题解 | #n的阶乘#

#include<iostream>
using namespace std;
int main(){
    long long int f[21];
    int n;
    cin>>n;
    f[0]=1;
    f[1]=1;
    for(int i=1;i<=n;++i){
        f[i] = f[i-1]*i;
    }
    cout<<f[n];
    return 0;
}