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;
}
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