CodeGraderBench/CodeGraderBench huggingface dataset:
Problem: Write a C++ program to find all prime numbers from 2 to n (n taken as input from the user) following the procedure given below:Create an array of consecutive integers from 2 through n: (2, 3, 4, ..., n).Initially, let p equal 2, the smallest prime number. Enumerate the multiples of p by counting in increments of p from 2p to n, and mark them in the array as non-prime (these will be 2p, 3p, 4p, ...; the p itself should not be marked). Find the smallest number in the array greater than p that is not marked. If there was no such number, stop. Otherwise, let p now be equal to this new number (which will be the next prime number), and repeat from step 3.When the algorithm terminates, the numbers remaining not marked in the array are all the primes below n. Your program should print out all prime numbers from 2 to n. You do not need to use functions in this task (you may if you wish to). 

Input Format - Integer n.
Output Format - Print out the prime numbers with a space in between. After every five prime numbers print a newline character for ease of reading. 

Constraints
n >= 2 and n <= 500

Student submission:
#include<iostream>
using namespace std;

void arrin(int a[] , int n)
{
	for(int i = 0; i < n ; i++)
	{
		a[i] = 2 + i;
	}
}

void arrout(int a[] , int n)
{ 	int count =0;
        for(int i = 0; i < n ; i++)
        {
     	  		if (a[i] != 0)
        		{	
      	        		 count = count + 1;
      	        		 cout << a[i] << " ";
      	        		 if( count%5 == 0)
      	        		 {
      	        		 	cout << endl;
      	        		 }
      			}
        }
}

int pnew(int a[] , int n , int p)
{
	for(int i = 0; i < (n-1) ; i++)
	{
		if(a[i] > p)
		{
			p = a[i];
			break;
		}
		else if(a[i] == p)
		{
			p = 0;	
		}
	}
	return p;
}

void mark(int a[] , int n , int p)
{
	int num,m;
	num = n/p;
	for(int i = 2 ;  i < (num+1) ; i++)
	{
		m = i*p;
		for(int j = 0; j < (n-1) ; j++)
		{
			if (a[j] == m)
			{
				a[j] = 0;
			}
		}
	}
}

void prime(int a[] , int n)
{
	int p = 2;
	while(p != 0)
	{
		mark(a,n,p);
		p = pnew(a,n,p);
	}
}

int main()
{
	int n;
	cin >> n;
	int a[n-1];
	arrin(a,n-1);
	prime(a,n);
	arrout(a,n-1);
}

Grading criterion: Creation of array of consecutive integers
Rating A: Correctly Done
Rating B: Not done
Rating C: 0
Rating D: 0
Assigned grade: A
Problem: Write a C++ program to find all prime numbers from 2 to n (n taken as input from the user) following the procedure given below:Create an array of consecutive integers from 2 through n: (2, 3, 4, ..., n).Initially, let p equal 2, the smallest prime number. Enumerate the multiples of p by counting in increments of p from 2p to n, and mark them in the array as non-prime (these will be 2p, 3p, 4p, ...; the p itself should not be marked). Find the smallest number in the array greater than p that is not marked. If there was no such number, stop. Otherwise, let p now be equal to this new number (which will be the next prime number), and repeat from step 3.When the algorithm terminates, the numbers remaining not marked in the array are all the primes below n. Your program should print out all prime numbers from 2 to n. You do not need to use functions in this task (you may if you wish to). 

Input Format - Integer n.
Output Format - Print out the prime numbers with a space in between. After every five prime numbers print a newline character for ease of reading. 

Constraints
n >= 2 and n <= 500

Student submission:
#include <iostream>
using namespace std;
int main()   {
	int array_A[2]={2,3,5,7,....n}
	for(int i=0;i<2,i++)
	{
		cout<<"prime number"<<array;
		cout<<\n;
	}
	return 0;
}

Grading criterion: Creation of array of consecutive integers
Rating A: Correctly Done
Rating B: Not done
Rating C: 0
Rating D: 0
Assigned grade: B
Problem: Write a C++ program to find all prime numbers from 2 to n (n taken as input from the user) following the procedure given below:Create an array of consecutive integers from 2 through n: (2, 3, 4, ..., n).Initially, let p equal 2, the smallest prime number. Enumerate the multiples of p by counting in increments of p from 2p to n, and mark them in the array as non-prime (these will be 2p, 3p, 4p, ...; the p itself should not be marked). Find the smallest number in the array greater than p that is not marked. If there was no such number, stop. Otherwise, let p now be equal to this new number (which will be the next prime number), and repeat from step 3.When the algorithm terminates, the numbers remaining not marked in the array are all the primes below n. Your program should print out all prime numbers from 2 to n. You do not need to use functions in this task (you may if you wish to). 

Input Format - Integer n.
Output Format - Print out the prime numbers with a space in between. After every five prime numbers print a newline character for ease of reading. 

Constraints
n >= 2 and n <= 500

Student submission:
#include<iostream>
using namespace std;
int linearSearch (int *A, int n, int q){ //search for i
for (int i = 0; i < n; i++){
if (q == A[i]) return i;
}
return 0; 
}
int main(){
int n,j;
cin>>n;
int a[n-1];
a[0]=2;
for(int i=1; i<n-1; i++)
a[i]=a[i-1]+1; //array created
for(int p=2; p<n; p++){
for (int i=2; i*p<n+1; i++){
int k;
k=i*p;
j= linearSearch(a,n,k); //this a[j] needs to be deleted from array
a[j]=0; // dont know how to delete from array so transformiong to 0
}

}
for(int i=0; i<n-2; i++)
 cout<<a[i];
}

Grading criterion: Creation of array of consecutive integers
Rating A: Correctly Done
Rating B: Not done
Rating C: 0
Rating D: 0
Assigned grade: A
Problem: Write a C++ program to find all prime numbers from 2 to n (n taken as input from the user) following the procedure given below:Create an array of consecutive integers from 2 through n: (2, 3, 4, ..., n).Initially, let p equal 2, the smallest prime number. Enumerate the multiples of p by counting in increments of p from 2p to n, and mark them in the array as non-prime (these will be 2p, 3p, 4p, ...; the p itself should not be marked). Find the smallest number in the array greater than p that is not marked. If there was no such number, stop. Otherwise, let p now be equal to this new number (which will be the next prime number), and repeat from step 3.When the algorithm terminates, the numbers remaining not marked in the array are all the primes below n. Your program should print out all prime numbers from 2 to n. You do not need to use functions in this task (you may if you wish to). 

Input Format - Integer n.
Output Format - Print out the prime numbers with a space in between. After every five prime numbers print a newline character for ease of reading. 

Constraints
n >= 2 and n <= 500

Student submission:
#include<iostream>
using namespace std;
int main(){
int n;
cin>>n;
int A[n-1];
for(int i=0;i<n-1;i++){
	A[i]=i+2;	
	}
int p=A[0];
int q=0;
for(int j=0;j<n-1;j++){
	if(A[j]!=0){p=A[j];
	cout<<p<<" ";
	q++;
	}
	if (q==5){q=0;cout<<"\n";}
for(int i=j+1;i<n-1;i++){
	if(A[i]%p==0){
		A[i]=0;
		}
	}
	}
}

Grading criterion: Creation of array of consecutive integers
Rating A: Correctly Done
Rating B: Not done
Rating C: 0
Rating D: 0
Assigned grade: A
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
#include<iostream>
using namespace std;
struct Memo {

   static const int NMAX = 200;
   bool filled[NMAX] = {};
   int memo[NMAX];
   bool has(unsigned i) { return filled[i]; }
   int get(unsigned i) { return memo[i]; }
   void add(unsigned i, int val) {
       memo[i] = val;
       filled[i] = true;
   }
};
int cal1(int arr[] , int n,int i, int sum, int memo[], int k){
    if(i>=n){
        return sum;
    }
    if(memo[k]!=-1){
        return memo[k];
    }
    memo[k]=sum+arr[i];
    sum=sum+arr[i];
    return cal1(arr,n,i+2,sum,memo,k+1);
}
int cal2(int arr[] , int n,int i, int sum, int memo[], int k){
    if(i>=n){
        return sum;
    }
    if(memo[k]!=-1){
        return memo[k];
    }
    memo[k]=sum+arr[i];
    sum=sum+arr[i];
    return cal2(arr,n,i+2,sum,memo,k+1);
}

int main(){
    Memo m;
    for(int i=0;i<201;i++){
        m.memo[i]=-1;
        }
     int n;
     cin>>n;
     int arr[n];
     for(int i=0;i<n;i++){
        cin>>arr[i];
     }
     int sum1=cal1(arr,n,0,0,m.memo,0);
     for(int i=0;i<201;i++){
        m.memo[i]=-1;
     }
     int sum2=cal2(arr,n,1,0,m.memo,0);
     if(sum1>sum2){
        cout<<sum1<<endl;
     }else{
        cout<<sum2<<endl;
     }

 



    return 0;
}

Grading criterion: Declaration of variables and array
Rating A: correctly declares the variable n and array/vector

Rating B: Some issue. 

Rating C: Major issue. 
Rating D: 0
Assigned grade: A
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
#include<iostream>
using namespace std;

struct Memo {

   static const int NMAX = 200;
   bool filled[NMAX] = {};
   int memo[NMAX];
   bool has(unsigned i) { return filled[i]; }
   int get(unsigned i) { return memo[i]; }
   void add(unsigned i, int val) {
       memo[i] = val;
       filled[i] = true;
   }
};


int main (){
int N;
cin>>N;
cout << 11;
}

Grading criterion: Declaration of variables and array
Rating A: correctly declares the variable n and array/vector

Rating B: Some issue. 

Rating C: Major issue. 
Rating D: 0
Assigned grade: B
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
#include<iostream>
using namespace std;

int main()
{
    int N;
    cin>>N;
    int A[N];
    for(int i=0;i<N;i++)
    {
        cin>>A[i];
    }
    
    return 0;
}

Grading criterion: Declaration of variables and array
Rating A: correctly declares the variable n and array/vector

Rating B: Some issue. 

Rating C: Major issue. 
Rating D: 0
Assigned grade: A
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
#include <iostream>
using std::cin;
using std::cout;

struct Memo {

   static const int NMAX = 200;
   bool filled[NMAX] = {};
   int memo[NMAX];
   bool has(unsigned i) { return filled[i]; }
   int get(unsigned i) { return memo[i]; }
   void add(unsigned i, int val) {
       memo[i] = val;
       filled[i] = true;
   }
};

int Collect(int A[], int k, int n){
    if(k>=n) return 0;
    int total1=0, total2=0;
    for(int i=k+1;i<n;i++){
        total1 += A[i] + Collect(A, i+2, n);
        total2 += Collect(A,i+1,n);
    }
    if(total1>total2) return total1;
    else return total2;
}

int main(){
    int n, t;
    int A[100];
    cin >> n;
    for(int i=0;i<n;i++){
        cin >> t;
        A[i] = t;
    }
    cout << Collect(A,0,n);
}

Grading criterion: Declaration of variables and array
Rating A: correctly declares the variable n and array/vector

Rating B: Some issue. 

Rating C: Major issue. 
Rating D: 0
Assigned grade: A
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
#include<iostream>
using namespace std;
int total1=0 ;
int j=0 ;
int theft1(int j, int n, int a[]){
   if(j>=n) return total1;
   else return a[j]+theft1(j+2,n,a) ;
}
int total2=0 ;
int k=1 ;
int theft2(int k, int n, int a[]){
   if(k>=n) return total2;
   else return a[k]+theft2(k+2,n,a) ;
}
int main(){
    int n;
    cin >> n ;
    int a[n];
    for(int i=0;i<n;i++){
        
        cin >> a[i];
    }
    int c= theft1(j,n,a);
    int b= theft2(k,n,a);
    if(c>=b)
     cout << c;
     else cout << b;
}

Grading criterion: Declaration of variables and array
Rating A: correctly declares the variable n and array/vector

Rating B: Some issue. 

Rating C: Major issue. 
Rating D: 0
Assigned grade: A
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
#include<iostream>
using namespace std;
int Collect(int collect[],int n){
    
}
int main(){
    int n;
    cin>>n;
    int collect[n];
    for(int i=0;i<n;i++){
        cin>>collect[i];
    }
}

Grading criterion: Declaration of variables and array
Rating A: correctly declares the variable n and array/vector

Rating B: Some issue. 

Rating C: Major issue. 
Rating D: 0
Assigned grade: A
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
using namespace std;
#include<iostream>
 void swap(int &a,int &b)
 {
    int c;
    c=a;
    a=b;
    b=c;
 }
int main()
{
    int n,k;
    cin>>n>>k;
    swap(n,k);
    cout<<n<<k;
     
}

Grading criterion: Declaration of variables and array
Rating A: correctly declares the variable n and array/vector

Rating B: Some issue. 

Rating C: Major issue. 
Rating D: 0
Assigned grade: C
Problem: Title: House Robbers

Problem Description

A robber is planning to rob houses along a street. Each house has a certain amount of money. However, the robber cannot rob two adjacent houses because doing so will trigger the alarm system.

Your task is to determine the maximum amount of money the robber can steal without robbing two adjacent houses.

You must solve this problem using recursion and/or memoization.



Your Task

1. Read an integer n representing the number of houses.
2. Read n integers into an array A[], where A[i] represents the money in the i-th house.
3. Implement a recursive function to decide at each house whether to rob it or skip it.
4. Use a base condition to stop recursion.
5. Use memoization to optimize repeated calculations.
6. Return the maximum money that can be robbed and print it.



Rules

- If you rob a house, you cannot rob the next adjacent house.
- At each index, you must decide:
  - Rob the current house and move to i+2, OR
  - Skip the current house and move to i+1.
- The answer is the maximum value obtained from these choices.



Input Format

- An integer n.
- n integers representing the money in each house.



Output Format

- Print a single integer representing the maximum money that can be robbed.


Assumptions on Input

- 1 <= n <= 10^5  
- 0 <= A[i] <= 10^9

Student submission:
#include<simplecpp>
#include<vector>
main_program{
    int N; cin>>N;
    int A[N];
    for(int i=0; i<N; i++){
        cin>>A[i];
    }
    vector<int> E; vector<int> O; int sumE=0, sumO=0;
     for(int i=0; i<N){
        E.push_back(A[i]);
        i=i+2;
     }
     for(int i=1; i<N){
        O.push_back(A[i]);
        i=i+2;
     }
     for(int i=0; i<E.size(); i++)
      {sumE=sumE+E.(begin()+i);}
     for(int i=0; i<O.size(); i++)
      {sumO=sumO+O.(begin()+i);}
     if(sumE>sumO){cout<<sumE<<endl;}
      else{cout<<sumO<<endl;}
}

Grading criterion: Declaration of variables and array
Rating A: correctly declares the variable n and array/vector

Rating B: Some issue. 

Rating C: Major issue. 
Rating D: 0
Assigned grade: A