题解 | #2的n次方计算#
2的n次方计算
https://www.nowcoder.com/practice/35a1e8b18658411388bc1672439de1d9
#include <iostream>
using namespace std;
int main() {
int n;
int sum=1;
cin>>n;
//for(int i=0;i<n;i++){
// sum*=2;
//}
cout<<(1<<n);
}
1<<n 就是2的n次方
