判断一个整数的N进制是否为回文

#include <stdio.h>
#include <stdlib.h>

int circle(int n,int d);

int main(int argc,char argv[]){

	int num = 0,d = 0;
	printf("请输入数和进制：");
	scanf("%d %d",&num,&d);
	printf("%d的%d进制是回文吗？%d\n",num,d,circle2(num,d));
	//exit(0);
}

/**
  *判断n的d进制数是否为回文
  *@return 0不是回文，1是回文
  */
int circle(int n,int d){

	int i = 0,size = 0;
	int a[10] = {0};
	int iscircle = 1;
	while(n){
		if(n % d != 0){
			a[i++] = n % d;
		}else{
			a[i++] = 0;
		}
		n = n/d;
	}
	size = i;
	for(i = 0;i <= size-1;i++,size--){
		if(a[i] != a[size-1]){
			iscircle = 0;
			break;
		}
	}
	return iscircle;
}

/**
  *方式二
  */
int circle2(int n,int d){

	int s = 0,m = n;
	while(m){
		s = s*d + m%d;
		m /= d;
	}
	return s == n;
}