Question

编写程序，计算2的N次幂(使用while循环)

Answer 1

#include<stdio.h>
void main(){
int i=1,n=0,count=0;
printf("请输入N：");
scanf("%d",&n);
if(n==0){
i=1;
printf("2的%d次幂是：%d\n",n,i);
}else if(n>0){
while(count<=n){
count++;
i*=2;
}
printf("2的%d次幂是：%d\n",n,i);
}else{
count=(n*(-1));
while(count!=0){
count--;
i*=2;
}
printf("2的%d次幂是：%lf\n",n,(1.0/i));
}
}

Answer 2

#include<stdio.h>
void main()
{
int i=1,n=0,count=0;
printf("请输入N：");
scanf("%d",&n);
if(n==0)
{
i=1;
printf("2的%d次幂是：%d\n",n,i);
}
else if(n>0)
{
while(count<=n)
{
count++;
i*=2;
}
printf("2的%d次幂是：%d\n",n,i);
}
else
{
count=(n*(-1));
while(count!=0)
{
count--;
i*=2;
}
printf("2的%d次幂是：%lf\n",n,(1.0/i));
}
}