Coding Challenges <C++>

What will this program output?

c++
#include <iostream>

unsigned long long factorial(int n) {
    if (n == 0 || n == 1) {
        return 1;
    }
    return n * factorial(n - 1);
}

int sumOfDigits(unsigned long long n) {
    int sum = 0;
    while (n != 0) {
        sum += n % 10;
        n /= 10;
    }
    return sum;
}

int main() {
    unsigned long long fact = factorial(10);
    std::cout << sumOfDigits(fact) << std::endl;
    return 0;
}

c++
#include <iostream>

unsigned long long factorial(int n) {
    if (n == 0 || n == 1) {
        return 1;
    }
    return n * factorial(n - 1);
}

int sumOfDigits(unsigned long long n) {
    int sum = 0;
    while (n != 0) {
        sum += n % 10;
        n /= 10;
    }
    return sum;
}

int main() {
    unsigned long long fact = factorial(10);
    std::cout << sumOfDigits(fact) << std::endl;
    return 0;
}

Solution:

|| I'm sorry, my post will be long. I just want to follow the code process Inside factorial(10) fac(10) -> ret 10 * fac(9) -> 10 * 362,880 = 3,628,800...

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2 Replies

Flo•3mo ago

Spoiler: 27 ...is that correct?

Solution

John Eric•3mo ago

I'm sorry, my post will be long. I just want to follow the code process Inside factorial(10) fac(10) -> ret 10 * fac(9) -> 10 * 362,880 = 3,628,800 fac(9) -> ret 9 * fac(8) -> 9 * 40,320 = 362,880 fac(8) -> ret 8 * fac(7) -> 8 * 5,040 = 40,320 fac(7) -> ret 7 * fac(6) -> 7 * 720 = 5,040 fac(6) -> ret 6 * fac(5) -> 6 * 120 = 720 fac(5) -> ret 5 * fac(4) -> 5 * 24 = 120 fac(4) -> ret 4 * fac(3) -> 4 * 6 = 24 fac(3) -> ret 3 * fac(2) -> 3 * 2 = 6 fac(2) -> ret 2 * fac(1) -> 2 * 1 = 2 fac(1) -> ret 1 Inside sumOfDigits(3,628,800) sum = 0 + 0 + 8 + 8 + 2 + 6 + 3 sum = 27 That would be 27. Again I'm sorry for the long post 😅

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Coding Challenges <C++>