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Commit a79dce4

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feat: add solutions to lc problem: No.2140 (doocs#4316)
No.2140.Solving Questions With Brainpower
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‎solution/2100-2199/2140.Solving Questions With Brainpower/README.md‎

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### 方法一:记忆化搜索
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我们设计一个函数 $dfs(i),ドル表示从第 $i$ 个问题开始解决,能够获得的最高分数。那么答案就是 $dfs(0)$。
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我们设计一个函数 $\textit{dfs}(i),ドル表示从第 $i$ 个问题开始解决,能够获得的最高分数。那么答案就是 $\textit{dfs}(0)$。
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函数 $dfs(i)$ 的计算方式如下:
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函数 $\textit{dfs}(i)$ 的计算方式如下:
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- 如果 $i \geq n,ドル表示已经解决完所有问题,返回 0ドル$;
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- 否则,设第 $i$ 个问题的分数为 $p,ドル需要跳过的问题数为 $b,ドル那么 $dfs(i) = \max(p + dfs(i + b + 1), dfs(i + 1))$。
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- 否则,设第 $i$ 个问题的分数为 $p,ドル需要跳过的问题数为 $b,ドル那么 $\textit{dfs}(i) = \max(p + \textit{dfs}(i + b + 1), \textit{dfs}(i + 1))$。
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为了避免重复计算,我们可以使用记忆化搜索的方法,用一个数组 $f$ 记录所有已经计算过的 $dfs(i)$ 的值。
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为了避免重复计算,我们可以使用记忆化搜索的方法,用一个数组 $f$ 记录所有已经计算过的 $\textit{dfs}(i)$ 的值。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 是问题的数量。
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int n = questions.size();
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long long f[n];
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memset(f, 0, sizeof(f));
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function<long long(int)> dfs = [&](int i) -> long long {
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auto dfs = [&](this auto&& dfs, int i) -> long long {
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if (i >= n) {
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return 0;
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}
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn most_points(questions: Vec<Vec<i32>>) -> i64 {
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let n = questions.len();
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let mut f = vec![-1; n];
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fn dfs(i: usize, questions: &Vec<Vec<i32>>, f: &mut Vec<i64>) -> i64 {
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if i >= questions.len() {
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return 0;
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}
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if f[i] != -1 {
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return f[i];
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}
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let p = questions[i][0] as i64;
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let b = questions[i][1] as usize;
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f[i] = (p + dfs(i + b + 1, questions, f)).max(dfs(i + 1, questions, f));
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f[i]
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}
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dfs(0, &questions, &mut f)
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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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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn most_points(questions: Vec<Vec<i32>>) -> i64 {
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let n = questions.len();
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let mut f = vec![0; n + 1];
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for i in (0..n).rev() {
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let p = questions[i][0] as i64;
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let b = questions[i][1] as usize;
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let j = i + b + 1;
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f[i] = f[i + 1].max(p + if j > n { 0 } else { f[j] });
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}
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f[0]
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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/2100-2199/2140.Solving Questions With Brainpower/README_EN.md‎

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<!-- solution:start -->
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### Solution 1: Memoization Search
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### Solution 1: Memoization
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We design a function $dfs(i),ドル which represents the maximum score that can be obtained starting from the $i$-th problem. Therefore, the answer is $dfs(0)$.
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We design a function $\textit{dfs}(i),ドル which represents the maximum score that can be obtained starting from the $i$-th question. The answer is $\textit{dfs}(0)$.
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The calculation method of the function $dfs(i)$ is as follows:
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The function $\textit{dfs}(i)$ is calculated as follows:
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- If $i \geq n,ドル it means that all problems have been solved, return 0ドル$;
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- Otherwise, let the score of the $i$-th problem be $p,ドル and the number of problems to skip be $b$, then $dfs(i) = \max(p + dfs(i + b + 1), dfs(i + 1))$.
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- If $i \geq n,ドル it means all questions have been solved, so return 0ドル$;
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- Otherwise, let the score of the $i$-th question be $p,ドル and the number of questions to skip be $b$. Then, $\textit{dfs}(i) = \max(p + \textit{dfs}(i + b + 1), \textit{dfs}(i + 1))$.
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To avoid repeated calculations, we can use the method of memoization search, using an array $f$ to record the values of all already computed $dfs(i)$.
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To avoid repeated calculations, we can use memoization by storing the values of $\textit{dfs}(i)$ in an array $f$.
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The time complexity is $O(n),ドル and the space complexity is $O(n)$. Here, $n$ is the number of problems.
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The time complexity is $O(n),ドル and the space complexity is $O(n)$, where $n$ is the number of questions.
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<!-- tabs:start -->
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int n = questions.size();
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long long f[n];
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memset(f, 0, sizeof(f));
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function<long long(int)> dfs = [&](int i) -> long long {
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auto dfs = [&](this auto&& dfs, int i) -> long long {
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if (i >= n) {
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return 0;
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}
@@ -201,6 +201,32 @@ function mostPoints(questions: number[][]): number {
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn most_points(questions: Vec<Vec<i32>>) -> i64 {
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let n = questions.len();
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let mut f = vec![-1; n];
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fn dfs(i: usize, questions: &Vec<Vec<i32>>, f: &mut Vec<i64>) -> i64 {
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if i >= questions.len() {
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return 0;
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}
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if f[i] != -1 {
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return f[i];
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}
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let p = questions[i][0] as i64;
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let b = questions[i][1] as usize;
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f[i] = (p + dfs(i + b + 1, questions, f)).max(dfs(i + 1, questions, f));
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f[i]
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}
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dfs(0, &questions, &mut f)
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
@@ -305,6 +331,24 @@ function mostPoints(questions: number[][]): number {
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn most_points(questions: Vec<Vec<i32>>) -> i64 {
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let n = questions.len();
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let mut f = vec![0; n + 1];
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for i in (0..n).rev() {
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let p = questions[i][0] as i64;
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let b = questions[i][1] as usize;
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let j = i + b + 1;
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f[i] = f[i + 1].max(p + if j > n { 0 } else { f[j] });
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}
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f[0]
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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/2100-2199/2140.Solving Questions With Brainpower/Solution.cpp‎

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int n = questions.size();
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long long f[n];
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memset(f, 0, sizeof(f));
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function<longlong(int)> dfs = [&](int i) -> long long {
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autodfs = [&](thisauto&& dfs, int i) -> long long {
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if (i >= n) {
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return 0;
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}
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impl Solution {
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pub fn most_points(questions: Vec<Vec<i32>>) -> i64 {
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let n = questions.len();
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let mut f = vec![-1; n];
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fn dfs(i: usize, questions: &Vec<Vec<i32>>, f: &mut Vec<i64>) -> i64 {
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if i >= questions.len() {
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return 0;
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}
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if f[i] != -1 {
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return f[i];
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}
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let p = questions[i][0] as i64;
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let b = questions[i][1] as usize;
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f[i] = (p + dfs(i + b + 1, questions, f)).max(dfs(i + 1, questions, f));
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f[i]
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}
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dfs(0, &questions, &mut f)
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}
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}
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impl Solution {
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pub fn most_points(questions: Vec<Vec<i32>>) -> i64 {
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let n = questions.len();
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let mut f = vec![0; n + 1];
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for i in (0..n).rev() {
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let p = questions[i][0] as i64;
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let b = questions[i][1] as usize;
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let j = i + b + 1;
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f[i] = f[i + 1].max(p + if j > n { 0 } else { f[j] });
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}
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f[0]
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}
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}

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