C语言编程数位的调换#include<stdio.h>int main(void){\x05int n,m,i,count=0,a,s=0;\x05scanf("%d",&n);\x05m=n;\x05while(n!=0)\x05{n=n/10;count++;} /*计算位数*/\x05\x05for(i=0;i<count

C语言编程数位的调换#include<stdio.h>int main(void){\x05int n,m,i,count=0,a,s=0;\x05scanf("%d",&n);\x05m=n;\x05while(n!=0)\x05{n=n/10;count++;} /*计算位数*/\x05\x05for(i=0;i<count
C语言编程数位的调换
#include<stdio.h>
int main(void)
{
\x05int n,m,i,count=0,a,s=0;
\x05scanf("%d",&n);
\x05m=n;
\x05while(n!=0)
\x05{n=n/10;count++;}                                     /*计算位数*/
\x05
\x05for(i=0;i<count;i++)
\x05{
\x05\x05a=m%10;
\x05\x05s=s+a*pow(10,(count-i));                      /*首尾调换*/
\x05\x05m=m/10;
\x05}
\x05printf("%d",s);
\x05return 0;
}

为什么会多出一位?

C语言编程数位的调换#include<stdio.h>int main(void){\x05int n,m,i,count=0,a,s=0;\x05scanf("%d",&n);\x05m=n;\x05while(n!=0)\x05{n=n/10;count++;} /*计算位数*/\x05\x05for(i=0;i<count
s = s + a * pow(10, count - i - 1);
//                            ^^^  要再减去1才对