题解 | #n的阶乘#
n的阶乘
https://www.nowcoder.com/practice/97be22ee50b14cccad2787998ca628c8
#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; }