PTA 乙级 1079 延迟的回文数

代码实现：

#include<stdio.h>
#include<string.h>
#define X 1010

void summation(char *num1, char *num2)
{
	int len1 = strlen(num1), len2 = strlen(num2), len = X - 1;
	char result[X] = { 0 };

	for (int i = 0; i < len; i++)
	{
		int n1 = (i < len1 ? num1[len1 - i - 1] - '0' : 0);
		int n2 = (i < len2 ? num2[len2 - i - 1] - '0' : 0);
		int n = n1 + n2;
		result[len - i - 1] += n;
		result[len - i - 2] += result[len - i - 1] / 10;
		result[len - i - 1] %= 10;
	}
	for (int i = 0; i < len; i++)
		result[i] += '0';

	for(int i=0;i<len;i++)
		if (result[i] != 0 && result[i] != '0')
		{
			strcpy(num1, result + i);
			break;
		}
}

int isPalindromic(char *num)
{
	int len = strlen(num);
	for (int i = 0, j = len - 1; i <= j; i++, j--)
		if (num[i] != num[j])
			return 0;
	return 1;
}

int main()
{
	char num[X];
	int cnt = 0;

	gets(num);
	while (isPalindromic(num) == 0 && cnt < 10)
	{
		char num2[X];
		for (int i = strlen(num) - 1; i >= 0; i--)
			num2[strlen(num) - 1 - i] = num[i];
		num2[strlen(num)] = 0;
		printf("%s + %s = ", num, num2);
		summation(num, num2);
		puts(num);
		cnt++;
	}
	if (cnt < 10)
		printf("%s is a palindromic number.", num);
	else
		printf("Not found in 10 iterations.");

	return 0;
}