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 ways(int a[],int n){
    for(int i=0;i<n;i++){
        for(int j=0;j<a[i];j++){

        }
        
    }
 
    return w;
}
int main(){
    int n,a[n];
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>a[i];
    }
    int Totalways=ways(a,n);
    cout<<Totalways;

    return 0;
}

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)
Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls
Rating B: Some issue with memoization

Rating C: Memoization not used

Rating D: Major issiue
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(int x[],int n)
{
    unsigned long int way = 0;
    int first = x[0];
    if(n == 1) return 1;
    bool all = true;
    for(int i=0;i<n-1;i++)
    {
        if(x[i]!=1) all = false;
    }
    if (all) return 1;
    for (int i=1;i<=first;i++)
    {
        int y[n-i];
        for(int k=0;k<n-i;k++)
        {
            y[n-i-k-1] = x[n-k-1];
        }
        way += ways(y,n-i);
    }
    return way;

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

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)
Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls
Rating B: Some issue with memoization

Rating C: Memoization not used

Rating D: Major issiue
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 numberPaths(int arr[],int len,int n, Memo &memo)
{
    if (len==1)
    return 1;
    else if (memo.has(len)) 
    return memo.get(len);
    int k=0;
    for (int j=1;j<=arr[len-1];j++)
    {
        if ((j+len)<n)
        k++;
    }
    memo.add(len,(k+ numberPaths(arr,len-1,n,memo)));
    return (k+ numberPaths(arr,len-1,n,memo));
    
   }

int main()
{
int n,paths;Memo m;
cin>>n;
int jumps[n];
for(int i=0;i<n;i++)
{
    cin>>jumps[i];
}
cout<<numberPaths(jumps,n,n,m);
}

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)
Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls
Rating B: Some issue with memoization

Rating C: Memoization not used

Rating D: Major issiue
Assigned grade: A
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 std::cin;
using std::cout;
using std::endl;

int main()
{
    int n;
    cin>>n;
    int jumps[n];
    if(n==4)
    {
        for(int i=0;i<4;i++)
        {
            cin>>jumps[i];
        }
            
            cout<<"2";
        
    }
    if(n==5)
    {
        for(int i=0;i<5;i++)
        {
            cin>>jumps[i];
        }
        
        
            cout<<"4";
        
    }
    if(n==6)
    {
        for(int i=0;i<6;i++)
        {
            cin>>jumps[i];
        }
       
            cout<<"1";
        
    }
    if(n==40)
    {
        for(int i=0;i<40;i++)
        {
            cin>>jumps[i];
        }
       
            cout<<"277663958";
        
    }
    
}

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)
Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls
Rating B: Some issue with memoization

Rating C: Memoization not used

Rating D: Major issiue
Assigned grade: D
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<<simplecpp>>
using std;{
int=(i<n;i+j
 {int= n>>0;
assume that;
  input format;
  int=2<=n<=50)
  no of distinct;
 p1<<Pk=n;
 k, 1 pt+1-pt}
 }
 struct memo {
    const int nmax=100;
   int memo nmax;
   then memo i value is nmax;
   memo i=nmax;}

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)
Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls
Rating B: Some issue with memoization

Rating C: Memoization not used

Rating D: Major issiue
Assigned grade: D
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 distinctPaths(int A[], int x, int b=0){
        int i = b;
        int j = 1;
        int s = 0;
        for(;j<=A[i]&&(i+j)<=x-1;j++){
                if((i+j)==x-1){ s++;continue;}
                s = s+ distinctPaths(A,x,i+j);
                };
        return s;
}

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

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)
Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls
Rating B: Some issue with memoization

Rating C: Memoization not used

Rating D: Major issiue
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>
int main(){
    int n;
    std::cin<<n;
    int jumps[n]
}

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)
Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls
Rating B: Some issue with memoization

Rating C: Memoization not used

Rating D: Major issiue
Assigned grade: D
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 {
    int l;
    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 memoL(unsigned n, Memo& memo) {
    if (n<=1) return n;
if (memo.has(n)) return memo.get(n);
int ans=0;
for (int i=n-1;i>=0;i--)
ans=ans+ memoL(i,memo);

memo.add(n,ans);
return ans;
}

 int main()
{ 
 int m;
 cin>>m;
 Memo meme;
 cout<<((memoL(m,meme))/2)  ; 
 }

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)
Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls
Rating B: Some issue with memoization

Rating C: Memoization not used

Rating D: Major issiue
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>
#include<vector>


using namespace std;
int main(){
    int n;
    cin>>n;
int m;
    cin>>m;


int p;
    cin>>p;
    cout<<""<<n<<endl;

    int jumps[n];


    for(int i=0;i<n;i++)
    for(int j=0;j>=1 && j<=jumps[i] ;j++){
      m=i+j;
      p= m<=n;


        cout<<""<< p <<endl;

        return 0;
        
    }
    







    return 0;
}

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)
Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls
Rating B: Some issue with memoization

Rating C: Memoization not used

Rating D: Major issiue
Assigned grade: D
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<simplecpp>;
int main();
{
    void;
        1<=j<=i;
        i+j<=n;
        i;i+1;i++;
        print("the total number of distinct paths");{

        0 = p1 << pk=n;
        t<k;1 <= pt+1-pt<= pt;
        printf("a single integer n is the length of an array");

        static const int NMAX = 100;
        bool filled[NMAX] = {};
        int memo[NMAX];
        bool has (unsinged i);
        int get (unsigned i);
        void add (unsigned i, int val);{
            memo[i] = val;
            filled[i] = true;
        }
        2<= n <=50;
        0 < i < 8;
        
    }

return 0;
}

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)
Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls
Rating B: Some issue with memoization

Rating C: Memoization not used

Rating D: Major issiue
Assigned grade: D
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 totalnumberofdistinctpath(int n, int jump[])
{
    if(n == 1) return 0;
    if(n==2) return 1;


    return (totalnumberofdistinctpath(n-1,jump) + totalnumberofdistinctpath(n-2,jump));
}

int main(){
    int n,  jump[100];
    cin >>n;
    for(int i = 0; i< n; i++)
    {
        cin >> jump[i];
    }
    int ans = totalnumberofdistinctpath(n,jump);
    cout << ans << endl;
    
    return 0;
}

Grading criterion: Memoization: The code may vary (used the code provided by the instructor or completely implemented by students)
Rating A: Correctly uses memoization by storing and retrieving results for each index and avoiding redundant recursive calls
Rating B: Some issue with memoization

Rating C: Memoization not used

Rating D: Major issiue
Assigned grade: D