题解 | #n的阶乘#
n的阶乘
https://www.nowcoder.com/practice/97be22ee50b14cccad2787998ca628c8
#include<bits/stdc++.h>
using namespace std;
long jiecheng(int n) {//返回值需要为long,否则13之后的数会因为结果过大而溢出
if (n == 0 || n == 1) {
return 1;
} else {
return n * jiecheng(n - 1);
}
}
int main() {
int n;
cin >> n;
long result = jiecheng(n);
cout << result;
}
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