The judgement of Palindrome number (three methods)

Recently done a little about palindrome number of the summary.
First of all, write a number of methods for the determination of palindrome numbers.
The concept of palindrome numbers: that is, given a number, this number is the same as the read and reverse. For example: 121,1221 is a palindrome number, 123,1231 is not a palindrome number.
method One:
Try out situations and deal with small numbers. Use mathematical methods. The number of palindrome numbers entered is x<10^9,int storage, or the number of X<10^18,long long stored, the number of the range is not large. Here is an int storage case.

#include <stdio.h>
int main ()
{
    int x,newed,t,n;
    while (scanf ("%d", &x)!=eof)
    {
        newed=0;
        n=x;
        Do
        {
            newed=newed*10+x%10;
            x/=10;
        } while (x>0);
        if (n==newed)
            printf ("yes\n");
        else
            printf ("no\n");
    }
    return 0;
}

Method Two:
Try out scenarios and handle large numbers. Use string handling. Because palindrome number about center symmetry, as long as the more symmetrical number can.

#include <stdio.h>
#include <string.h>
int main ()
{
    int i,length,flag=1;
    Char a[100];
    Gets (a);
    Length=strlen (a);
    for (i=0;i<=length/2;i++) {
       if (A[i]!=a[length-i-1]) {
           flag=0;
           break;
       }
    }
    if (flag==1)
      printf ("yes");
    else
      printf ("no");
    return 0;
}

Method Three:
Try out scenarios and handle large numbers. The idea of using stacks. Similar to string handling, here is a comparison of stack elements with string characters, such as unequal, is no.

 #include <stdio.h> #include <string.h> #define STACKSIZE typedef struct {Char
    data[stacksize];//open stack for 100;
int top=0;
}seqstack;
    int main () {seqstack s;
    Char str[100];
    scanf ("%s", str);
    int Len=strlen (str);
    int i,flag=1;
    for (i=0;i<len/2;i++)//will be half the characters into the stack s.data[s.top++]=str[i]; if (len%2) i++;
    Odd-numbered automatically skip the middle of the number, such as 121, skip 2, compare 1 while (s.top)//equivalent to Emptystack (&s), to determine whether the stack empty.
        {///each popup character compares char temp=s.data[--s.top with corresponding character];
        if (Temp!=str[i]) {flag=0;
        }else{i++;
    } if (flag==0) {printf ("no\n");
    }else{printf ("yes\n");
return 0; }