题解 | #2的n次方计算#
2的n次方计算
https://www.nowcoder.com/practice/35a1e8b18658411388bc1672439de1d9
#include <stdio.h>
int main() {
int i;
int k=1;
scanf("%d",&i);//3
for(int j=1;j<=i;j++){
k*=2;//k=k*2; 2,4,8
}
printf("%d",k);
return 0;
}
