CodeGraderBench/CodeGraderBench huggingface dataset:
Problem: An iterative method, called the Babylonian method, for finding square roots, was known to the ancient Babylonians (1500 BC) and Greeks (100 AD) long before Newton invented his general procedure. Here's how it works. Suppose you are given any positive number S. To find the square root of S, do the following: Make an initial guess. Guess any positive number x0.Improve the guess. Apply the formula x1 = (x0 + S / x0) / 2. The number x1 is a better approximation to sqrt(S).Iterate until convergence. Apply the formula xn+1 = (xn + S / xn) / 2 until the process converges. Convergence is achieved when the absolute difference between xn+1  and xn falls below some user specified epsilon.Your job is to write a C++ program containing a function that determines the approximate square root of the number S (taken as input) up to some desired precision epsilon. To get credit, you must write a function.

Input Format - S should be greater than 0. Epsilon should be small in value. 
Output Format - Print the value of the square root up to 5 decimal places.  

Constraints
S > 0, 
epsilon <= 0.01.

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

double f(double S,double &x,double eps){
   double y;
   do{
       y=x;
       x=(x+S/x)/2;}
   while(abs(y-x)>eps);
   return x;}

int main(){
   double S,eps,x=5.000000;
   cin>>S>>eps;
  cout<<fixed;
  cout.precision(5);
   cout<<f(S,x,eps);
}

Grading criterion: Quality and number of comments
Rating A: Well Commented
Rating B: Ok Commented
Rating C: Poor or no comments
Rating D: 0
Assigned grade: C
Problem: An iterative method, called the Babylonian method, for finding square roots, was known to the ancient Babylonians (1500 BC) and Greeks (100 AD) long before Newton invented his general procedure. Here's how it works. Suppose you are given any positive number S. To find the square root of S, do the following: Make an initial guess. Guess any positive number x0.Improve the guess. Apply the formula x1 = (x0 + S / x0) / 2. The number x1 is a better approximation to sqrt(S).Iterate until convergence. Apply the formula xn+1 = (xn + S / xn) / 2 until the process converges. Convergence is achieved when the absolute difference between xn+1  and xn falls below some user specified epsilon.Your job is to write a C++ program containing a function that determines the approximate square root of the number S (taken as input) up to some desired precision epsilon. To get credit, you must write a function.

Input Format - S should be greater than 0. Epsilon should be small in value. 
Output Format - Print the value of the square root up to 5 decimal places.  

Constraints
S > 0, 
epsilon <= 0.01.

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



void sqrt(double s,double eps){
	cout<<fixed;
	cout.precision(5);

	double x1=s-1,x2=s;

	while(abs(x2-x1)>eps){
		x2=(x1+s/x1)/2;
		double temp=x1;
		x1=x2;
		x2=temp;
		
		

	}
	cout<<x1;
}

int main(){

double s,eps;
cin>>s>>eps;
sqrt(s,eps);

}

Grading criterion: Quality and number of comments
Rating A: Well Commented
Rating B: Ok Commented
Rating C: Poor or no comments
Rating D: 0
Assigned grade: C
Problem: An iterative method, called the Babylonian method, for finding square roots, was known to the ancient Babylonians (1500 BC) and Greeks (100 AD) long before Newton invented his general procedure. Here's how it works. Suppose you are given any positive number S. To find the square root of S, do the following: Make an initial guess. Guess any positive number x0.Improve the guess. Apply the formula x1 = (x0 + S / x0) / 2. The number x1 is a better approximation to sqrt(S).Iterate until convergence. Apply the formula xn+1 = (xn + S / xn) / 2 until the process converges. Convergence is achieved when the absolute difference between xn+1  and xn falls below some user specified epsilon.Your job is to write a C++ program containing a function that determines the approximate square root of the number S (taken as input) up to some desired precision epsilon. To get credit, you must write a function.

Input Format - S should be greater than 0. Epsilon should be small in value. 
Output Format - Print the value of the square root up to 5 decimal places.  

Constraints
S > 0, 
epsilon <= 0.01.

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

float square_root(int S, float eps, float x)
{

	float x1=(x + (S/x))/2;
	if((fabs(x1-x))>eps) return square_root(S,eps,x1);
	else return x1;

}

int main()
{
	int S;
	float eps;
	float x0=10.0; //a random positive number guess
	cin >> S >> eps;
	if((S>0) && (eps<=0.01))
	{
		cout << fixed;
		cout.precision(5);
		cout << square_root(S,eps,x0);
	}
}

Grading criterion: Quality and number of comments
Rating A: Well Commented
Rating B: Ok Commented
Rating C: Poor or no comments
Rating D: 0
Assigned grade: C
Problem: An iterative method, called the Babylonian method, for finding square roots, was known to the ancient Babylonians (1500 BC) and Greeks (100 AD) long before Newton invented his general procedure. Here's how it works. Suppose you are given any positive number S. To find the square root of S, do the following: Make an initial guess. Guess any positive number x0.Improve the guess. Apply the formula x1 = (x0 + S / x0) / 2. The number x1 is a better approximation to sqrt(S).Iterate until convergence. Apply the formula xn+1 = (xn + S / xn) / 2 until the process converges. Convergence is achieved when the absolute difference between xn+1  and xn falls below some user specified epsilon.Your job is to write a C++ program containing a function that determines the approximate square root of the number S (taken as input) up to some desired precision epsilon. To get credit, you must write a function.

Input Format - S should be greater than 0. Epsilon should be small in value. 
Output Format - Print the value of the square root up to 5 decimal places.  

Constraints
S > 0, 
epsilon <= 0.01.

Student submission:
#include<iostream>
using namespace std;
float rec_sqrt(float x,int n)
{
		if (n == 1) return 1;

		if (n == 0) return x;
		float temp = rec_sqrt(x,n);
		if ((n+1) == 0) return temp*temp;
		else return temp*temp*n;
		}
		int main()
		{
		float x,S; int n;
		cout<<"enter x and n:"; cin>>x>>n;
		S=rec_sqrt(x,n);
		cout<<"S= " <<S;
		}

Grading criterion: Quality and number of comments
Rating A: Well Commented
Rating B: Ok Commented
Rating C: Poor or no comments
Rating D: 0
Assigned grade: C
Problem: An iterative method, called the Babylonian method, for finding square roots, was known to the ancient Babylonians (1500 BC) and Greeks (100 AD) long before Newton invented his general procedure. Here's how it works. Suppose you are given any positive number S. To find the square root of S, do the following: Make an initial guess. Guess any positive number x0.Improve the guess. Apply the formula x1 = (x0 + S / x0) / 2. The number x1 is a better approximation to sqrt(S).Iterate until convergence. Apply the formula xn+1 = (xn + S / xn) / 2 until the process converges. Convergence is achieved when the absolute difference between xn+1  and xn falls below some user specified epsilon.Your job is to write a C++ program containing a function that determines the approximate square root of the number S (taken as input) up to some desired precision epsilon. To get credit, you must write a function.

Input Format - S should be greater than 0. Epsilon should be small in value. 
Output Format - Print the value of the square root up to 5 decimal places.  

Constraints
S > 0, 
epsilon <= 0.01.

Student submission:
#include <iostream>

using namespace std ;
const double E = 0.01 ;
void sqrt(int s, int n, int x0);
	 while (((n+1)-n) <= E); {
		 {
		 n=((x0 + s) / x0) / 2 ;
		 n = n + 1;
		 precision(5)
     return;
			 }
	 }
int main() {
	
	 cin >> s >> x0 ;
	 sqrt(s, x0) ;
	 cout >> n ;
	 return 0 ;
}

Grading criterion: Quality and number of comments
Rating A: Well Commented
Rating B: Ok Commented
Rating C: Poor or no comments
Rating D: 0
Assigned grade: C
Problem: An iterative method, called the Babylonian method, for finding square roots, was known to the ancient Babylonians (1500 BC) and Greeks (100 AD) long before Newton invented his general procedure. Here's how it works. Suppose you are given any positive number S. To find the square root of S, do the following: Make an initial guess. Guess any positive number x0.Improve the guess. Apply the formula x1 = (x0 + S / x0) / 2. The number x1 is a better approximation to sqrt(S).Iterate until convergence. Apply the formula xn+1 = (xn + S / xn) / 2 until the process converges. Convergence is achieved when the absolute difference between xn+1  and xn falls below some user specified epsilon.Your job is to write a C++ program containing a function that determines the approximate square root of the number S (taken as input) up to some desired precision epsilon. To get credit, you must write a function.

Input Format - S should be greater than 0. Epsilon should be small in value. 
Output Format - Print the value of the square root up to 5 decimal places.  

Constraints
S > 0, 
epsilon <= 0.01.

Student submission:
#include<iostream>
using namespace std;
float f(int S,float epsilon){
int n=100;
float p;
float A[n+2];
A[0]=1;
for (int i=1;i<n+2;i++){
cout << fixed; 
cout.precision(5);
float t=(float)S/(float)A[i-1];
cout << fixed; 
cout.precision(5);
float u=(float)A[i-1]+(float)t;
cout << fixed; 
cout.precision(5);
A[i]=(float)u/(float)2;
if ((A[i-1]-A[i])<=epsilon && (A[i-1]-A[i])>=-epsilon) {p=A[i-1];break;}
}
cout << fixed; 
cout.precision(5);
return p;
}
int main(){
int S;float epsilon;
cin>>S>>epsilon;
float o=f(S,epsilon);
cout << fixed;
cout.precision(5); 
cout<<o;
}

Grading criterion: Quality and number of comments
Rating A: Well Commented
Rating B: Ok Commented
Rating C: Poor or no comments
Rating D: 0
Assigned grade: C
Problem: An iterative method, called the Babylonian method, for finding square roots, was known to the ancient Babylonians (1500 BC) and Greeks (100 AD) long before Newton invented his general procedure. Here's how it works. Suppose you are given any positive number S. To find the square root of S, do the following: Make an initial guess. Guess any positive number x0.Improve the guess. Apply the formula x1 = (x0 + S / x0) / 2. The number x1 is a better approximation to sqrt(S).Iterate until convergence. Apply the formula xn+1 = (xn + S / xn) / 2 until the process converges. Convergence is achieved when the absolute difference between xn+1  and xn falls below some user specified epsilon.Your job is to write a C++ program containing a function that determines the approximate square root of the number S (taken as input) up to some desired precision epsilon. To get credit, you must write a function.

Input Format - S should be greater than 0. Epsilon should be small in value. 
Output Format - Print the value of the square root up to 5 decimal places.  

Constraints
S > 0, 
epsilon <= 0.01.

Student submission:
#include<simplecpp>
double sqt(double a,double epsilon);

main_program{
	double s;
	double eps,c;
	cin>>s>>eps;
	c=sqt(s,eps);	
	cout<<fixed;
	cout.precision(5);
	cout<<c;
	
}

double sqt(double a,double epsilon)
{
	double x[1000];
	int i;
	x[0]=100;
	for(i=1;i<1000;i++)
	{
	x[i]=(x[i-1]+ (a/x[i-1]))/2;
	if(fabs(x[i]-x[i-1])<epsilon)
	break;
	}
	return x[i];
}

Grading criterion: Quality and number of comments
Rating A: Well Commented
Rating B: Ok Commented
Rating C: Poor or no comments
Rating D: 0
Assigned grade: C
Problem: An iterative method, called the Babylonian method, for finding square roots, was known to the ancient Babylonians (1500 BC) and Greeks (100 AD) long before Newton invented his general procedure. Here's how it works. Suppose you are given any positive number S. To find the square root of S, do the following: Make an initial guess. Guess any positive number x0.Improve the guess. Apply the formula x1 = (x0 + S / x0) / 2. The number x1 is a better approximation to sqrt(S).Iterate until convergence. Apply the formula xn+1 = (xn + S / xn) / 2 until the process converges. Convergence is achieved when the absolute difference between xn+1  and xn falls below some user specified epsilon.Your job is to write a C++ program containing a function that determines the approximate square root of the number S (taken as input) up to some desired precision epsilon. To get credit, you must write a function.

Input Format - S should be greater than 0. Epsilon should be small in value. 
Output Format - Print the value of the square root up to 5 decimal places.  

Constraints
S > 0, 
epsilon <= 0.01.

Student submission:
#include<iostream>
#include<cmath>
using namespace std;
double root(double s,double eps){
double x0=1,x1=s;
while(true){
x1=(x0+s/x0)/2;
if(fabs(x0-x1)<=eps) {break;}
x0=x1;
}
return x1;}
int main(){
double s,eps;
cin>>s>>eps;
double k;
k=root(s,eps);
cout<<fixed;
cout.precision(5);
cout<<k;}

Grading criterion: Quality and number of comments
Rating A: Well Commented
Rating B: Ok Commented
Rating C: Poor or no comments
Rating D: 0
Assigned grade: C
Problem: An iterative method, called the Babylonian method, for finding square roots, was known to the ancient Babylonians (1500 BC) and Greeks (100 AD) long before Newton invented his general procedure. Here's how it works. Suppose you are given any positive number S. To find the square root of S, do the following: Make an initial guess. Guess any positive number x0.Improve the guess. Apply the formula x1 = (x0 + S / x0) / 2. The number x1 is a better approximation to sqrt(S).Iterate until convergence. Apply the formula xn+1 = (xn + S / xn) / 2 until the process converges. Convergence is achieved when the absolute difference between xn+1  and xn falls below some user specified epsilon.Your job is to write a C++ program containing a function that determines the approximate square root of the number S (taken as input) up to some desired precision epsilon. To get credit, you must write a function.

Input Format - S should be greater than 0. Epsilon should be small in value. 
Output Format - Print the value of the square root up to 5 decimal places.  

Constraints
S > 0, 
epsilon <= 0.01.

Student submission:
#include<iostream>
#include<cmath>
using namespace std;
double sqrt(int s,double e){
	double x0=1;
	double x1=x0+4;
	while(fabs(x1-x0)>=e){
		x0 = x1;
		x1= (x0+ s/x0)/2;
		 }
return x1;}





int main(){
int s;
cin>>s;

double  epsilon;
cin>>epsilon;
double p;
p= sqrt(s,epsilon);
cout<<fixed;
cout.precision(5); 
cout<<p;

}

Grading criterion: Quality and number of comments
Rating A: Well Commented
Rating B: Ok Commented
Rating C: Poor or no comments
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 numofways(int a);

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

}

Grading criterion: Accumulating total path
Rating A: Correctly accumulates the total path at each recursion step

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(int pos,int arr[],int n){
    if(pos==(n-1)) return 1;
    int total;
    for(int i=pos;i<=pos+arr[pos];i++){
        total+= ways(i,arr,n);
    }
    return total;


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

}

Grading criterion: Accumulating total path
Rating A: Correctly accumulates the total path at each recursion step

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 jumps[n];
    for(int i=0;i<n;i++)
     cin >> jumps[i];
     int d=0;
       for(int i=0;i<n;i++) {
       if(a[i]=n)
       d++;
     }
        for(int i=0;i<n;i++) {
            for(int j=1;)
        }

     

}

Grading criterion: Accumulating total path
Rating A: Correctly accumulates the total path at each recursion step

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 f(int n, int i , int jumps[]){
    if(i=n-2){
        return 1;
    }
    int my;
    if(jumps[i] > n-1-i){
      my = (n-1-i)*(n-i)/2 ;
    }
    else my = jumps[i] ;
    return f(n,i,jumps) = f(n,i+1,jumps) +  my;
}
int main(){
    int n;
    cin >> n;
    int jumps[n];
    for(int i=0;i<n;i++){
        cin >> jumps[i];
    }
return f(n,0,jumps) ;
}

Grading criterion: Accumulating total path
Rating A: Correctly accumulates the total path at each recursion step

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;

Grading criterion: Accumulating total path
Rating A: Correctly accumulates the total path at each recursion step

Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: C