题解 | #n的阶乘#

#include <iostream>
using namespace std;

long long  jiecheng(long long x){
    if(x == 1) return 1;
    else return x * jiecheng(x - 1);
}

int main() {
    long long  n;
    while(cin >> n){
        cout << jiecheng(n) << endl;
    }
}