Factorial digit sum (without BigInt) C++

Question

I am trying to figure out how to solve this problem (Project Euler):

n! means n × (n − 1) × ... × 3 × 2 × 1

For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800, and the sum of the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.

Find the sum of the digits in the number 100!

Using BigInt is not an option, I am trying to figure out how to implement a solution using only c++.

I thought maybe splitting the big numbers to an array of about 7 digit's long or something like that and then deal with them but i still cant figure out how to do this either..

thanks in advance!

Answer 1

Try this

#include "iostream"
#include "vector"

int n,remainder,sum;
int main ()
{   
    std::vector <int> digits(5000);    
    std::cin>>n;    
    digits[0]=1;
    digits[1]=1;
    for (int k=2;k<n+1;k++) {
        for (int i=1;i<=digits[0];i++) {
            digits[i]=digits[i]*k+remainder;
            remainder=0;
            if (digits[i]>9) {
                remainder=digits[i]/10;
                digits[i]%=10;
                if (i==digits[0])
                digits[0]++;
            }
        }
    }   
    for (int i=digits[0];i>=1;i--)
        sum+=digits[i];
    std::cout<<sum;
}