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;

void getJump(int jump[], int n){

}

int main(){
    int jump[100],n;
    cin>>n;
    getJump(jump,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 <simplecpp>
main_program{
    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 memo(unsigned i, Memo& memo) {
if memo.has(i) return memo.get(i);
int ans = ( 1 <= j <= jumps[i])? n : memo(i + j <= n,memo);
memo.add(n,ans);
return ans;
}

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>
using namespace std;

int totalPaths(int arr[], int n , int i){
int maxLength=arr[i];

   if(i==n-2) return 0;
   int numberOfWays=((arr[i])*(arr[i]-1))/2;

   int j=n-arr[i]-1;
   return j+totalPaths(arr,n,j);
}


   


int main(){
    int n;
    cin>>n;
    int jumps[n];
    for(int i=0; i<n; i++){
        cin>>jumps[i];
    }
    
cout<<totalPaths(jumps,n,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: 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 min(int a,int b)
{
    return (a>b)?b: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 sum=0;

void rec(int i,int jumps[], const int&n)
{
    if(i==(n-1))
    {
        ++sum;
        return;
    }
    for(int k=i+1;k<=min(n-1,i+jumps[i]);k++)
    {
        rec(k, jumps, n);
    }
}

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

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 countWays(int jumps[]){
    if()
}
int main(){
    int n;
    cin>>n;
    int jumps[n];
    for(int i=0;i<n;i++){
        cin>>jumps[i];
    }
    int count=countWays(jumps)
}

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 {
   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 way(int n,int i,int jumps[],Memo& memo){
    if(memo.has(n)){
        return memo.get(n);
    }
    int ans=0;
    if(n-i==1){
        return 1;
    }
    else if(i>n-2){
        return 0;
    }
    else{
        int j=i+1;
        int f= i+jumps[i];
        while(j<=f){
            ans+= way(n,j,jumps,memo);
            j+=1;
        }
        memo.add(n,ans);
    }
    return ans;
}
int main(){
    int n;
    cin>>n;
    int arr[n];
    Memo memo;
    for(int i=0;i<n;i++){
        cin>>arr[i];
    }
    cout<< way(n,0,arr,memo);
}

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>
using namespace std;

struct Memo {
   static const int NMAX = 100;
   bool filled[NMAX] = {};
   unsigned long long int memo[NMAX];
   bool has(unsigned i) { return filled[i]; }
   int get(unsigned i) { return memo[i]; }
   void add(unsigned i, unsigned long long int val) {
       memo[i] = val;
       filled[i] = true;
   }
};

unsigned long long int f(int n, Memo& memo, int arr[]){
    if (memo.has(n)) return memo.get(n);
    unsigned long long int ans =0;
    int k=min(8,n);
    for (int i=1;i<=k;i++){
        if (arr[n-i]>=i) {ans += f(n-i,memo,arr);}
    }
    memo.add(n,ans);
    return ans;
}

int main(){
    Memo memo;
    memo.add(0,1);
    memo.add(1,1);
    int n; cin >> n;
    int jumps[n];
    for (int i=0; i<n;i++) cin >> jumps[i];
    cout <<f(n,memo,jumps);
}

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 namespace std;

int main () {
    int n;
    cin >> n;
    int jumps [n];
    for (int a=0; a<i; a++){
        cin>> jumps [a];
    }
}

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 ways(int n ; int arr[n]; int i=0)
{
    int j=0;
    int sum=0;
    while (sum<=n)
    {
        if (i+j<n)
        {
            sum=arr[i]+arr[i+j];
            j++;
        };
        if (sum==n)
        a=a+1;
    };
    return a + ways(int n ; int arr[n]; i++)
};
int main()
{
    int n;
    cin>>n;
    int a[n];
    for(int i=0;i<n;i++)
    cin>>a[i];
    int no_of_ways = ways(n ; a[n]; int i=0);
    cout<<no_of_ways;
    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 std::cout; using std::cin; using std::endl;
int main() {
char c, vowels[] = {'a', 'e', 'i', 'o', 'u'}; // initialising explicitly
int N[5] = {}; // initialising all elements to 0
for(cin >> c; c >= 'a' && c <= 'z'; cin >> c)
for (int i=0; i<5; i++)
if(c == vowels[i]) {
N[i]++; break;
}
for (int i=0; i<5; i++)
cout << "#" << vowels[i] << " = " << N[i] << (i==4?"":", ");
cout << endl;
}

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;
 void path(int jumps[],int i,int target,int& sum){
    if(i==target){
        ++sum;
        return ;
    }
    if(i>target){
        return ;
        
    }
    int possible=jumps[i];
    for(int j=1;j<=possible && i+j<=target+1;j++){

     path(jumps,i+j,target,sum);
    }
    
}
int main(){
    int n;
    cin>>n;
    int jumps[n],sum=0;
    for(int i=0;i<n;i++){
        cin>>jumps[n];
    }
    path(jumps,0,n-1,sum);
    cout<<sum<<endl;
}

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