1079 延迟的回文数 (20分)

思路：这道题，我用了两个函数，第一个函数判断是否是回文数，第二个函数里面将传入的s进行反转（注意，PTA上的编译器不支持C中的反转函数strver())，再进行相加，再进行输出，最后返回相加后的结果，然后利用strcpy（）函数，将相加后的结果重新赋值给s

#include
#include
#include
int ishuiwen(char s[])
{
 int len = strlen(s)-1;
 for (int i = 0; i <= len ; i++)
  if (s[i] != s[len - i]) return 0;
 return 1;
}
char* zhuan(char s[])
{
 char s1[10000], s2[10000] , s3[10000];
 int k = 0,len=strlen(s),tmp=0,n=0,k1=0;
 for (int i = len - 1; i >= 0; i--)
  s1[k++] = s[i];
 s1[k] = '\0';
 for (int i = 0; i < len; i++)
 {
  int t = (s[i] - '0') + (s1[i] - '0')+n;
  s2[tmp++] = t % 10+'0';
  n = t / 10;
 }
 if (n > 0) s2[tmp++] = n + '0';
 s2[tmp] = '\0';
 for (int i = tmp - 1; i >= 0; i--)
  s3[k1++] = s2[i];
 s3[k1] = '\0';
 printf("%s + %s = %s\n", s, s1,s3);
 return s3;
}
int main()
{
 char s[10000];
 scanf("%s", s);
 if (ishuiwen(s)==1) printf("%s is a palindromic number.\n", s);
 else {
  int cot = 0;
  while (cot<10)
  {
   strcpy(s, zhuan(s));
   if (ishuiwen(s)==1)
   {
    printf("%s is a palindromic number.\n", s);
    break;
   }
   cot++;
  }
  if (cot == 10) printf("Not found in 10 iterations.\n");
 }
 system("pause");
 return 0;
}