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;

struct Memo {
    const static int N = 50;
    long int numForPos[N];
    bool isCalc[N] = {};

    bool has(int i) {
        return isCalc[i];
    }
    int get(int i) {
        return numForPos[i];
    }
    void add(int i, long int x) {
        numForPos[i] = x;
        isCalc[i] = true;
    }
};

long int numOfWays(const int n, const int jumps[], Memo& memo, int pos = 0) {
    long int num = 0;
    if(pos == n-1) return 1;
    if(pos > n-1) return 0;

    for(int i = 1; i <= jumps[pos]; i++) {
        if(!memo.has(pos+i)) memo.add(pos+i, numOfWays(n, jumps, memo, pos+i));
        num += memo.get(pos+i);
    }

    return num;
}

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

    Memo memo;
    cout << numOfWays(n, jumps, 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: 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 f(int arr[],int n,int k,int y){
    int d=0;
    if(n<=0)return 0;
    if(n==1)return 1;
    
for(int i=y;i<k;i++){
    
    for(int j=1;j<=arr[i];j++){
        
d=d+f(arr,n-j,k,j);
    
}}
return d;
}
int main(){
int n;
cin>>n;
int arr[n];
for(int i=0;i<n;i++){
    cin>>arr[n];
}
cout<<f(arr,n,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;
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 c=0;
void f(int arr[],int n,int pos)
{
    if(pos==(n-1))
    {
    c++;
    return;
    }
    if(pos>=n)
    return;
    for(int j=1;j<=arr[pos];j++)
    {
        f(arr,n,pos+j);
    }
}
int main(){
    int n;
    cin>>n;
    int arr[n];
    f(arr,n,0);
    cout<<c;
}

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 nways(int jumps[],int a,int n){
    if (a>=n-1){
        return 0;
    }
    int m=0;
    for(int i=0;i<=jumps[a];++i){
        m+=nways(jumps,a+jumps[i],n);
    }
    return m;
}
int main(){
    int n;
    cin>>n;
    int jumps[n];
    for (int i=0;i<n;++i){
        cin>>jumps[i];
    }
    cout<<nways(jumps,0,n)<<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: 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 func(int jumps[],int index, int ini,int fin){

if(ini==fin) return 1;
if(ini>fin) return 0;
int tot=0;
for(int i=1; i<=jumps[index];i++){
tot+= func(jumps,index+i,ini+i,fin);
}
//tot=sec;
return tot;
}

int main(){
int n;cin>>n;
int jumps[n];
for(int i=0;i<n;i++){
cin>>jumps[i];
}
int deb= func(jumps,0,0,n-1);
cout << deb<<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: 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 steps(int i, int n, int jumps[]){
    if(i==n) return 1;
    if(i>n) return 0;

    int c=0;

    for(int j=1; j<=jumps[i]; j++){
        c+= steps(i+j,n, jumps);
    }
    return c;
}
int main(){
    int n;
    cin>>n;
    int jumps[n];
    for(int i=0; i<n; i++){
        cin>>jumps[i];
    }
    cout<<steps(0,n, 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: 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 totalnumberofdistinctpaths(int n,int jumps[]){
    //Base case

if (n==2) return 1;
int ans2;
for (int i=0;i<n;i++){
    int ans2 = totalnumberofdistinctpaths(n-jumps[i],jumps);
}
    return (n-1)+ ans2;
}
int main(){
    int n;
    cin >> n;int jumps[100];
    for (int i=0;i<n;i++){
        cin >> jumps[i];
    }
    int ans=totalnumberofdistinctpaths(n,jumps);
    cout << ans;
    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: 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;
struct totalpath {
   static const int j = jumps[i];
   bool jumps[j] = {};
   int totalpath[j];
   bool has(unsigned i) { return jumps[i]; }
   int get(unsigned i) { return totalpath[i]; }
   void add(unsigned i, int val) {
       totalpath[i] = val;
       jumps[i] = true;
   }
};

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

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

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 path(int j[],int n,int d,int w=0,int idx=0){
    if(d==0) return (w+1);
    for(int i=idx;i<n;i++)
    { for(int p=i+1;p<=i+j[i];p++)
        { if((idx+p)<=(n-1))
            {int s=d-p;
            int id=idx+p;
            w=path(j,n,s,w,id);
        }
    }

}
return w;
}
int main(){
    int n;
    cin>>n;
    int jump[10];
    for(int i=0;i<n;i++)
    {
        cin>>jump[i];
    }
    int ways=path(jump,n,n-1);
    cout<<ways;
}

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>
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 func(int k,int a[]){
    int b=0;
    int x=;
    return for(int x;<a[x];i++){

        b=b+func(int k-i;)
    }

}
int main(){

    
}

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