CodeGraderBench/CodeGraderBench huggingface dataset:
Problem: accept n and r from the user, compute nCr, and print the result. A naive approach would be to calculate the factorial of n, factorial of r, and factorial of n-r and then finally compute the result as given in the formulae above. However, calculating the factorial of a large number can be tricky, and the result may overflow due to the limitation of the size of data type. you are free to use other data types, including long double. you are free to use other data types, including long double.

Constraints:1 <= r <= n1 <= n <= 35000

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

int main(){
long double n, r;
cin>>n;
long double nFactorial;

 repeat(n-1){
nFactorial = nFactorial*(n-1);
cout<<nFactorial<<endl;
n--;

 }
}

Grading criterion: Quality of variable names
Rating A: Good Variable Names
Rating B: Ok variable names
Rating C: poor variable names
Rating D: 0
Assigned grade: C
Problem: accept n and r from the user, compute nCr, and print the result. A naive approach would be to calculate the factorial of n, factorial of r, and factorial of n-r and then finally compute the result as given in the formulae above. However, calculating the factorial of a large number can be tricky, and the result may overflow due to the limitation of the size of data type. you are free to use other data types, including long double. you are free to use other data types, including long double.

Constraints:1 <= r <= n1 <= n <= 35000

Student submission:
#include<simplecpp>
main_program{
	int n,r;
	cin>>n>>r;
	float c;
	if((n>=1 && n<=35000)&&(r>=1 && r<=n)){
		int i=1;
		c=1;
		while(i<r){
			c=c*((n-(i-1))/i);
			i++;
		}
		c=((c*(n-(i-1)))/i);
		cout<<c;
	}
}

Grading criterion: Quality of variable names
Rating A: Good Variable Names
Rating B: Ok variable names
Rating C: poor variable names
Rating D: 0
Assigned grade: B
Problem: accept n and r from the user, compute nCr, and print the result. A naive approach would be to calculate the factorial of n, factorial of r, and factorial of n-r and then finally compute the result as given in the formulae above. However, calculating the factorial of a large number can be tricky, and the result may overflow due to the limitation of the size of data type. you are free to use other data types, including long double. you are free to use other data types, including long double.

Constraints:1 <= r <= n1 <= n <= 35000

Student submission:
#include<simplecpp>
main_program
{
    	long double n,r,c,x;
	cin>>n>>r;
	c=n-r;
	if(n>r && n>=1 && r>=1 && n<=35000)
	{
		long double d,a,b,j=1,i=1,k=1;
		a=n;b=r;d=n-r;
		while(a>i)
		{
			n=n*i;
			i=i+1;
		}
		while(b>j)
		{
			r=r*j;
			j=j+1;
		}
		while(d>k)
		{
			c=c*k;
			k=k+1;
		}
		x=n/(r*c);
		cout<<x;
		

	}
}

Grading criterion: Quality of variable names
Rating A: Good Variable Names
Rating B: Ok variable names
Rating C: poor variable names
Rating D: 0
Assigned grade: B
Problem: accept n and r from the user, compute nCr, and print the result. A naive approach would be to calculate the factorial of n, factorial of r, and factorial of n-r and then finally compute the result as given in the formulae above. However, calculating the factorial of a large number can be tricky, and the result may overflow due to the limitation of the size of data type. you are free to use other data types, including long double. you are free to use other data types, including long double.

Constraints:1 <= r <= n1 <= n <= 35000

Student submission:
#include<simplecpp>
main_program{
    long long int n,r,k=1,l;
    long double s=1.0;
    cin>>n>>r;
    l=n-r;
    if(l<r){
        r=l;
    }
    repeat(r){
        s=s*n/k;
        n--;k++;
    }
    cout<<s;
}

Grading criterion: Quality of variable names
Rating A: Good Variable Names
Rating B: Ok variable names
Rating C: poor variable names
Rating D: 0
Assigned grade: B
Problem: accept n and r from the user, compute nCr, and print the result. A naive approach would be to calculate the factorial of n, factorial of r, and factorial of n-r and then finally compute the result as given in the formulae above. However, calculating the factorial of a large number can be tricky, and the result may overflow due to the limitation of the size of data type. you are free to use other data types, including long double. you are free to use other data types, including long double.

Constraints:1 <= r <= n1 <= n <= 35000

Student submission:
#include<simplecpp>
main_program[
float n,r;
cin<<n<<r;
for(int num=1; num<=n; num++){
num *=num;
cout<<num;


}

Grading criterion: Quality of variable names
Rating A: Good Variable Names
Rating B: Ok variable names
Rating C: poor variable names
Rating D: 0
Assigned grade: C
Problem: accept n and r from the user, compute nCr, and print the result. A naive approach would be to calculate the factorial of n, factorial of r, and factorial of n-r and then finally compute the result as given in the formulae above. However, calculating the factorial of a large number can be tricky, and the result may overflow due to the limitation of the size of data type. you are free to use other data types, including long double. you are free to use other data types, including long double.

Constraints:1 <= r <= n1 <= n <= 35000

Student submission:
#include<simplecpp>
main_program{
	int n, r;
	long double a=1, b=1, c=1;
	cin>>n>>r;
	int i=1, j=1, k=1;
	repeat(n){
		a=a*i;
		i++;
	}
	repeat(r){
		b=b*j;
		j++;
	}
	
	repeat(n-r){
		c=c*k;
		k++;
	}
	  long double result=(a/b);
	  long double sum=result/c;
	cout<<sum<<endl;



}

Grading criterion: Quality of variable names
Rating A: Good Variable Names
Rating B: Ok variable names
Rating C: poor variable names
Rating D: 0
Assigned grade: B
Problem: You are tasked with creating a special numbering pattern for a triangular grid of side length n.

The triangle is formed using rows such that:
- The first row has n elements.
- The second row has n - 1 elements.
- The third row has n - 2 elements.
- ...
- The last row has 1 element.

Your job is to fill this triangular grid with integers starting from 01, following a unique diagonal numbering pattern.

Numbering Rules

1. Start from the top-left element of the triangle with 01.
2. Fill the numbers diagonally downward while moving to the right.
3. Each diagonal starts from the top row and moves down toward the bottom-left of the triangle.
4. Filling must stay within the valid triangular bounds.

Hint

In each row, the difference between two consecutive entries increases by 1 with each step. The same rule applies to each column.

Input

An integer n, representing the side length of the triangle.

Output

Print n lines.
The i-th line contains n - i + 1 integers corresponding to the i-th row of the triangle.

Assumptions on Input

1 <= n <= 13

Note

- Do not print any extra messages such as:
  cout << "Enter a number"
  cout << "The computed answer is"
- Use cout only to print the computed final output.
- Do not print unnecessary spaces.
- Do not hard-code the solution.
- If any test case passes by simply printing with cout and without logic, marks will not be awarded.

Student submission:
#include<simplecpp>
main_program {
        int n=0, r=1;
        cin>> n;
        int i=1, a=1, b=2, c=1, m=1;
        int row=n;
        while( row>0){
          while (r<=n) {
               cout<< i<< " ";
        i=i+a;
                a++;
               }
          cout<< endl;
          row--;
          n--;
          r=1;
          i=1;
          i=i+b;
          b=b+m;
          a=1;
          a=a+c;
          c=c+m;
          m++;
        }
}

Grading criterion: Accepting input
Rating A: correctly accepts an integer as input
Rating B: Some issue. 

Rating C: Major issue. 

Rating D: 0
Assigned grade: A
Problem: You are tasked with creating a special numbering pattern for a triangular grid of side length n.

The triangle is formed using rows such that:
- The first row has n elements.
- The second row has n - 1 elements.
- The third row has n - 2 elements.
- ...
- The last row has 1 element.

Your job is to fill this triangular grid with integers starting from 01, following a unique diagonal numbering pattern.

Numbering Rules

1. Start from the top-left element of the triangle with 01.
2. Fill the numbers diagonally downward while moving to the right.
3. Each diagonal starts from the top row and moves down toward the bottom-left of the triangle.
4. Filling must stay within the valid triangular bounds.

Hint

In each row, the difference between two consecutive entries increases by 1 with each step. The same rule applies to each column.

Input

An integer n, representing the side length of the triangle.

Output

Print n lines.
The i-th line contains n - i + 1 integers corresponding to the i-th row of the triangle.

Assumptions on Input

1 <= n <= 13

Note

- Do not print any extra messages such as:
  cout << "Enter a number"
  cout << "The computed answer is"
- Use cout only to print the computed final output.
- Do not print unnecessary spaces.
- Do not hard-code the solution.
- If any test case passes by simply printing with cout and without logic, marks will not be awarded.

Student submission:
#include<simplecpp>
main_program
{
        int a,b=0,c=2,n,x=1;
        cin>>n;
        for(int i=0;i<n;i++)
        {
                a=i+1;
                for(int j=0;j<n-i;j++)
                {
                        if(x<10)
                        {
                                cout<<0<<x<<' ';
                        }
                        else
                        {
                                cout<<x<<' ';
                        }
                        x=x+a;
                        a++;
                }
                cout<<endl;
                b=b+c;
                x=1+b;
                c++;
        }
}

Grading criterion: Accepting input
Rating A: correctly accepts an integer as input
Rating B: Some issue. 

Rating C: Major issue. 

Rating D: 0
Assigned grade: A
Problem: You are tasked with creating a special numbering pattern for a triangular grid of side length n.

The triangle is formed using rows such that:
- The first row has n elements.
- The second row has n - 1 elements.
- The third row has n - 2 elements.
- ...
- The last row has 1 element.

Your job is to fill this triangular grid with integers starting from 01, following a unique diagonal numbering pattern.

Numbering Rules

1. Start from the top-left element of the triangle with 01.
2. Fill the numbers diagonally downward while moving to the right.
3. Each diagonal starts from the top row and moves down toward the bottom-left of the triangle.
4. Filling must stay within the valid triangular bounds.

Hint

In each row, the difference between two consecutive entries increases by 1 with each step. The same rule applies to each column.

Input

An integer n, representing the side length of the triangle.

Output

Print n lines.
The i-th line contains n - i + 1 integers corresponding to the i-th row of the triangle.

Assumptions on Input

1 <= n <= 13

Note

- Do not print any extra messages such as:
  cout << "Enter a number"
  cout << "The computed answer is"
- Use cout only to print the computed final output.
- Do not print unnecessary spaces.
- Do not hard-code the solution.
- If any test case passes by simply printing with cout and without logic, marks will not be awarded.

Student submission:
#include<simplecpp>
main_program{
    int n=1,
    cin>>k
    cout<<n<<n+1<<n+2<<(k*(k+1))/2;
    }

Grading criterion: Accepting input
Rating A: correctly accepts an integer as input
Rating B: Some issue. 

Rating C: Major issue. 

Rating D: 0
Assigned grade: A
Problem: You are tasked with creating a special numbering pattern for a triangular grid of side length n.

The triangle is formed using rows such that:
- The first row has n elements.
- The second row has n - 1 elements.
- The third row has n - 2 elements.
- ...
- The last row has 1 element.

Your job is to fill this triangular grid with integers starting from 01, following a unique diagonal numbering pattern.

Numbering Rules

1. Start from the top-left element of the triangle with 01.
2. Fill the numbers diagonally downward while moving to the right.
3. Each diagonal starts from the top row and moves down toward the bottom-left of the triangle.
4. Filling must stay within the valid triangular bounds.

Hint

In each row, the difference between two consecutive entries increases by 1 with each step. The same rule applies to each column.

Input

An integer n, representing the side length of the triangle.

Output

Print n lines.
The i-th line contains n - i + 1 integers corresponding to the i-th row of the triangle.

Assumptions on Input

1 <= n <= 13

Note

- Do not print any extra messages such as:
  cout << "Enter a number"
  cout << "The computed answer is"
- Use cout only to print the computed final output.
- Do not print unnecessary spaces.
- Do not hard-code the solution.
- If any test case passes by simply printing with cout and without logic, marks will not be awarded.

Student submission:
#include<simplecpp>
main_program{
	int a=01;
	int n;
	int b=01;
	int f=01;
	int c=02;
	int d=03;
	cin>>n;
	while(n>0){
		repeat(n){
			if(a<10){
				cout<<"0"<<a<<" ";}
			else
				cout<<a<<" ";
			a=a+b;
			b++;
		}
		n=n-1;
		cout<<endl;
		a=01;
		b=01;
		b=b+f;
		f++;
		a=a+c;
		c=c+d;
		d++;
		}
}

Grading criterion: Accepting input
Rating A: correctly accepts an integer as input
Rating B: Some issue. 

Rating C: Major issue. 

Rating D: 0
Assigned grade: A
Problem: You are tasked with creating a special numbering pattern for a triangular grid of side length n.

The triangle is formed using rows such that:
- The first row has n elements.
- The second row has n - 1 elements.
- The third row has n - 2 elements.
- ...
- The last row has 1 element.

Your job is to fill this triangular grid with integers starting from 01, following a unique diagonal numbering pattern.

Numbering Rules

1. Start from the top-left element of the triangle with 01.
2. Fill the numbers diagonally downward while moving to the right.
3. Each diagonal starts from the top row and moves down toward the bottom-left of the triangle.
4. Filling must stay within the valid triangular bounds.

Hint

In each row, the difference between two consecutive entries increases by 1 with each step. The same rule applies to each column.

Input

An integer n, representing the side length of the triangle.

Output

Print n lines.
The i-th line contains n - i + 1 integers corresponding to the i-th row of the triangle.

Assumptions on Input

1 <= n <= 13

Note

- Do not print any extra messages such as:
  cout << "Enter a number"
  cout << "The computed answer is"
- Use cout only to print the computed final output.
- Do not print unnecessary spaces.
- Do not hard-code the solution.
- If any test case passes by simply printing with cout and without logic, marks will not be awarded.

Student submission:
#include<simplecpp>

main_program{
        int n;
        cin >> n;
        int i=0;
        repeat(n){
        cout << 1+i;
        i++;
        }}

Grading criterion: Accepting input
Rating A: correctly accepts an integer as input
Rating B: Some issue. 

Rating C: Major issue. 

Rating D: 0
Assigned grade: A
Problem: You are tasked with creating a special numbering pattern for a triangular grid of side length n.

The triangle is formed using rows such that:
- The first row has n elements.
- The second row has n - 1 elements.
- The third row has n - 2 elements.
- ...
- The last row has 1 element.

Your job is to fill this triangular grid with integers starting from 01, following a unique diagonal numbering pattern.

Numbering Rules

1. Start from the top-left element of the triangle with 01.
2. Fill the numbers diagonally downward while moving to the right.
3. Each diagonal starts from the top row and moves down toward the bottom-left of the triangle.
4. Filling must stay within the valid triangular bounds.

Hint

In each row, the difference between two consecutive entries increases by 1 with each step. The same rule applies to each column.

Input

An integer n, representing the side length of the triangle.

Output

Print n lines.
The i-th line contains n - i + 1 integers corresponding to the i-th row of the triangle.

Assumptions on Input

1 <= n <= 13

Note

- Do not print any extra messages such as:
  cout << "Enter a number"
  cout << "The computed answer is"
- Use cout only to print the computed final output.
- Do not print unnecessary spaces.
- Do not hard-code the solution.
- If any test case passes by simply printing with cout and without logic, marks will not be awarded.

Student submission:
#include<simplecpp>


main_program{

int n,m=1,d = 1,g = 1,h = 1, i = 1;
cin >> n;

repeat(n){
    repeat(n){
     if (m < 10){cout << "0"<< m << " ";}
     else {
      cout << m << " ";}
     
     m = m + d;
     d = d + 1; }
     
  n = n - 1;
  h = h + 1;
  g = g + h;
  m = g;
  i = i +1;
  d = i;
  
     cout << endl;
     }
}

Grading criterion: Accepting input
Rating A: correctly accepts an integer as input
Rating B: Some issue. 

Rating C: Major issue. 

Rating D: 0
Assigned grade: A
Problem: You are tasked with creating a special numbering pattern for a triangular grid of side length n.

The triangle is formed using rows such that:
- The first row has n elements.
- The second row has n - 1 elements.
- The third row has n - 2 elements.
- ...
- The last row has 1 element.

Your job is to fill this triangular grid with integers starting from 01, following a unique diagonal numbering pattern.

Numbering Rules

1. Start from the top-left element of the triangle with 01.
2. Fill the numbers diagonally downward while moving to the right.
3. Each diagonal starts from the top row and moves down toward the bottom-left of the triangle.
4. Filling must stay within the valid triangular bounds.

Hint

In each row, the difference between two consecutive entries increases by 1 with each step. The same rule applies to each column.

Input

An integer n, representing the side length of the triangle.

Output

Print n lines.
The i-th line contains n - i + 1 integers corresponding to the i-th row of the triangle.

Assumptions on Input

1 <= n <= 13

Note

- Do not print any extra messages such as:
  cout << "Enter a number"
  cout << "The computed answer is"
- Use cout only to print the computed final output.
- Do not print unnecessary spaces.
- Do not hard-code the solution.
- If any test case passes by simply printing with cout and without logic, marks will not be awarded.

Student submission:
#include<simplecpp>
main_program{
int n=0,a=1;
cin>>n;
for(int k=n;k>0;k--){
int i=1+(a-1)*a/2;
a++;
for (int j=1; j<=k;j++){;
int c=0;
c=i;
i++;
cout<<c<<" ";}
cout<<endl;
}
}

Grading criterion: Accepting input
Rating A: correctly accepts an integer as input
Rating B: Some issue. 

Rating C: Major issue. 

Rating D: 0
Assigned grade: A
Problem: You are tasked with creating a special numbering pattern for a triangular grid of side length n.

The triangle is formed using rows such that:
- The first row has n elements.
- The second row has n - 1 elements.
- The third row has n - 2 elements.
- ...
- The last row has 1 element.

Your job is to fill this triangular grid with integers starting from 01, following a unique diagonal numbering pattern.

Numbering Rules

1. Start from the top-left element of the triangle with 01.
2. Fill the numbers diagonally downward while moving to the right.
3. Each diagonal starts from the top row and moves down toward the bottom-left of the triangle.
4. Filling must stay within the valid triangular bounds.

Hint

In each row, the difference between two consecutive entries increases by 1 with each step. The same rule applies to each column.

Input

An integer n, representing the side length of the triangle.

Output

Print n lines.
The i-th line contains n - i + 1 integers corresponding to the i-th row of the triangle.

Assumptions on Input

1 <= n <= 13

Note

- Do not print any extra messages such as:
  cout << "Enter a number"
  cout << "The computed answer is"
- Use cout only to print the computed final output.
- Do not print unnecessary spaces.
- Do not hard-code the solution.
- If any test case passes by simply printing with cout and without logic, marks will not be awarded.

Student submission:
#include<simplecpp>
main_program{
    int n;cin>>n;
    int i=1;int j=1;
    while(i<=n){
        while(j<=n){
            cout<<((j*j+j)*0.5+(i*j-j)+0.5*(i*i-3*i+2));j++;
        }
        i++;
    }
}

Grading criterion: Accepting input
Rating A: correctly accepts an integer as input
Rating B: Some issue. 

Rating C: Major issue. 

Rating D: 0
Assigned grade: A
Problem: You are tasked with creating a special numbering pattern for a triangular grid of side length n.

The triangle is formed using rows such that:
- The first row has n elements.
- The second row has n - 1 elements.
- The third row has n - 2 elements.
- ...
- The last row has 1 element.

Your job is to fill this triangular grid with integers starting from 01, following a unique diagonal numbering pattern.

Numbering Rules

1. Start from the top-left element of the triangle with 01.
2. Fill the numbers diagonally downward while moving to the right.
3. Each diagonal starts from the top row and moves down toward the bottom-left of the triangle.
4. Filling must stay within the valid triangular bounds.

Hint

In each row, the difference between two consecutive entries increases by 1 with each step. The same rule applies to each column.

Input

An integer n, representing the side length of the triangle.

Output

Print n lines.
The i-th line contains n - i + 1 integers corresponding to the i-th row of the triangle.

Assumptions on Input

1 <= n <= 13

Note

- Do not print any extra messages such as:
  cout << "Enter a number"
  cout << "The computed answer is"
- Use cout only to print the computed final output.
- Do not print unnecessary spaces.
- Do not hard-code the solution.
- If any test case passes by simply printing with cout and without logic, marks will not be awarded.

Student submission:
#include<simplecpp>
main_program{
    int n,i=1,j=1,sum=1;
    cin >> n;
    for(int i=1;i<=n;i++){
        int a=sum;
        j=i;
        while(j<=n){
            if(a<10){
                cout << "0" << a << " ";
            }
            else    cout << a << " ";
            a=a+j;
            j++;
        }
        sum=sum+(i+1);
        cout << endl;
    }
}

Grading criterion: Accepting input
Rating A: correctly accepts an integer as input
Rating B: Some issue. 

Rating C: Major issue. 

Rating D: 0
Assigned grade: A