CodeGraderBench/CodeGraderBench huggingface dataset:
Problem: You are given a game consisting of n positions arranged in a line from index 0 to n-1. From each position i, you are given a number jumps[i] which represents how far you are allowed to jump forward from that position.

Your task is to compute how many valid ways exist to reach the last position starting from position 0, following the rules below.

Rules

1. You start from position 0.
2. From position i, you may move to any position j such that:
   j > i and j - i <= jumps[i].
3. You may only move forward.
4. A path is considered valid if you can reach the last position (n-1) using the allowed jumps.

The program uses a recursive function to count all such valid paths.



Input Format

- An integer n representing the number of positions.
- n integers representing the array jumps, where jumps[i] indicates the maximum jump length from position i.



Output Format

- Print the total number of valid ways to reach the last position starting from index 0.



Constraints

- 1 <= n <= 1000  
- 0 <= jumps[i] <= n  



Explanation

The recursive function works backward and checks whether a position can reach the end based on the jump limits. If a position can reach the target, it contributes to the total count of paths.

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

int num(int n, int arr[], int i=0){
    int count(int n, int st, int en, int j){
        if (n<0) return 0;
        if (n==1){
            return 1;
        }
        if (n==0 || n<0){
            return 0;
        }
        int sum = 0;
        for (int k=0;k<j; k++){
        
        }
    }
}

Grading criterion: Recursion call
Rating A: Correctly sets up the recursive call, with valid parameters and accumulates paths for all valid jumps

Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: C
Problem: You are given a game consisting of n positions arranged in a line from index 0 to n-1. From each position i, you are given a number jumps[i] which represents how far you are allowed to jump forward from that position.

Your task is to compute how many valid ways exist to reach the last position starting from position 0, following the rules below.

Rules

1. You start from position 0.
2. From position i, you may move to any position j such that:
   j > i and j - i <= jumps[i].
3. You may only move forward.
4. A path is considered valid if you can reach the last position (n-1) using the allowed jumps.

The program uses a recursive function to count all such valid paths.



Input Format

- An integer n representing the number of positions.
- n integers representing the array jumps, where jumps[i] indicates the maximum jump length from position i.



Output Format

- Print the total number of valid ways to reach the last position starting from index 0.



Constraints

- 1 <= n <= 1000  
- 0 <= jumps[i] <= n  



Explanation

The recursive function works backward and checks whether a position can reach the end based on the jump limits. If a position can reach the target, it contributes to the total count of paths.

Student submission:
#include<iostream>
#include<vector>
using namespace std;
struct Memo{
    bool filled[50]={};
    unsigned long long int A[50]={};
    bool has(int i){
        if(filled [i]){return true;}
        else {return false;}

    }
    unsigned long long int get(int i){
        if (filled[i]){return A[i];}
        else return 0;
    }
    void add(int i,unsigned long long int n){
        if(filled[i]){return;}
        else{filled[i]=true;A[i]=n;}
    } 
};
unsigned long long int find(int n,const vector<int> jumps,Memo& memo){
if(n==0){return 1;}
if(memo.has(n)){return memo.get(n);}
else{
    unsigned long long int sum=0;
    for(int i=0;i<n;i++){
        if(jumps[i]>=(n-i)){
            sum+=find(i,jumps,memo);
        }
    }
    memo.add(n,sum);
    return sum;
}
    }


int main(){
    int n;cin>>n;vector<int> jumps;
    for(int i=0;i<n;i++){
        int x;cin>>x;cin.ignore();jumps.push_back(x);
    }
     Memo memo;
    cout<<find(n-1,jumps,memo)<<endl; 
}

Grading criterion: Recursion call
Rating A: Correctly sets up the recursive call, with valid parameters and accumulates paths for all valid jumps

Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: B
Problem: You are given a game consisting of n positions arranged in a line from index 0 to n-1. From each position i, you are given a number jumps[i] which represents how far you are allowed to jump forward from that position.

Your task is to compute how many valid ways exist to reach the last position starting from position 0, following the rules below.

Rules

1. You start from position 0.
2. From position i, you may move to any position j such that:
   j > i and j - i <= jumps[i].
3. You may only move forward.
4. A path is considered valid if you can reach the last position (n-1) using the allowed jumps.

The program uses a recursive function to count all such valid paths.



Input Format

- An integer n representing the number of positions.
- n integers representing the array jumps, where jumps[i] indicates the maximum jump length from position i.



Output Format

- Print the total number of valid ways to reach the last position starting from index 0.



Constraints

- 1 <= n <= 1000  
- 0 <= jumps[i] <= n  



Explanation

The recursive function works backward and checks whether a position can reach the end based on the jump limits. If a position can reach the target, it contributes to the total count of paths.

Student submission:
#include<iostream>
using namespace std;
int tnodp(int a){
    
}
struct Memo {
   static const int NMAX = 100;
   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;
    int jumps[n];
    for(int i=0;i<n;i++){
    cin>>jumps[i];
    cout<<tnodp(n)<<endl;
    cout<<jumps[n-1];
}
}

Grading criterion: Recursion call
Rating A: Correctly sets up the recursive call, with valid parameters and accumulates paths for all valid jumps

Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: C
Problem: You are given a game consisting of n positions arranged in a line from index 0 to n-1. From each position i, you are given a number jumps[i] which represents how far you are allowed to jump forward from that position.

Your task is to compute how many valid ways exist to reach the last position starting from position 0, following the rules below.

Rules

1. You start from position 0.
2. From position i, you may move to any position j such that:
   j > i and j - i <= jumps[i].
3. You may only move forward.
4. A path is considered valid if you can reach the last position (n-1) using the allowed jumps.

The program uses a recursive function to count all such valid paths.



Input Format

- An integer n representing the number of positions.
- n integers representing the array jumps, where jumps[i] indicates the maximum jump length from position i.



Output Format

- Print the total number of valid ways to reach the last position starting from index 0.



Constraints

- 1 <= n <= 1000  
- 0 <= jumps[i] <= n  



Explanation

The recursive function works backward and checks whether a position can reach the end based on the jump limits. If a position can reach the target, it contributes to the total count of paths.

Student submission:
#include<iostream>
#include<iostream>
#include<vector>

int now (int n;vector<int>jump;intp){
    int step
    if (p=n-1){
        return 1 ;
    }
    if (p=n-2){
        return 2;
    }
    step = now(n,jump,p+1);
    return step;
}
   

int main(){
    int n;
    cin>>n;
    vector<int> jumps(n);
    cin>>n;
    for(i=0,i<n,i++){
        cin>> jumps[i];
    }
    cout<<now(n,jump 0);
    
}

Grading criterion: Recursion call
Rating A: Correctly sets up the recursive call, with valid parameters and accumulates paths for all valid jumps

Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: B
Problem: You are given a game consisting of n positions arranged in a line from index 0 to n-1. From each position i, you are given a number jumps[i] which represents how far you are allowed to jump forward from that position.

Your task is to compute how many valid ways exist to reach the last position starting from position 0, following the rules below.

Rules

1. You start from position 0.
2. From position i, you may move to any position j such that:
   j > i and j - i <= jumps[i].
3. You may only move forward.
4. A path is considered valid if you can reach the last position (n-1) using the allowed jumps.

The program uses a recursive function to count all such valid paths.



Input Format

- An integer n representing the number of positions.
- n integers representing the array jumps, where jumps[i] indicates the maximum jump length from position i.



Output Format

- Print the total number of valid ways to reach the last position starting from index 0.



Constraints

- 1 <= n <= 1000  
- 0 <= jumps[i] <= n  



Explanation

The recursive function works backward and checks whether a position can reach the end based on the jump limits. If a position can reach the target, it contributes to the total count of paths.

Student submission:
#include<iostream>
using namespace std;
struct Memo {
   static const int NMAX = 100;
   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 rec(int a,int ar[],int copy,int i=0){
    if(a==0)
    return -1;
    int num=ar[i++];
    for(int i=1;i<=num;i++){
        if(i+a>=copy+2)
        return 0;
        return 1+rec(a-i,ar,copy,i);
    }
}
int main(){
    Memo obj;
    int n;
    cin>>n;
    int copy=n;
    int jumps[n];
    for(int i=0;i<n;i++){
        cin>>jumps[i];
    }
    cout<<rec(n,jumps,copy);
}

Grading criterion: Recursion call
Rating A: Correctly sets up the recursive call, with valid parameters and accumulates paths for all valid jumps

Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: C
Problem: You are given a game consisting of n positions arranged in a line from index 0 to n-1. From each position i, you are given a number jumps[i] which represents how far you are allowed to jump forward from that position.

Your task is to compute how many valid ways exist to reach the last position starting from position 0, following the rules below.

Rules

1. You start from position 0.
2. From position i, you may move to any position j such that:
   j > i and j - i <= jumps[i].
3. You may only move forward.
4. A path is considered valid if you can reach the last position (n-1) using the allowed jumps.

The program uses a recursive function to count all such valid paths.



Input Format

- An integer n representing the number of positions.
- n integers representing the array jumps, where jumps[i] indicates the maximum jump length from position i.



Output Format

- Print the total number of valid ways to reach the last position starting from index 0.



Constraints

- 1 <= n <= 1000  
- 0 <= jumps[i] <= n  



Explanation

The recursive function works backward and checks whether a position can reach the end based on the jump limits. If a position can reach the target, it contributes to the total count of paths.

Student submission:
#include<iostream>
using namespace std;
struct Memo {
   static const int NMAX = 100;
   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 dispath(int jumps[],int n,int ways,int i){
    if(i==n){
        return 0;
    }
    if(i<n){
        for(int k=(i+1);k<(i+jumps[i]);k++){
        return dispath(jumps,n,ways+1,k);
        }
    }
    return ways+2;
}
int main(){
    int n;
    cin>>n;
    int jumps[n];
    for(int i=0;i<n;i++){
        cin>>jumps[i];
    }
    cout<<dispath(jumps,n,0,0);
}

Grading criterion: Recursion call
Rating A: Correctly sets up the recursive call, with valid parameters and accumulates paths for all valid jumps

Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: B
Problem: You are given a game consisting of n positions arranged in a line from index 0 to n-1. From each position i, you are given a number jumps[i] which represents how far you are allowed to jump forward from that position.

Your task is to compute how many valid ways exist to reach the last position starting from position 0, following the rules below.

Rules

1. You start from position 0.
2. From position i, you may move to any position j such that:
   j > i and j - i <= jumps[i].
3. You may only move forward.
4. A path is considered valid if you can reach the last position (n-1) using the allowed jumps.

The program uses a recursive function to count all such valid paths.



Input Format

- An integer n representing the number of positions.
- n integers representing the array jumps, where jumps[i] indicates the maximum jump length from position i.



Output Format

- Print the total number of valid ways to reach the last position starting from index 0.



Constraints

- 1 <= n <= 1000  
- 0 <= jumps[i] <= n  



Explanation

The recursive function works backward and checks whether a position can reach the end based on the jump limits. If a position can reach the target, it contributes to the total count of paths.

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

int main(){    
    int n;
    cin>>n;
    int jump[n];
    for(int i=0; i<n; i++){
        cin>>jump[i];
    }
    int count=0;
    for(int i=0; i<n; i++){
        for(int j=i; j<n; j++){
            if(i+jump[i]<(n-1-i)){
                count++;
            }
        }
    }
    cout<<count;
}

Grading criterion: Recursion call
Rating A: Correctly sets up the recursive call, with valid parameters and accumulates paths for all valid jumps

Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: C
Problem: You are given a game consisting of n positions arranged in a line from index 0 to n-1. From each position i, you are given a number jumps[i] which represents how far you are allowed to jump forward from that position.

Your task is to compute how many valid ways exist to reach the last position starting from position 0, following the rules below.

Rules

1. You start from position 0.
2. From position i, you may move to any position j such that:
   j > i and j - i <= jumps[i].
3. You may only move forward.
4. A path is considered valid if you can reach the last position (n-1) using the allowed jumps.

The program uses a recursive function to count all such valid paths.



Input Format

- An integer n representing the number of positions.
- n integers representing the array jumps, where jumps[i] indicates the maximum jump length from position i.



Output Format

- Print the total number of valid ways to reach the last position starting from index 0.



Constraints

- 1 <= n <= 1000  
- 0 <= jumps[i] <= n  



Explanation

The recursive function works backward and checks whether a position can reach the end based on the jump limits. If a position can reach the target, it contributes to the total count of paths.

Student submission:
#include<iostream>
using namespace std;
int ways=0;
int calcways(int n, int jumps[]);
int n;
int main(){
    
    cin>>n;
    int jumps[n];
    cin>>jumps[n];
    for(int i=0;i<n;i++){
        int inp;
        cin>>inp;
        jumps[i]=inp;
    }
    
    cout<<calcways(n, jumps)-1;
  
    

    
}

int calcways(int m, int jumps[]){
    if (m==0){
        return ways;
    }
    for(int i=0;i<m;i++){
        if(m-1-(jumps[i])>=0){
            ways++;
            ways=ways+calcways(i+1, jumps);

        }
    }
    
        
    

}

Grading criterion: Recursion call
Rating A: Correctly sets up the recursive call, with valid parameters and accumulates paths for all valid jumps

Rating B: Some issue
Rating C: Major issue
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:
/*
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). 
*/

#include <iostream>
using namespace std;

int main(){
int n, p;

cin>>n;
p=2;

int a1[n-1];
bool a2[n-1];

for(int i=0; i<n-1; i++){
	a1[i]=i+2;
	a2[i]=true;
}
int k=p;
while(n-p*p>=0){
	k=p;
	for(int i=2; n-p*i>=0; i++){
		a2[p*i-2]=false;
	}
	for(int i=n-2; i+2>k; i--){
		if(a2[i]) p=a1[i];
	}
}

int j=1;
cout<<2;

for(int i=1; i<n-1; i++){
	if(j%5==0 && a2[i]){
		cout<<'\n'<<a1[i];
		j++;
	}
	else if(a2[i]){
		cout<<" "<<a1[i];
		j++;
	}
}
}

Grading criterion: Comments
Rating A: Well Commented
Rating B: Ok Commented
Rating C: Poor or No Comments
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 prime(int a){
	int k;
int n;
int j;
int array[n];
for(int i=2; i<k;i++){
   for( int j=1; i<k;i++){
	   array[j]=j*i;}
}	
for(int i=2; i<a;i++){
	if(a%i==0)
	{return 0;}}	
return 1;}
int main(){
int x;
	int k=0;
cin>>x;
	for(int i=2 ;i<=x;i++){
if (prime(i)==1){cout<<i<<" ";
		 k=k+1;
	if(k%5==0){
		cout<<endl;}
		}}
}

Grading criterion: Comments
Rating A: Well Commented
Rating B: Ok Commented
Rating C: Poor or No Comments
Rating D: 0
Assigned grade: C
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 arr[n-1];
	for(int i=0;i<=n-2;i++){
		arr[i]=i+2;
	}
	
	for(int i=0;i<=n-2;i++){
		int k=1;
		for(int j=2;j<arr[i];j++){
			k=0;
		    if(arr[i]%j==0){
				break;
			}
			else{
				k=1;
			}
		}
		if(k==1){
			cout<<arr[i]<<" ";
		}
          if(i%9==0 && i!=0 && i!=18 && i!= 36 && i!=54){
			cout<<endl;
		 }
	}
	
	return 0;
}

Grading criterion: Comments
Rating A: Well Commented
Rating B: Ok Commented
Rating C: Poor or No Comments
Rating D: 0
Assigned grade: C
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 arr[n-1];
for(int i=0; i<n-1; i++){
arr[i]=i+2;
}
for(int i=2; i<=n; i++){
for(int k=2; k<i; k++){
if(arr[i-2]%k!=0){
continue;

}

else{

}
}
}
cout<<"2 3 5 7 11 13 17 19 23 19 31 37 41 43 47 53 ";
return 0;
}

Grading criterion: Comments
Rating A: Well Commented
Rating B: Ok Commented
Rating C: Poor or No Comments
Rating D: 0
Assigned grade: C