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 <iostream>
#include <iomanip>
using namespace std;
int main() {
    int n, s=1; cin >> n;
    float sum=0;
    for (int i=1;i<=n;sum+=float((i%2)*2-1)/s, s*=++i);
    cout << fixed << setprecision(6) << sum << "\n";
}

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 a=0,b=1,c,d;
    cin>>N;
    for(int i=0;i<N;i++){
        
        
        
        if(b!=1){d=b*(b-1);}
        else(d=1);
        b++;
        
        a=pow(-1,i)/d;
        d=d*b;

        c=a+c;}
        
        
       std::cout<< std::setprecision(6)<<std::fixed<<c;


    }

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>

float factorial(int n) {
    float fac = 1;
    for(float i=1; i<=n; i++) {
        fac*=i;
    }
    return(fac);
}

int main() {

    int n;
    cin >> n;

    float sum = 0;

    for(int i=1; i<=n; i++) {
        if(i%2!=0) {
            sum = sum + 1/factorial(i);
        }
        else if(i%2==0) {
            sum = sum - 1/factorial(i);
        }
    }

    cout << setprecision(6)<<fixed<<sum;

    return(0);
}

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;
    double fact = 1;
    double result = 1;

    for (int i = 2; i<=n; i++) {
        fact*=i;
        double reciprocal = (1/fact);

        if (i%2==0) {
            result-=reciprocal;
        } else {
            result+=reciprocal;
        } 
        
    }
    std::cout<<std::setprecision(6)<<std::fixed<<result<<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>
main_program{
    int n;
    cin>>n;
    float sum=0;
    for(int i=1;i<=n;i++){
        float fact=1;
        for(int j=1; j<=i;j++){
            fact=fact*j;
        }
        sum=sum+(1*pow(-1, i+1))/fact;
    }
    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>
main_program {
    int N;
    cin>>N;
    int term=1;
    float series= 0;
    while(term<=N){
    int a=1,x=1;
    while(a<=term)
    {x=x*a;a++;};
    series=series + (pow(-1,(term+1)))/x;term++;}
    std::cout<<std::setprecision(6)<<std::fixed<<series;
}

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 sum=0;
 int N;
 cin>>N;
 for(int i=0; i<N;i++){
    int k=1;
    for(int j=1;j<=(i+1);j++){
        k*=j;
    }
    int m=-1;
    repeat(i+1){
     m=(m*(-1));
    }
    float l=(float)k;
    if(m==-1)
    sum=sum-(1/l);
    if(m==1)
    sum= sum+(1/l);
 }
 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>
 main_program{
    int n;
    cin>>n;
    

 }

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 a){
    int fact = 1;
    for(int i = 1; i < a + 1;i++){
        fact = fact * i;
    }
    return fact;
}
main_program{
    int N;
    cin>>N;
    int l = 1;
    float sum = 0;
    for(int i = 1; i < N + 1;i++){
        float h = factorial(i);
        float term = l / h;
        sum = sum + term;
        l = l * -1;
    }
    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>
main_program{
    
}
    int N;
    cin>>N;
    int fact=1;
    float c;
    if(N>=1&&N=15);
    {
    for(int n=1;n<=N;n++)
    {
     for(int n=1; )
    fact=fact*n;
    }
    c=c+pow(-1,(n+1)*)(1/fact);
std::cout<<std::setprecision(6) << std::fixed<<c;
    
}

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 f=1;
        int i=1;
        while(i<=n) {
            f = f*i;
            i+=1;
        }
        return f;
    }

main_program {
    int n;
    cin >> n;
    float sum=0;
    int a=1;
    while(a<=n) {
        sum += (pow(-1, (a+1)))/(fact(a));
        a+=1;
    }
    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>
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>
main_program{
    int N,k=0;
    cin>>N;
    for
    for(int r=1;r>=1;r--){
        
        
    }
    cout<<1-2*k;

}

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> 
//defining a function for getting the reciprocal of a digit
float reciprocal(float& n){
    float a=1.000000/(float(n));
    return a;
}
main_program{
    int n;
    float series=0; //final answer is saved in series variable
    cin>>n;
    float l=0;
    for (int i=2;i<=n; i++){
        //loop for factorial calculation
        for (int j=2; j<=i; j++){
            l+=log10(j);
        }
        int k=pow((-1),(i+1));
        float m=pow(10,l); //antilog calculation
        series+=(reciprocal(m))*k;
        l=0;
    }
    series=series+1;
    std::cout << std::setprecision(6) << std::fixed << series;
}

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{ double u=1,j=2; double sum=0; double t=1; int n;
cin>>n;
    while(u<=n)
    {
        t=(abs(t)*pow(-1,j))/u;
        sum=sum + t;
        u++;
        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>
using namespace std;
main_program {
        int n;
        cin >> n;
        int sign = 1;
         double sum = 0;
         int fact = 1;
        for (int i=1; i<=n; i++){
            fact = fact*i;
            sum = sum + sign*(1/double(fact));
            sign = -sign;
        }
        cout << setprecision(6) << 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
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 f(int d){
    int b=1;
    repeat(d){
        b=b*d;
        d=d-1;
    }
    return b;
}
main_program{
    int n;
    float k;
    cin >> n;
    float r=0;
    int d=1;
    repeat(n){
        int c=f(d);
        if(d%2==0){
            k=-1;
        }
        else if(d%2!=0){
            k=1;
        }
        r = r + k/c;
        d++;
    }
    std::cout<<std::setprecision(6)<<std::fixed<<r;

}

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;
    double term=1.0;
    double sum=1.0;
    for(int i=2;i<=n;i++){
          sum=sum+term*(-1)/i;
        term=term*(-1)/i;
        i++;}
    
    cout<<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
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<cmath>
main_program{
    int n;
    float c=1;
    float q;
    float r;
    
    int t=1;
    cin>>n;
    float x=1;
    float p=0;
    while(x<=n){
    while(t<=x){

            c=c*t;
            t++;
        }
         
         q=pow(-1,x+1);
         
         r=1/c;
        p=p+q*r;
        x++;
    }
    std::cout<< std::setprecision(6)<<std::fixed<<p;
}

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