C++ Sum of factorial -Request 2nd code - c++

Question

I would like to analyze the complexity of my code algorithm.Therefore,i must have 2 different programs giving the same functions to allow me to start off.
Currently this is my own code.
I'm not sure if it is allowed that i would like to have someone that could volunteer his own way code to compute summation of factorial for me as the 2nd program code.
Preferrably a nested loop.
#include <iostream>
using namespace std;
int main()
{
int val;
int i;
int a = 0;
int c = 1;
cout << "Please enter a number: ";
cin >> val;
cout << endl;
for (i = 1; i <= val; i++)
{
c = c * i;
a = a + c;
}
cout << "The sum of the factorials is " << a << endl;
system("pause");
return 0;
}

Answer 1

#include <iostream>
using namespace std;
int main()
{
int val;
cout << "Please enter a number: ";
cin >> val;
cout << endl;
static const int results[] = {
0, 1, 3, 9, 33, 153, 873, 5913, 46233, 409113,
4037913, 43954713, 522956313
};
cout << "The sum of the factorials is " << results[val < 0 ? 0 : val] << endl;
system("pause");
return 0;
}
Note that I replicated the defect in the original program which causes it to return the incorrect value if the user enters 0.
This alternate version assumes 32-bit integers because it takes advantage of overflow behavior. Extending to 64-bit integers is left as an exercise.

Answer 2

I do not understand what you do with another nested way but i hope this can help...
#include <iostream>
using namespace std;
int main()
{
int val;
int i;
int a = 0;
int c = 1;
cout << "Please enter a number: ";
cin >> val;
cout << endl;
for (i = 1; i <= val; i++){
c *= i;
a += c;
}
int c2=1;
for (i = val; i > 1; i--){
c2*=i;
c2++;
}
cout << "The sum of the factorials is " << a << endl;
cout << "The sum of the factorials is " << c2 << endl;
system("pause");
return 0;
}

Answer 3

#include <iostream>
using namespace std;
int main()
{
int suma = 0;
int n = 0;
cout << "Sum of factorials\n";
cout << "-------------------------------\n";
cout << "Insert number of n: ";
cin >> n;
int i = 1;
while (i <= n)
{
int factorial = 1;
for(int j=1; j<=i; j++)
{
factorial = factorial * j;
}
suma += factorial;
i++;
}
cout << "Sum of factorials is: " << suma;
system("pause");
return 0;
}