CodeGraderBench/CodeGraderBench huggingface dataset:
Problem: You are given an infinite alternating series:
1-1/2!+1/3!-1/4!+1/5!  ... 
Your task is to compute the sum of the first N terms of this series. The series alternates in sign, with each term involving the reciprocal of a factorial.

Example: If n = 1, the n terms of the series is just 1. Hence, answer is 1.000000

Input Format: A single positive integer N where 1 ≤ N ≤ 15.

Output Format
	A single floating-point number representing the sum of the first N terms of the series, rounded to exactly 6 decimal places. 
	To print exactly six digits after the decimal point, write std::cout << std::setprecision(6) << std::fixed << ans;. Here we assume that the answer is stored in variable ans.

Student submission:
#include<simplecpp>
#include<iomanip>

double fact(int n)
{
    int i;
    double factorial =1.0;
    for(i=1; i<=n; i++)
    {
        factorial *= i;
    }
    return factorial;
}
main_program
{
    int N;
    cin >> N;
    float ans=1.0;
    if(N>=1 && N<=15)
    {
    if(N==1)
        ans = 1;
    else
    {
    for(int i=2; i<=N; i++)
    {
        if(i%2 == 0)
        ans -= (1.0/fact(i));
        else if(i%2 != 0)
        ans += (1.0/fact(i));
    }
    }
    std::cout << std::setprecision(6) << std::fixed << ans;
    }
}

Grading criterion: Alternating signs
Rating A: correctly changes the sign +ve to -ve and vice versa
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: You are given an infinite alternating series:
1-1/2!+1/3!-1/4!+1/5!  ... 
Your task is to compute the sum of the first N terms of this series. The series alternates in sign, with each term involving the reciprocal of a factorial.

Example: If n = 1, the n terms of the series is just 1. Hence, answer is 1.000000

Input Format: A single positive integer N where 1 ≤ N ≤ 15.

Output Format
	A single floating-point number representing the sum of the first N terms of the series, rounded to exactly 6 decimal places. 
	To print exactly six digits after the decimal point, write std::cout << std::setprecision(6) << std::fixed << ans;. Here we assume that the answer is stored in variable ans.

Student submission:
#include <simplecpp>
#include <iomanip>

main_program {
	int N;
	cin >> N;
	
	float ans;
	
	float term = -1; // so that the first term is 1
	for (int i = 1; i <= N; i++) {
		term *= -1.0 / i;
		ans += term;
	}
	
	cout << setprecision(6) << fixed << ans << endl; // no need for std:: since <simplecpp> has 'using namespace std'
	
}

Grading criterion: Alternating signs
Rating A: correctly changes the sign +ve to -ve and vice versa
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: You are given an infinite alternating series:
1-1/2!+1/3!-1/4!+1/5!  ... 
Your task is to compute the sum of the first N terms of this series. The series alternates in sign, with each term involving the reciprocal of a factorial.

Example: If n = 1, the n terms of the series is just 1. Hence, answer is 1.000000

Input Format: A single positive integer N where 1 ≤ N ≤ 15.

Output Format
	A single floating-point number representing the sum of the first N terms of the series, rounded to exactly 6 decimal places. 
	To print exactly six digits after the decimal point, write std::cout << std::setprecision(6) << std::fixed << ans;. Here we assume that the answer is stored in variable ans.

Student submission:
#include <simplecpp>
#include <iomanip>

int main () {
    int N;
    cin >> N;
    float ans=0.0;
    for (int fac=1, mult = 1, it=1; it<=N; ++it, mult *= -1) {
        fac *= it; 
        ans = ans + mult*(1.0/(fac));
    }
    std::cout << std::setprecision(6) << std::fixed << ans;
}

Grading criterion: Alternating signs
Rating A: correctly changes the sign +ve to -ve and vice versa
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: You are given an infinite alternating series:
1-1/2!+1/3!-1/4!+1/5!  ... 
Your task is to compute the sum of the first N terms of this series. The series alternates in sign, with each term involving the reciprocal of a factorial.

Example: If n = 1, the n terms of the series is just 1. Hence, answer is 1.000000

Input Format: A single positive integer N where 1 ≤ N ≤ 15.

Output Format
	A single floating-point number representing the sum of the first N terms of the series, rounded to exactly 6 decimal places. 
	To print exactly six digits after the decimal point, write std::cout << std::setprecision(6) << std::fixed << ans;. Here we assume that the answer is stored in variable ans.

Student submission:
#include <simplecpp>
#include<iomanip>
double factorial(double n){
        double f = 1; double i =1;
        repeat(n){
            f = f*i; i = i +1;
        }
        return f;
    }
main_program{
    double n; cin>>n; double sum = 0; double i = 0;
    
    double j =1;
    repeat(n){
        double k = factorial(j);
        sum = sum + (pow(-1,i)/k);
        j = j+1; i =i +1;
    }
    std::cout<<std::setprecision(6)<<std::fixed<<sum;
}

Grading criterion: Alternating signs
Rating A: correctly changes the sign +ve to -ve and vice versa
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: You are given an infinite alternating series:
1-1/2!+1/3!-1/4!+1/5!  ... 
Your task is to compute the sum of the first N terms of this series. The series alternates in sign, with each term involving the reciprocal of a factorial.

Example: If n = 1, the n terms of the series is just 1. Hence, answer is 1.000000

Input Format: A single positive integer N where 1 ≤ N ≤ 15.

Output Format
	A single floating-point number representing the sum of the first N terms of the series, rounded to exactly 6 decimal places. 
	To print exactly six digits after the decimal point, write std::cout << std::setprecision(6) << std::fixed << ans;. Here we assume that the answer is stored in variable ans.

Student submission:
#include<simplecpp>
#include<iomanip>
main_program{
int N;
cin>>N;
double ans=0,term=-1;
for(double i=1;i<=N;i++){
    term*=(-1/i);
ans+=term;
}
std::cout<<std::setprecision(6)<<std::fixed<<ans;

}

Grading criterion: Alternating signs
Rating A: correctly changes the sign +ve to -ve and vice versa
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: You are given an infinite alternating series:
1-1/2!+1/3!-1/4!+1/5!  ... 
Your task is to compute the sum of the first N terms of this series. The series alternates in sign, with each term involving the reciprocal of a factorial.

Example: If n = 1, the n terms of the series is just 1. Hence, answer is 1.000000

Input Format: A single positive integer N where 1 ≤ N ≤ 15.

Output Format
	A single floating-point number representing the sum of the first N terms of the series, rounded to exactly 6 decimal places. 
	To print exactly six digits after the decimal point, write std::cout << std::setprecision(6) << std::fixed << ans;. Here we assume that the answer is stored in variable ans.

Student submission:
#include<simplecpp>
main_program
{
    int n;
    cin>>n;
    int s=0;
    for(int i=1;i<=n;i++)
    {
        
        int f=1;
        while(f<=i)
        {
            f*=i;
            f++;
        }

    }



}

Grading criterion: Alternating signs
Rating A: correctly changes the sign +ve to -ve and vice versa
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: C
Problem: You are given an infinite alternating series:
1-1/2!+1/3!-1/4!+1/5!  ... 
Your task is to compute the sum of the first N terms of this series. The series alternates in sign, with each term involving the reciprocal of a factorial.

Example: If n = 1, the n terms of the series is just 1. Hence, answer is 1.000000

Input Format: A single positive integer N where 1 ≤ N ≤ 15.

Output Format
	A single floating-point number representing the sum of the first N terms of the series, rounded to exactly 6 decimal places. 
	To print exactly six digits after the decimal point, write std::cout << std::setprecision(6) << std::fixed << ans;. Here we assume that the answer is stored in variable ans.

Student submission:
#include <simplecpp>

main_program{
        float n;
        cin>>n;
        float m=0.0;
        float k=1.0;
        for(int i=1; i<=n; i++){
               float x=pow(-1,i-1);
                k=k*i;
                m+=x/k;
        }
        cout<<m<<endl;
}

Grading criterion: Alternating signs
Rating A: correctly changes the sign +ve to -ve and vice versa
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: You are given an infinite alternating series:
1-1/2!+1/3!-1/4!+1/5!  ... 
Your task is to compute the sum of the first N terms of this series. The series alternates in sign, with each term involving the reciprocal of a factorial.

Example: If n = 1, the n terms of the series is just 1. Hence, answer is 1.000000

Input Format: A single positive integer N where 1 ≤ N ≤ 15.

Output Format
	A single floating-point number representing the sum of the first N terms of the series, rounded to exactly 6 decimal places. 
	To print exactly six digits after the decimal point, write std::cout << std::setprecision(6) << std::fixed << ans;. Here we assume that the answer is stored in variable ans.

Student submission:
#include<simplecpp>
#include<iomanip>
int fact(int n)
{  int facta=1;
    for(int i=1;i<=n;i++)
     facta=facta*i;
    return facta;
}
main_program
{   
    int N ;double sum=0;
    cin>>N;
    for(int j=1;j<=N;j++)
    {   int c=j+1;
        sum=sum + pow((-1),c)/fact(j);
  }
std::cout<<std::setprecision(6)<<std::fixed<<sum;

}

Grading criterion: Alternating signs
Rating A: correctly changes the sign +ve to -ve and vice versa
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: You are given an infinite alternating series:
1-1/2!+1/3!-1/4!+1/5!  ... 
Your task is to compute the sum of the first N terms of this series. The series alternates in sign, with each term involving the reciprocal of a factorial.

Example: If n = 1, the n terms of the series is just 1. Hence, answer is 1.000000

Input Format: A single positive integer N where 1 ≤ N ≤ 15.

Output Format
	A single floating-point number representing the sum of the first N terms of the series, rounded to exactly 6 decimal places. 
	To print exactly six digits after the decimal point, write std::cout << std::setprecision(6) << std::fixed << ans;. Here we assume that the answer is stored in variable ans.

Student submission:
#include <simplecpp>
#include <iomanip>
float factorial(int n) {
    for (int product=1, a=1; a<=n; a++) {
        product=product*a;
        return product;
    }
}
main_program {
int n; cin >> n;
float sum1=0, sum2=0;
if (n%2==0) {
for(int a=1; a<=(n/2); a++) {
    float x= 1/(factorial)(2a-1);
    sum1+=x;
}
for(int a; sum2; a<=(n/2); a++) {
    float y= 1/(factorial)(2a);
    sum2+=y;
}
float ans= sum1-sum2;
std::cout << std::setprecision(6) << std:: fixed << ans;
}
}

Grading criterion: Alternating signs
Rating A: correctly changes the sign +ve to -ve and vice versa
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: You are given an infinite alternating series:
1-1/2!+1/3!-1/4!+1/5!  ... 
Your task is to compute the sum of the first N terms of this series. The series alternates in sign, with each term involving the reciprocal of a factorial.

Example: If n = 1, the n terms of the series is just 1. Hence, answer is 1.000000

Input Format: A single positive integer N where 1 ≤ N ≤ 15.

Output Format
	A single floating-point number representing the sum of the first N terms of the series, rounded to exactly 6 decimal places. 
	To print exactly six digits after the decimal point, write std::cout << std::setprecision(6) << std::fixed << ans;. Here we assume that the answer is stored in variable ans.

Student submission:
#include<simplecpp>
#include<iomanip>
main_program{
    int N;
    float a=1;
    float sum=0;
    cin>>N;
    for(int i=0;i<N;i++){
        
        sum=sum+(a/(i+1));
        a=((a)*(-1))/(i+1);
    }
    cout<<setprecision(6)<<fixed<<sum;
}

Grading criterion: Alternating signs
Rating A: correctly changes the sign +ve to -ve and vice versa
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: You are given an infinite alternating series:
1-1/2!+1/3!-1/4!+1/5!  ... 
Your task is to compute the sum of the first N terms of this series. The series alternates in sign, with each term involving the reciprocal of a factorial.

Example: If n = 1, the n terms of the series is just 1. Hence, answer is 1.000000

Input Format: A single positive integer N where 1 ≤ N ≤ 15.

Output Format
	A single floating-point number representing the sum of the first N terms of the series, rounded to exactly 6 decimal places. 
	To print exactly six digits after the decimal point, write std::cout << std::setprecision(6) << std::fixed << ans;. Here we assume that the answer is stored in variable ans.

Student submission:
#include <simplecpp>
#include<iomanip>

int main(){
    int N;
    cin>>N;
    float i,x,n;
    i=1;x=2;n=2;
    repeat(N-1){
        i=i+1/n*pow(-1,x-1);
        n=x*(x+1);x++;
    }
    std::cout<<std::setprecision(6)<<std::fixed<<i;
}

Grading criterion: Alternating signs
Rating A: correctly changes the sign +ve to -ve and vice versa
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: You are given an infinite alternating series:
1-1/2!+1/3!-1/4!+1/5!  ... 
Your task is to compute the sum of the first N terms of this series. The series alternates in sign, with each term involving the reciprocal of a factorial.

Example: If n = 1, the n terms of the series is just 1. Hence, answer is 1.000000

Input Format: A single positive integer N where 1 ≤ N ≤ 15.

Output Format
	A single floating-point number representing the sum of the first N terms of the series, rounded to exactly 6 decimal places. 
	To print exactly six digits after the decimal point, write std::cout << std::setprecision(6) << std::fixed << ans;. Here we assume that the answer is stored in variable ans.

Student submission:
#include<simplecpp>
#include<iomanip>
int main()
{
    int n, factorial=1;
    float ans=0;
    cin>>n;
    for (int i=1;i<=n;i++)
    {
        factorial*=i;
        ans=ans+pow(-1,i+1)/factorial;
    }
   std:: cout<< std::setprecision(6)<<std::fixed<<ans; 
}

Grading criterion: Alternating signs
Rating A: correctly changes the sign +ve to -ve and vice versa
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: You are given an infinite alternating series:
1-1/2!+1/3!-1/4!+1/5!  ... 
Your task is to compute the sum of the first N terms of this series. The series alternates in sign, with each term involving the reciprocal of a factorial.

Example: If n = 1, the n terms of the series is just 1. Hence, answer is 1.000000

Input Format: A single positive integer N where 1 ≤ N ≤ 15.

Output Format
	A single floating-point number representing the sum of the first N terms of the series, rounded to exactly 6 decimal places. 
	To print exactly six digits after the decimal point, write std::cout << std::setprecision(6) << std::fixed << ans;. Here we assume that the answer is stored in variable ans.

Student submission:
#include <simplecpp>
#include <iomanip>

double factorial(int a){
    int x = 1;
    for(int i = 1; i <= a; i++){
        x *= i;
    }
    return x;
}

main_program {
    int n;
    cin >> n;

    double x = 1;
    if(n > 1){
        for(int i = 2; i <= n; i++){
            double f = 1 / factorial(i);
            if(i % 2 == 0){
                x -= f;
            } else {
                x += f;
            }
        }
    }
    cout << setprecision(6) << fixed << x;
}

Grading criterion: Alternating signs
Rating A: correctly changes the sign +ve to -ve and vice versa
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: You are given an infinite alternating series:
1-1/2!+1/3!-1/4!+1/5!  ... 
Your task is to compute the sum of the first N terms of this series. The series alternates in sign, with each term involving the reciprocal of a factorial.

Example: If n = 1, the n terms of the series is just 1. Hence, answer is 1.000000

Input Format: A single positive integer N where 1 ≤ N ≤ 15.

Output Format
	A single floating-point number representing the sum of the first N terms of the series, rounded to exactly 6 decimal places. 
	To print exactly six digits after the decimal point, write std::cout << std::setprecision(6) << std::fixed << ans;. Here we assume that the answer is stored in variable ans.

Student submission:
#include<simplecpp>
#include<iomanip>
main_program{
    float s=0,n;
    cin>>n;
    for( float a=1,t=1; a<=n ; a++)
    {
        t=t/a;
        s=s+(pow((-1),a+1))*t ;

    }
 std::cout<< std::setprecision(6) << std::fixed << s ;
}

Grading criterion: Alternating signs
Rating A: correctly changes the sign +ve to -ve and vice versa
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: You are given an infinite alternating series:
1-1/2!+1/3!-1/4!+1/5!  ... 
Your task is to compute the sum of the first N terms of this series. The series alternates in sign, with each term involving the reciprocal of a factorial.

Example: If n = 1, the n terms of the series is just 1. Hence, answer is 1.000000

Input Format: A single positive integer N where 1 ≤ N ≤ 15.

Output Format
	A single floating-point number representing the sum of the first N terms of the series, rounded to exactly 6 decimal places. 
	To print exactly six digits after the decimal point, write std::cout << std::setprecision(6) << std::fixed << ans;. Here we assume that the answer is stored in variable ans.

Student submission:
#include<simplecpp>
#include<iomanip>

main_program
{
    int N;
    float ans;
    cin>>N;
    for(int i=1;i<=N;i++)
    do()
    {
    std::cout<<std::setprecision(6)<<std::fixed<<ans;
    }

}

Grading criterion: Alternating signs
Rating A: correctly changes the sign +ve to -ve and vice versa
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: C
Problem: You are given an infinite alternating series:
1-1/2!+1/3!-1/4!+1/5!  ... 
Your task is to compute the sum of the first N terms of this series. The series alternates in sign, with each term involving the reciprocal of a factorial.

Example: If n = 1, the n terms of the series is just 1. Hence, answer is 1.000000

Input Format: A single positive integer N where 1 ≤ N ≤ 15.

Output Format
	A single floating-point number representing the sum of the first N terms of the series, rounded to exactly 6 decimal places. 
	To print exactly six digits after the decimal point, write std::cout << std::setprecision(6) << std::fixed << ans;. Here we assume that the answer is stored in variable ans.

Student submission:
#include<simplecpp>
#include<math.h>
#include<iomanip>
    int factorial(int x) {
        int fac=1;
    for(int j=1; j<=x; j++) {
        fac = fac*j;
    }
        return fac;
    }
    main_program {
    int N;
    cin>>N;
    float ans=0;
    for(int i=1; i<=N; i++) {
        ans=ans-(pow(-1,i)/factorial(i));
    }
    std::cout<< std::setprecision(6)<< std::fixed << ans;
}

Grading criterion: Alternating signs
Rating A: correctly changes the sign +ve to -ve and vice versa
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: You are given an infinite alternating series:
1-1/2!+1/3!-1/4!+1/5!  ... 
Your task is to compute the sum of the first N terms of this series. The series alternates in sign, with each term involving the reciprocal of a factorial.

Example: If n = 1, the n terms of the series is just 1. Hence, answer is 1.000000

Input Format: A single positive integer N where 1 ≤ N ≤ 15.

Output Format
	A single floating-point number representing the sum of the first N terms of the series, rounded to exactly 6 decimal places. 
	To print exactly six digits after the decimal point, write std::cout << std::setprecision(6) << std::fixed << ans;. Here we assume that the answer is stored in variable ans.

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

 int fact(int x){
    int f=1;   
    for(int i=1 ;i<=x; i++){
        f*=i ;
    }
    return  f;
 }
 int main(){
     int n;
     cin>> n ;
     float sum=0;
     
     for ( int i=1 ;i<=n ;i++){
        if( n%2==0){
            fact(i);
            int a=1/ (fact(i));
            sum+=a ;
        }
        if( n%2!=0){
            fact(i);
            int a= 1/(fact(i));
            sum-=a;
        }
        cout << sum ;
     }
      cout << endl;
    }

Grading criterion: Alternating signs
Rating A: correctly changes the sign +ve to -ve and vice versa
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: B
Problem: You are given an infinite alternating series:
1-1/2!+1/3!-1/4!+1/5!  ... 
Your task is to compute the sum of the first N terms of this series. The series alternates in sign, with each term involving the reciprocal of a factorial.

Example: If n = 1, the n terms of the series is just 1. Hence, answer is 1.000000

Input Format: A single positive integer N where 1 ≤ N ≤ 15.

Output Format
	A single floating-point number representing the sum of the first N terms of the series, rounded to exactly 6 decimal places. 
	To print exactly six digits after the decimal point, write std::cout << std::setprecision(6) << std::fixed << ans;. Here we assume that the answer is stored in variable ans.

Student submission:
#include<simplecpp>
#include<iomanip>
//;
main_program{
    int N;
    cin>>N;
  
int n=1;
float Sn=0;
  while(n<=N){
        int i=1;
        int p=i;
        while(i<=n){
            
            p=p*i;
            i++;
           
        }
        

        float T = (pow(-1,n+1))/p;// term nth is defined by this exp
     Sn = Sn + T;
// Now, we are defining the sum of n terms
n++;

}
std::cout<< std::setprecision(6) << std::fixed << Sn;
}

Grading criterion: Alternating signs
Rating A: correctly changes the sign +ve to -ve and vice versa
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: You are given an infinite alternating series:
1-1/2!+1/3!-1/4!+1/5!  ... 
Your task is to compute the sum of the first N terms of this series. The series alternates in sign, with each term involving the reciprocal of a factorial.

Example: If n = 1, the n terms of the series is just 1. Hence, answer is 1.000000

Input Format: A single positive integer N where 1 ≤ N ≤ 15.

Output Format
	A single floating-point number representing the sum of the first N terms of the series, rounded to exactly 6 decimal places. 
	To print exactly six digits after the decimal point, write std::cout << std::setprecision(6) << std::fixed << ans;. Here we assume that the answer is stored in variable ans.

Student submission:
#include<simplecpp>
main_program {
    int n,i=1,z=2;
    float y=0,x=1;
    cin>>n;
    while(i<=n){
        x=x*i;
        y=y+pow(-1,z)*(1/x);
        i++;
        z++;
        } 
        cout<<y;
        
}

Grading criterion: Alternating signs
Rating A: correctly changes the sign +ve to -ve and vice versa
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A