# Appeal Regarding Plagiarism Flag: Submission [351054331]↵
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**To the Administration/Judge,**↵
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I am writing to formally appeal the plagiarism flag on my submission **[351054331]** for problem **2170F**. While I acknowledge that my solution coincides with a list of other users, I want to state clearly and unequivocally that I wrote this code independently.↵
↵
The reason so many people (including myself) wrote nearly identical code is that this problem requires a very specific, standard optimization technique known as **"Offline Queries with Rightmost Index DP."**↵
↵
Here is a detailed breakdown of exactly why my code ended up looking like others:↵
↵
### 1. The "Offline Query" Boilerplate↵
Since the problem asks about ranges, the standard, most efficient method is to process the queries "offline." This involves iterating through the array once (from index $1$ to $N$) and handling all queries ending at the current index $R$.↵
↵
My code does exactly this:↵
↵
> vector<Query> qryAt[MAXN];↵
> // ...↵
> for (int r = 1; r <= n; r++) { ... }↵
↵
This is not unique logic; it is the standard boilerplate for offline processing. Almost every C++ competitive programmer uses a vector of structs or pairs to store the queries in this manner.↵
↵
### 2. The "Rightmost Index" Trick↵
The core difficulty of this problem is checking if a subset exists within the range $[L, R]$. The textbook trick for this is to avoid storing a boolean and instead store the **rightmost index**.↵
↵
The logic is defined as follows:↵
* $dp[k][x]$ = the largest index $i$ such that we can form XOR sum $x$ using $k$ elements ending at $i$.↵
↵
The validation check then becomes standard:↵
If $dp[k][target] \ge L$, then the subset is valid.↵
↵
Once you know this trick, there is essentially only one way to write the `if` statement. It is not copied; it is a direct implementation of the required logic.↵
↵
### 3. The Loop Structure (Knapsack Style)↵
You will notice my inner loop iterates backwards:↵
↵
> for (int k = 12; k >= 2; k--) { ... }↵
↵
I wrote it this way because it functions exactly like the **1D Knapsack DP**. You must iterate backwards to ensure you do not use the current element $a[r]$ more than once for the same subset size. This is not a stylistic choice—if I had looped forwards, the logic would be incorrect.↵
↵
### 4. Constants and Variables↵
* **4096 and 13:** These are not random numbers. The problem constraints limit values to 4096 and the subset size to 12. I simply defined constants `MAXV` and `MAXK` to match the problem statement.↵
* **Variable Names:** `n`, `q`, `l`, `r`, `x` are simply the variable names provided in the input section of the problem statement.↵
↵
### Conclusion↵
When a problem has tight constraints and relies on a specific "textbook" optimization (Rightmost Index DP), the solution space is extremely small. There are only so many ways to write three nested loops and a max function.↵
↵
The similarity is due to **"convergent evolution"** on a standard algorithm, not plagiarism. I request that you review the code with this context in mind.
↵
**To the Administration/Judge,**↵
↵
I am writing to formally appeal the plagiarism flag on my submission **[351054331]** for problem **2170F**. While I acknowledge that my solution coincides with a list of other users, I want to state clearly and unequivocally that I wrote this code independently.↵
↵
The reason so many people (including myself) wrote nearly identical code is that this problem requires a very specific, standard optimization technique known as **"Offline Queries with Rightmost Index DP."**↵
↵
Here is a detailed breakdown of exactly why my code ended up looking like others:↵
↵
### 1. The "Offline Query" Boilerplate↵
Since the problem asks about ranges, the standard, most efficient method is to process the queries "offline." This involves iterating through the array once (from index $1$ to $N$) and handling all queries ending at the current index $R$.↵
↵
My code does exactly this:↵
↵
> vector<Query> qryAt[MAXN];↵
> // ...↵
> for (int r = 1; r <= n; r++) { ... }↵
↵
This is not unique logic; it is the standard boilerplate for offline processing. Almost every C++ competitive programmer uses a vector of structs or pairs to store the queries in this manner.↵
↵
### 2. The "Rightmost Index" Trick↵
The core difficulty of this problem is checking if a subset exists within the range $[L, R]$. The textbook trick for this is to avoid storing a boolean and instead store the **rightmost index**.↵
↵
The logic is defined as follows:↵
* $dp[k][x]$ = the largest index $i$ such that we can form XOR sum $x$ using $k$ elements ending at $i$.↵
↵
The validation check then becomes standard:↵
If $dp[k][target] \ge L$, then the subset is valid.↵
↵
Once you know this trick, there is essentially only one way to write the `if` statement. It is not copied; it is a direct implementation of the required logic.↵
↵
### 3. The Loop Structure (Knapsack Style)↵
You will notice my inner loop iterates backwards:↵
↵
> for (int k = 12; k >= 2; k--) { ... }↵
↵
I wrote it this way because it functions exactly like the **1D Knapsack DP**. You must iterate backwards to ensure you do not use the current element $a[r]$ more than once for the same subset size. This is not a stylistic choice—if I had looped forwards, the logic would be incorrect.↵
↵
### 4. Constants and Variables↵
* **4096 and 13:** These are not random numbers. The problem constraints limit values to 4096 and the subset size to 12. I simply defined constants `MAXV` and `MAXK` to match the problem statement.↵
* **Variable Names:** `n`, `q`, `l`, `r`, `x` are simply the variable names provided in the input section of the problem statement.↵
↵
### Conclusion↵
When a problem has tight constraints and relies on a specific "textbook" optimization (Rightmost Index DP), the solution space is extremely small. There are only so many ways to write three nested loops and a max function.↵
↵
The similarity is due to **"convergent evolution"** on a standard algorithm, not plagiarism. I request that you review the code with this context in mind.
