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{
    
}
    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: Output
Rating A: correctly prints the output
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: Output
Rating A: correctly prints the output
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: Output
Rating A: correctly prints the output
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: Output
Rating A: correctly prints the output
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: Output
Rating A: correctly prints the output
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: Output
Rating A: correctly prints the output
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: Output
Rating A: correctly prints the output
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: Output
Rating A: correctly prints the output
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: Output
Rating A: correctly prints the output
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: Output
Rating A: correctly prints the output
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 <iomanip>
#include <simplecpp>
main_program{
    float N;
    cin >> N;
    float ans=0;
    float fac=1.0;
    for (int i=1; i<=N; i++) {
        for(int j=1; j<=i; j++) {
        fac*=j;
    }
    if(i%2==1)
    ans=ans+1/(fac);
    else
    ans=ans-1/(fac);
    fac=1;
    }
    std::cout << std::setprecision(6)<<std::fixed << ans;
}

Grading criterion: Output
Rating A: correctly prints the output
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: A sequence of characters ‘(‘ and ‘)’ is said to be balanced or valid, if it can be obtained from a valid mathematical expression by erasing all other symbols. For instance, from a mathematical expression like (1+2)*((3+4)*(1+2)) yields a balanced sequence ()(()()). On the other hand, the sequence (() or the sequence )( is not balanced. Write a C++ program to read a sequence of characters consisting of characters ‘(‘ and ‘)’ and identify if it is VALID or INVALID. To check whether a sequence is valid, it is enough to keep track of the count of ‘(‘s and ‘)’s seen at any point in the sequence, when scanning the sequence left to right. You can use a few examples to understand when a sequence becomes invalid. 
Input Format
A non-negative integer n, followed by a sequence of n characters consisting of ‘(’ and ‘)’  

Output Format
VALID N
or
INVALID N 

where N = openCount - closeCount, where openCount is the number of  ‘(’ symbols and  closeCount is the number of  ‘)’ symbols in the sequence.

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

int main(){
    int N, openCnt, balance, closeCnt;
    bool validity = true;
    cin >> N;
    openCnt = 0;
    closeCnt = 0;
    int unmatchedOpen = 0;
    for(int i = 0; i < N; i++){
        char C;
        cin >> C;
        if(C == '('){
            openCnt++;
            unmatchedOpen++;
        }
        else if(C == ')'){
            closeCnt++;
            if(unmatchedOpen != 0){
                unmatchedOpen--;
            }
        }
        balance = openCnt-closeCnt;
    }
    if(balance != 0){
        cout << "INVALID " << balance;
    }
    else{
        if(unmatchedOpen != 0){
            cout << "INVALID " << balance;
        }
        else{
            cout << "VALID " << balance;
        }
    }



}

Grading criterion: Declaration of variables and accepting input
Rating A: correctly declared and accepts input for the variable n and char for accepting paranthesis
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: A sequence of characters ‘(‘ and ‘)’ is said to be balanced or valid, if it can be obtained from a valid mathematical expression by erasing all other symbols. For instance, from a mathematical expression like (1+2)*((3+4)*(1+2)) yields a balanced sequence ()(()()). On the other hand, the sequence (() or the sequence )( is not balanced. Write a C++ program to read a sequence of characters consisting of characters ‘(‘ and ‘)’ and identify if it is VALID or INVALID. To check whether a sequence is valid, it is enough to keep track of the count of ‘(‘s and ‘)’s seen at any point in the sequence, when scanning the sequence left to right. You can use a few examples to understand when a sequence becomes invalid. 
Input Format
A non-negative integer n, followed by a sequence of n characters consisting of ‘(’ and ‘)’  

Output Format
VALID N
or
INVALID N 

where N = openCount - closeCount, where openCount is the number of  ‘(’ symbols and  closeCount is the number of  ‘)’ symbols in the sequence.

Student submission:
#include<simplecpp>
main_program{
	int n,c=0;
	char s;
	cin>>n;
	bool e=true;
	for(int i=0;i<n;i++){
		cin>>s;
		if(s == '(')
			c++;
		if(s == ')')
			c--;
		if(c<0) e=false;
	}
	if(c !=0) e=false;
	if(e) cout<<"VALID ";
	else cout<<"INVALID ";
	cout<<c;
}

Grading criterion: Declaration of variables and accepting input
Rating A: correctly declared and accepts input for the variable n and char for accepting paranthesis
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: A sequence of characters ‘(‘ and ‘)’ is said to be balanced or valid, if it can be obtained from a valid mathematical expression by erasing all other symbols. For instance, from a mathematical expression like (1+2)*((3+4)*(1+2)) yields a balanced sequence ()(()()). On the other hand, the sequence (() or the sequence )( is not balanced. Write a C++ program to read a sequence of characters consisting of characters ‘(‘ and ‘)’ and identify if it is VALID or INVALID. To check whether a sequence is valid, it is enough to keep track of the count of ‘(‘s and ‘)’s seen at any point in the sequence, when scanning the sequence left to right. You can use a few examples to understand when a sequence becomes invalid. 
Input Format
A non-negative integer n, followed by a sequence of n characters consisting of ‘(’ and ‘)’  

Output Format
VALID N
or
INVALID N 

where N = openCount - closeCount, where openCount is the number of  ‘(’ symbols and  closeCount is the number of  ‘)’ symbols in the sequence.

Student submission:
#include<simplecpp>
main_program{
	int n;
	int count1=0,count2=0,order=0;
	cin>>n;
	char ch;
	repeat(n)
	{
		cin>>ch;
		if(ch=='(')
			count1++;
		if(ch==')')
			count2++;
		if(count2>count1)
		{
			order++;
		}
	}
	if(order!=0)
		cout<<"INVALID "<<count1-count2;
	else if(count1==count2)
	{
		cout<<"VALID 0";}
	else
		cout<<"INVALID "<<count1-count2;
	return 0;
}

Grading criterion: Declaration of variables and accepting input
Rating A: correctly declared and accepts input for the variable n and char for accepting paranthesis
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: A sequence of characters ‘(‘ and ‘)’ is said to be balanced or valid, if it can be obtained from a valid mathematical expression by erasing all other symbols. For instance, from a mathematical expression like (1+2)*((3+4)*(1+2)) yields a balanced sequence ()(()()). On the other hand, the sequence (() or the sequence )( is not balanced. Write a C++ program to read a sequence of characters consisting of characters ‘(‘ and ‘)’ and identify if it is VALID or INVALID. To check whether a sequence is valid, it is enough to keep track of the count of ‘(‘s and ‘)’s seen at any point in the sequence, when scanning the sequence left to right. You can use a few examples to understand when a sequence becomes invalid. 
Input Format
A non-negative integer n, followed by a sequence of n characters consisting of ‘(’ and ‘)’  

Output Format
VALID N
or
INVALID N 

where N = openCount - closeCount, where openCount is the number of  ‘(’ symbols and  closeCount is the number of  ‘)’ symbols in the sequence.

Student submission:
#include<simplecpp>
main_program
{   
    bool validity= true;
    int n;
    char c;
    int N=0;
    cin>>n;
    repeat(n)
    {
        cin>>c;
        if(c=='(')
        N++;
        else if(c==')'&&N<=0)
        {
            validity = false;
            N--;
        }
        else if(c==')')
        N--;
    }
    if(validity)
    {
        if(N==0)
        cout<<"VALID "<<N;
        else
        cout<<"INVALID "<<N;
    

    }
    else cout<<"INVALID "<<N;
}

Grading criterion: Declaration of variables and accepting input
Rating A: correctly declared and accepts input for the variable n and char for accepting paranthesis
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: A sequence of characters ‘(‘ and ‘)’ is said to be balanced or valid, if it can be obtained from a valid mathematical expression by erasing all other symbols. For instance, from a mathematical expression like (1+2)*((3+4)*(1+2)) yields a balanced sequence ()(()()). On the other hand, the sequence (() or the sequence )( is not balanced. Write a C++ program to read a sequence of characters consisting of characters ‘(‘ and ‘)’ and identify if it is VALID or INVALID. To check whether a sequence is valid, it is enough to keep track of the count of ‘(‘s and ‘)’s seen at any point in the sequence, when scanning the sequence left to right. You can use a few examples to understand when a sequence becomes invalid. 
Input Format
A non-negative integer n, followed by a sequence of n characters consisting of ‘(’ and ‘)’  

Output Format
VALID N
or
INVALID N 

where N = openCount - closeCount, where openCount is the number of  ‘(’ symbols and  closeCount is the number of  ‘)’ symbols in the sequence.

Student submission:
#include<simplecpp>
main_program {
    int n;
    string input;
    cin>>n>>input;
    
    repeat (n) {
        //read char
    if (char==')') {
        int a;
        a++;
    }

    else if (char=='(') {
        int b;
        b++;
    }
}

    int c=b-a;
    if (char1=='(' && a==b) cout<<"VALID "<<c<<endl;
    if (a!=b || char1==')') cout<<"INVALID "<<c<<endl;
    
}

Grading criterion: Declaration of variables and accepting input
Rating A: correctly declared and accepts input for the variable n and char for accepting paranthesis
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: A
Problem: A sequence of characters ‘(‘ and ‘)’ is said to be balanced or valid, if it can be obtained from a valid mathematical expression by erasing all other symbols. For instance, from a mathematical expression like (1+2)*((3+4)*(1+2)) yields a balanced sequence ()(()()). On the other hand, the sequence (() or the sequence )( is not balanced. Write a C++ program to read a sequence of characters consisting of characters ‘(‘ and ‘)’ and identify if it is VALID or INVALID. To check whether a sequence is valid, it is enough to keep track of the count of ‘(‘s and ‘)’s seen at any point in the sequence, when scanning the sequence left to right. You can use a few examples to understand when a sequence becomes invalid. 
Input Format
A non-negative integer n, followed by a sequence of n characters consisting of ‘(’ and ‘)’  

Output Format
VALID N
or
INVALID N 

where N = openCount - closeCount, where openCount is the number of  ‘(’ symbols and  closeCount is the number of  ‘)’ symbols in the sequence.

Student submission:
#include<simplecpp>
main_program {
    int x;
    cin>>x;
    repeat(n){
    }
}

Grading criterion: Declaration of variables and accepting input
Rating A: correctly declared and accepts input for the variable n and char for accepting paranthesis
Rating B: Some issue
Rating C: Major issue
Rating D: 0
Assigned grade: C