19 @@
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class Solution {
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public int distributeCandies(int n, int limit) {
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if (n > 3 * limit) {
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return 0;
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}
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long ans = comb2(n + 2);
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if (n > limit) {
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ans -= 3 * comb2(n - limit + 1);
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}
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if (n - 2 >= 2 * limit) {
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ans += 3 * comb2(n - 2 * limit);
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}
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return (int) ans;
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}
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private long comb2(int n) {
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return 1L * n * (n - 1) / 2;
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}
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}
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class Solution:
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def distributeCandies(self, n: int, limit: int) -> int:
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if n > 3 * limit:
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return 0
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ans = comb(n + 2, 2)
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if n > limit:
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ans -= 3 * comb(n - limit + 1, 2)
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if n - 2 >= 2 * limit:
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ans += 3 * comb(n - 2 * limit, 2)
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return ans
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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,14 @@
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function distributeCandies(n: number, limit: number): number {
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const comb2 = (n: number) => (n * (n - 1)) / 2;
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if (n > 3 * limit) {
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return 0;
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}
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let ans = comb2(n + 2);
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if (n > limit) {
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ans -= 3 * comb2(n - limit + 1);
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}
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if (n - 2 >= 2 * limit) {
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ans += 3 * comb2(n - 2 * limit);
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}
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return ans;
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}

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