Find sum of digits in factorial of a number in C++

Suppose, we have a number n, our task is to find the sum of digits in then!. Consider n = 5, then n! = 120. So the result will be 3.

To solve this problem, we will create a vector to store factorial digits and initialize it with 1. Then multiply 1 to n one by one to the vector. Now sum all the elements in the vector and return the sum

Example

 Live Demo

#include<iostream> #include<vector> using namespace std; void vectorMultiply(vector<int> &v, int x) {    int carry = 0, res;    int size = v.size();    for (int i = 0 ; i < size ; i++) {       int res = carry + v[i] * x;       v[i] = res % 10;       carry = res / 10;    }    while (carry != 0) {       v.push_back(carry % 10);       carry /= 10;    } } int digitSumOfFact(int n) {    vector<int> v;    v.push_back(1);    for (int i=1; i<=n; i++)       vectorMultiply(v, i);    int sum = 0;    int size = v.size();    for (int i = 0 ; i < size ; i++)       sum += v[i];    return sum; } int main() {    int n = 40;    cout << "Number of digits in " << n << "! is: " << digitSumOfFact(n); }

Output

Number of digits in 40! is: 189