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<simplecpp>

main_program{

	long double n, r;
	cin >> n >> r;
	
	long double deno = 1;

	for(int times = n - r; times != 0; times--){
	
		deno *= times;
	
	}

	long double nume = 1;

	for(; n != r; n--){
	
		nume *= n;
	
	}

	cout << nume / deno << 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: 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;
        cin>>n>>r;
        if (r<=n/2){
                long double p=1;
                long double cnt=0;
                repeat(r){
                        p*=(n-cnt)/(cnt + 1);
                        cnt++;
                }
                cout<<p;
        }
        else {
                long double P=1;
                long double CNT=0;
                repeat(n-r){
                        P*=(n-CNT)/(CNT + 1);
                        CNT++;
                }
                cout<<P;
        }
}

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;
	cin>>n>>r;
	long double a,x=1,y=1;
	a=n-r;
	if(r>a){
		for(n;n>r;n--){
			x=x*n;
		}
		for(int i=1; i<=a;i++){
			y=y*i;
		}
		cout<<x/y;
	}
	else{
		for(n;n>a;n--){
			x=x*n;
		}
		for(int i;i<=r;i++){
			y=y*i;
		}
		cout<<x/y;
	}
}

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, factorial1 = 1, factorial2 = 1, n_minus_r;
	cin >> n >> r;
	n_minus_r = n-r;
	if (n >= 1 && n <= 35000 && r >= 1 && n >= r ) {
		repeat (n_minus_r - 1){
			factorial1 = factorial1 * (n_minus_r);
			n_minus_r --;
			}
		while  (r < n){
			factorial2 = factorial2 * (r+1);
			r ++;
		}
		cout << factorial2/factorial1;
	}
	else{
		cout << "invalid input";
	}
}

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: A
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;
	double  p,k,z;
	    p=n;
	    k=r;
	    z=1;
	    while(p>0&&k>0){
                z=z*p/k;
	        p--;
	        k--;}
 
	    cout<<z;
}

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,a;
	cin>>n>>r;

	if(n>=1&&n<=35000;){
		for(a=n;n--;)
	n = n*a;
	
}
//	while(r>=1&&r<=n){
//		r*=r-1;}
//	while(n>=1&&n<=35000){
//		(n-1)*=(n-2);
//	}
	cout<<n<<endl;
//	cout<<r<<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: 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{
	long double n,r;
	cin >> n >> r;
	
	int p;
	long double q=1;
	p=n-r;
	for(int a =p;a>=n; a++){
		q= q*p;
	}

			
		


	long double j;
	long double factoriala=1;
	j=n-r;
	for(int p=1;p<=j;p++){
		factoriala= factoriala*p;
	}
	cout <<q/factoriala;
}

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,i,j,nCr=1;
cin>>n>>r;
i = r;
j = n;

while(i>=1){

	nCr = nCr*(j/i);
	j = j - 1;
	i = i - 1;

}

cout<<nCr;
}

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;
	cin>>n>>r;
	int i=0;
	long double s=1;
	long double num=n,den=r;
	/*while(i<=r-1)
	{
		s=s*((num)/(den));
		num = n-i;
		den= r-i;
		i++;
		cout<<i<<"\n"<<s<<"\n";
	}*/
	while(i<=r-1)
	{
		num=num*(n-i);
		den=den*(r-i);
		i++;
		cout<<i<<"\n"<<num<<"\n";

	}
	s=num/den;


	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: 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;

	long double product =1.0;

	for(long double i = 0; i < r; i++){
		product *= (n-i)/(r-i);
	}

	cout << product;

}

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;
	long double r;
	long double denominator=1;
	long double numerator;
	long double product;
	long double i=1;

	cin >> n >> r;

	product = n;

	while (i<n-r){
		product = product*(n-i);
		i = i+1;
	}

	numerator = product;

	for(int k=1;k<=n-r;k++){
		denominator = denominator*k;
	}

	cout<<numerator/denominator;










	
}

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: A
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:
k

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{
        long double n,r;
        cin>>n>>r;
        long double m=1;
        long double q=1;
        if(r>n/2){r=n-r;}
        while(q<=r){
                m=m*((n-q+1)/q);
                q=q+1;

        }
        cout<<m;
}

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{
	int n,r;
	cin>>n>>r;
	int c=n-r;
	long double prod1=1,prod2=1;
	if(c>(n/2)){
	repeat(r){
		prod1*=n;
		n--;
	}
	repeat(r){
		prod2*=r;
		r--;
	}
	}
	else{
	repeat(c){
		prod1*=n;
		n--;
	}
	repeat(c){
		prod2*=c;
		c--;
	}
	}
	cout<<prod1/prod2<<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: 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;
	int k=min(r,(n-r));
        int q=k;
	//int i=n;
	//int j=k;
	double result=1;
	repeat(q)
	{
		result*=n*1.0/k;
		n--;
		k--;
	}
	cout<<result;
}

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; int r;
	cin>>n>>r;
	long double num=1;long double den=1;int n_=n;int i=1;
	repeat(n-r){num=num*n_;n_--;}
	repeat(n-r){den=den*i;i++;}
	cout<<num/den;
}

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{

int n,r;
cin>>n>>r;

long double n1=n;

long double r1=r;
long double fact=1.0;
long double i1=1.0;
for(int i=1; i<=r;i++)
{
	i1=i;
	fact=fact*((n1-r1+i1)/i1);

}
cout<<fact;





}

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{
int n, r;
cin >> n >> r;
long double num = 1;
long double denom = 1;
if(r > n-r){
for(int i = n; i > r; i--){
num *= i;
}
for(int j = 1; j <=n-r; j++){
denom *= j;
}
cout << num/denom;
}

else{
for(int i = n; i > n-r; i--){
num *= i;
}
for(int j = 1; j <= r; j++){
denom *= j;}
cout << num/denom;
}
}

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;
	cin >> n >> r;
	long double ans = 1;
	if(r <= n - r){
		for(long long int i = 1; i <= r; i++){
			ans *= (n + 1 - i);
			ans /= i;
		}
	}else{
		for(long long int i = 1; i <= n - r; i++){
			ans *= (n + 1 - i);
			ans /= i;
		}
	}
	cout << ans << "\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: 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 {

	int n, r;
	cin >> n >> r;


	long double ans = 1;
	int ctr=1;
	int numy=n-r+1;

	repeat(r) {
		ans = ans * (numy*1.0/ctr*1.0);
		numy++;
		ctr++;

	}

	cout << ans;




}

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;
cin>>n>>r;
long double x=1;
int i=r+1,j=1;
while(j<=(n-r )){

x=x/j;
j++;

}

while(i<=n ){

x=x*i;
i++;

}

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{
	int n,r;
	long double p,i,j,k,a=1.00,b=1.00,c=1.00,result;

       	
	cin>>n>>r;
	p=n-r;
	for(i=1;i<=n;i++)
	{ a=a*i;}
	  for(j=1;j<=r;j++)
	  { b=b*j;
	  }
	  for(k=1;k<=p;k++)
	  { c=c*k;
	  }
	  result=a/(b*c);
	  cout<<result;
}

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 c=1;

	cin >> n >> r;

	if(n*0.5<=r )
		r=n-r;

	repeat(r){
		c=c*n*1.0/r;
		n--;
	       	r--;
	}
	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: 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{
	long double  n, r;
	cin >> n >> r;
	long double tot = 1;
	if(r<n/2){
		r = n-r;}
	for(long double i=0;i<(n-r);i++){
		tot *= (n-i)/(i+1);}
	cout << tot;
}

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