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