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>
#include <math.h>
int main() {
int n = 0 ;
double m ;
scanf("%d",&n);
m = (double)n ;
if (n>=0&&n<=31) {
printf("%.0lf",pow(2.0,m));
}
return 0;
}
