Large Factorial

• Time : $O(log{10}(n!) * n)$
• Space : $O(n)$

string mul(string A, string B) {
    string res = string(A.length() + B.length() + 1, '0');
    reverse(begin(A), end(A));
    reverse(begin(B), end(B));
    for(int i = 0; i < A.length(); i++) {
        int carry = 0;
        for(int j = 0; j < B.length(); j++) {
            int now = res[i + j] - '0' + (A[i] - '0') * (B[j] - '0') + carry;
            carry = now / 10;
            res[i + j] = (now % 10) + '0';
        }
        int p = i + B.length();
        while(carry) {
            int now = res[p] - '0' + carry;
            carry = now / 10;
            res[p] = (now % 10) + '0';
            p += 1;
        }
    }
    
    
    while(res.back() == '0') res.pop_back();
    reverse(begin(res), end(res));
    return res;
}
string Solution::solve(int A) {
    string res = "1";
    for(int i = 2; i <= A; i++) res = mul(res, to_string(i));
    return res;
}