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>

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: Accumulating sum
Rating A: correctly keeps accumulating the sum of the series
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: Accumulating sum
Rating A: correctly keeps accumulating the sum of the series
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: Accumulating sum
Rating A: correctly keeps accumulating the sum of the series
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;
    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: Accumulating sum
Rating A: correctly keeps accumulating the sum of the series
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: Accumulating sum
Rating A: correctly keeps accumulating the sum of the series
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>
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: Accumulating sum
Rating A: correctly keeps accumulating the sum of the series
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: Accumulating sum
Rating A: correctly keeps accumulating the sum of the series
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: Accumulating sum
Rating A: correctly keeps accumulating the sum of the series
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: Accumulating sum
Rating A: correctly keeps accumulating the sum of the series
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: Accumulating sum
Rating A: correctly keeps accumulating the sum of the series
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: Accumulating sum
Rating A: correctly keeps accumulating the sum of the series
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: Accumulating sum
Rating A: correctly keeps accumulating the sum of the series
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: Accumulating sum
Rating A: correctly keeps accumulating the sum of the series
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;

for(int j=1; j<=N; j--)  
        {
            float f=0.0;
            f=f+log10(j);
        
    float d=pow(10,f);
    float sum=0.0;
    sum=sum+(pow(-1,j-1))/d;

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

Grading criterion: Accumulating sum
Rating A: correctly keeps accumulating the sum of the series
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<iostream>
#include<iomanip>
int factorial(int n);
int factorial(int n){
    int c;
    if(n>=1){
        c=n*(factorial(n-1));
    }
    return c;
}
int main(){
         int n,f=0;cin>>n;
      for(int i=0,k;i<n;i++){
        k=factorial(i);
        if(n%2==1){
            f=(1/k)+f;
        }
        else {
            f=-(1/k)+f;
        }
         
      }
      std::cout<<std::setprecision(6)<<std::fixed<<f;

}

Grading criterion: Accumulating sum
Rating A: correctly keeps accumulating the sum of the series
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<iomanip>
int factorial(int n){
    int fac=1;
    for(int i=2;i<=n;i++){
        fac *= i;
    }
    return fac;
}
main_program{
    int n;
    cin>>n;
    if(n==1){
        float ans =1;
        std::cout<<std::setprecision(6)<<std::fixed<<ans;
    }
    else{
        float sum = 1;
    for(int i=2;i<=n;i++){
        float value = 1.0/(factorial(i));
        sum += (pow(-1,i))*(-value);
    }
    std::cout<<std::setprecision(6)<<std::fixed<<sum;
    }
    
}

Grading criterion: Accumulating sum
Rating A: correctly keeps accumulating the sum of the series
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) {
    int f = 1;
    for (int i = 1; i <= n; ++i) {
            f = f*i;
    }
    return f;
}

float recip(int n) {
    float r = 1/(factorial(n));
    return r;
}

main_program {
    int n;
    cin >> n;
    float ans = 0;
    for (int i = 1; i <= n; ++i) {
        if (i%2 == 0) {
            ans = ans - recip(i);
        }
        else {
            ans = ans + recip(i);
        }
    }
    std::cout << std::setprecision(6) << std::fixed << ans;
}

Grading criterion: Accumulating sum
Rating A: correctly keeps accumulating the sum of the series
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>
#include<iostream>
void factorial(int N){
    for (k=1;k<=N;k++){
    //defining a function for factorial
                factorial*=k
}
}
int N;
cin>>N;
float sum;
float summ;
main_program{
    for (i=2; i<=N; i=i+2){
        sum+=1/factorial(i)
}
    for (j=1;j<=N; j=j+2){
        summ+=1/factorial(j)
    }
    float ans;
    ans=(sum-summ)
}
std::cout<<std::setprecision(6)<<std::fixed<<ans;

Grading criterion: Accumulating sum
Rating A: correctly keeps accumulating the sum of the series
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<iomanip>
main_program 
{  int n;
float Sum=0,f=1,i=1;
   cin>>n;
    while(i<=n)
   {
      Sum+=f;
      f=f/-(i+1);
      i++;
   }
 std:: cout << std ::setprecision(6)<<std :: fixed<< Sum<<endl;
}

Grading criterion: Accumulating sum
Rating A: correctly keeps accumulating the sum of the series
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A