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>
#include<math.h>
int factorial(int x){
    int product=1;
    for(int i=1;i<=x;i++){
        product=product*i;
    }
    return product;
}
main_program{
int N;
cin>>N;
float ans=0.0000000;
for(int i=1;i<=N;i++){
ans= ans+ pow(-1,i-1)*1/factorial(i);
}
std::cout<<std::setprecision(6)<<std::fixed<<ans;
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>
int factorial(int n){
    int y=1;
    for(int i=1;i<=n;i++){
        y=y*i;
    } return y;
}
main_program{
    int N;
    cin>>N;
    float x=0;
    for(int i=1;i<=N; i++){
       float d=1;
    
        repeat(i+1){
            d=d*(-1);
        }
        float z= factorial(i);
        x = (x + d/z);
        

    } 
    std::cout << std::setprecision(6) << std::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>

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

main_program
{
    int n; cin>>n;

    float sum=0;

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

        else
        {
            sum -= (1/factorial(i));
        }
    }
    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;
    double sum = 0;

    for (int i = 1; i <= n; i++){
        
        int fact = 1;
        for (int x = 1; x <= i; x++){ fact = fact*x;}
        if (i %2 == 0){
            sum -= 1.0/fact;
        }
        else {
            sum += 1.0/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 fact=1;
    float c;
    if(N>=1&&N<=15)
    {
    for(int n=1;n<=N;n++)
    {
    for(int i=1;i<=n;i++)
    {
        fact=fact*i;
    }
    c=c+(pow(-1,(n+1)))*(pow(fact,-1));
    }
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: 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 <cmath>
#include <iomanip>

int fact(int n);

main_program{
    int n;
    cin>>n;
    double sum=0.0;
    for(int i=1;i<=n;i++)
    {
        sum+=(pow(-1,(i-1))/fact(i));
    }
    cout<<fixed<<setprecision(6)<<sum<<endl;
}

int fact(int n){
    int prod=1;
    for(int i=1;i<=n;i++)
    {
        prod*=i;
    }
    return prod;
}

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 fact(int num){
    int pro=1;
    for(int i=1; i<=num; i+=1){
        pro*=i;
    }
    return pro;
}

main_program{
    float sum=0, sign=1, i=1;
    int n;
    cin >> n;
    while(i<=n){
        sum=sum+ (sign*(1/fact(i)));
        sign*=(-1);
        i+=1;
    }
    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
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(float a){
    float b=1.0;
    float product =1.0;
    while (b<=a){
        product=product*b;
        b=b+1;
    }
    return(product);
}
int main(){
    float n;
    cin>>n;
    float i=1.0;
    float sum=0.0;
    while (i<=n){
        float x=1.0/factorial(i);
        float z=0.0;
        int s=int(i)%10;

        if (s==2){
            sum=sum-x;
        }
        else if(s==4){
            sum=sum-x;
        }
        else if(s==6){
            sum=sum-x;
        }
        else if(s==8){
            sum=sum-x;
        }
        else if(s==0){
            sum=sum-x;
        }
        else{
            sum=sum+x;
        }
        i=i+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<iostream>
#include<iomanip>


float factorial(float n){
    float a=1;
    for(;n>=1;n-=1){
        a=a*n;
    }
    return a;
}

int main(){
    int n;float a=1;int d=1;float r=2;
    std::cin>>n;
    for(;n>1;n-=1){
        if(d%2==1){
            a=a-(1.0/factorial(r));
        }
        if(d%2==0){
            a=a+(1.0/factorial(r));

        }
        d+=1;
        r+=1;

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

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;
    double ans=0,num1,num2,num3;
    cin>>N;
    for(int p=1;p<=N;p++){
        double prod=1;
        for(int q=p;q>0;q--){
            prod=prod*q;
        }
     num1=1/prod;
     num2=pow(-1,p-1);
      num3=num1*num2;
      ans=ans+num3;
    }
    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{
    

    }


}

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>
main_program{
    cout << "Enter the Integer N";
    int N;
    cin >> N;
    for("i = N;(-1)^N * 1/N!");
    cout << "The Answer of the given question is as follows";
    cin >> "i==" << 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: 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<iostream>
#include<iomanip>
using namespace std;
int main(){
        float term = 1;
        float out = 0;
        int n; cin >> n;
        for (int i = 1; i<=n; i++){
                out += term;
                term = -term/(i+1);
        }
        cout << setprecision(6) << fixed << out;
}

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 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: 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 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: 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 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: 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(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: 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 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: 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 s=0;
    for(int i=1;i<=n;i++)
    {
        
        int f=1;
        while(f<=i)
        {
            f*=i;
            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