大整数运算（三）（PAT）A1024.Palindromic Number

题目描述：

A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers.

Non-palindromic numbers can be paired with palindromic ones via a series of operations. First, the non-palindromic number is reversed and the result is added to the original number. If the result is not a palindromic number, this is repeated until it gives a palindromic number. For example, if we start from 67, we can obtain a palindromic number in 2 steps: 67 + 76 = 143, and 143 + 341 = 484.

Given any positive integer N, you are supposed to find its paired palindromic number and the number of steps taken to find it.

我的解题模板

代码：

#include
#include
#include
using namespace std;
struct bign{
	int d[1000];
	int len;
	bign(){
		memset(d,0,sizeof(d));
		len=0;
	}
};

bign change(char str[]){
	bign a;
	a.len=strlen(str);
	for(int i=0;i=0;i--){
		printf("%d",a.d[i]);
	}
	printf("\n");
}

bool Judge(bign a){//判断是否回文
	for(int i=0;i<=a.len/2;i++){
		if(a.d[i]!=a.d[a.len-1-i]){
			return false;//对称位置不等，则一定不回文
		}
	}
	return true;//回文
}

bign add(bign a,bign b){//高精度a+b
	bign c;
	int carry=0;//carry是进位
	for(int i=0;i