题解 | #n的阶乘#

#include <iostream>

using namespace std;
long long multi(long long n) {
    long long res = 1;
    if (n == 1) return n;
    return n * multi(n - 1);
}
int main() {
    long long x;
    while (scanf("%d", &x) != EOF) {
        cout << multi(x);
    }
    return 0;
}