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 n = 0;
scanf("%d",&n);
int m = pow(2,n); //pow(a,b)等于a的b次方
printf("%d",m);
return 0;
}
2的n次方计算
https://www.nowcoder.com/practice/35a1e8b18658411388bc1672439de1d9
#include <stdio.h>
int main()
{
int n = 0;
scanf("%d",&n);
int m = pow(2,n); //pow(a,b)等于a的b次方
printf("%d",m);
return 0;
}
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