A1024 Palindromic Number [回文+大整数加法]

//找质数，找能除的质数
#include<iostream>
#include<cstring>
#include<algorithm>
#include<string>
#include<time.h>
#include<math.h>
using namespace std;
struct bign
{
	int d[1001];
	int length;
	bign()
	{
		memset(d, 0, sizeof(d));
		length = 0;
	}
};
bign change(char str[])
{
	bign a;
	a.length = strlen(str);
	for (int i = 0; i < a.length; i++)
	{
		a.d[i] = str[a.length - i - 1] - '0';
	}
	return a;
}
bign add(bign a, bign b)
{
	bign c;
	int carry = 0;
	for (int i = 0; i < a.length||i<b.length; i++)
	{
		int temp = a.d[i] +b.d[i] + carry;
		c.d[c.length++] = temp % 10;
		carry = temp / 10;
	}
	if (carry != 0)
	{
		c.d[c.length++] = carry;
	}
	return c;
}
bool judge(bign a)
{
	for (int i = 0; i <= a.length / 2; i++)
	{
		if (a.d[i] != a.d[a.length - 1 - i])
			return false;
	}
	return true;
}
void print(bign a)
{
	for (int i = a.length - 1; i >= 0; i--)
	{
		cout << a.d[i];
	}
}
int main()
{
	char c[1001];
	int count = 0, m;
	cin >> c >> m; 
	bign a = change(c);
	while (!judge(a)&&count<m)
	{
		bign b = a;
		reverse(b.d, b.d + b.length);
		a = add(a, b);
		count++;
	}
	print(a);
	cout << endl << count;

}